Chapter 6 Resource Masters - commackschools.org 6 Worksheets.pdf · Chapter 6 Resource Masters Chapter Resources Student-Built Glossary (pages 1–2) These masters are a student study

Chapter 6 Resource Masters

Bothell, WA • Chicago, IL • Columbus, OH • New York, NY

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CONSUMABLE WORKBOOKS Many of the worksheets contained in the Chapter Resource Masters booklets are available as consumable workbooks in both English and Spanish.

ISBN10 ISBN13Study Guide and Intervention Workbook 0-07-660292-3 978-0-07-660292-6Homework Practice Workbook 0-07-660291-5 978-0-07-660291-9

Spanish VersionHomework Practice Workbook 0-07-660294-X 978-0-07-660294-0

Answers For Workbooks The answers for Chapter 6 of these workbooks can be found in the back of this Chapter Resource Masters booklet.

ConnectED All of the materials found in this booklet are included for viewing, printing, and editing at connected.mcgraw-hill.com.

Spanish Assessment Masters (MHID: 0-07-660289-3, ISBN: 978-0-07-660289-6) These masters contain a Spanish version of Chapter 6 Test Form 2A and Form 2C.

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All rights reserved. The contents, or parts thereof, may be reproduced in print form for non-profit educational use with Glencoe Algebra 1, provided such reproductions bear copyright notice, but may not be reproduced in any form for any other purpose without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, network storage or transmission, or broadcast for distance learning.

Send all inquiries to:McGraw-Hill Education8787 Orion PlaceColumbus, OH 43240

ISBN: 978-0-07-660280-3MHID: 0-07-660280-X

Printed in the United States of America.

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Teacher’s Guide to Using the Chapter 6 Resource Masters ..............................................iv

Chapter Resources Chapter 6 Student-Built Glossary ...................... 1Chapter 6 Anticipation Guide (English) ............. 3Chapter 6 Anticipation Guide (Spanish) ............ 4

Lesson 6-1Graphing Systems of EquationsStudy Guide and Intervention ............................ 5Skills Practice .................................................... 7Practice ............................................................. 8Word Problem Practice ...................................... 9Enrichment ...................................................... 10Graphing Calculator Activity .............................11

Lesson 6-2SubstitutionStudy Guide and Intervention .......................... 12Skills Practice .................................................. 14Practice ........................................................... 15Word Problem Practice .................................... 16Enrichment ...................................................... 17

Lesson 6-3Elimination Using Addition and SubtractionStudy Guide and Intervention .......................... 18Skills Practice .................................................. 20Practice ........................................................... 21Word Problem Practice .................................... 22Enrichment ...................................................... 23

Lesson 6-4Elimination Using MultipcationStudy Guide and Intervention .......................... 24Skills Practice .................................................. 26Practice ........................................................... 27Word Problem Practice .................................... 28Enrichment ...................................................... 29

Lesson 6-5Applying Systems of Linear EquationsStudy Guide and Intervention .......................... 30Skills Practice .................................................. 32Practice ........................................................... 33Word Problem Practice .................................... 34Enrichment ...................................................... 35

Lesson 6-6Systems of InequalitiesStudy Guide and Intervention .......................... 36Skills Practice .................................................. 38Practice ........................................................... 39Word Problem Practice .................................... 40Enrichment ...................................................... 41Graphing Calculator ......................................... 42Spreadsheet .................................................... 43Student Recording Sheet ............................... 45Rubric for Scoring Extended Response .......... 46Chapter 6 Quizzes 1 and 2 ............................. 47Chapter 6 Quizzes 3 and 4 ............................. 48Chapter 6 Mid-Chapter Test ............................ 49Chapter 6 Vocabulary Test ............................... 50Chapter 6 Test, Form 1 .................................... 51Chapter 6 Test, Form 2A ................................. 53Chapter 6 Test, Form 2B ................................. 55Chapter 6 Test, Form 2C ................................. 57Chapter 6 Test, Form 2D ................................. 59Chapter 6 Test, Form 3 .................................... 61Chapter 6 Extended-Response Test ................ 63Standardized Test Practice .............................. 64Unit 2 Test ........................................................ 67

Answers ........................................... A1–A32

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iv

Teacher’s Guide to Using the Chapter 6 Resource Masters

Chapter Resources

Student-Built Glossary (pages 1–2) These masters are a student study tool that presents up to twenty of the key vocabulary terms from the chapter. Students are to record definitions and/or examples for each term. You may suggest that students highlight or star the terms with which they are not familiar. Give this to students before beginning Lesson 6-1. Encourage them to add these pages to their mathematics study notebooks. Remind them to complete the appropriate words as they study each lesson.

Anticipation Guide (pages 3–4) This master, presented in both English and Spanish, is a survey used before beginning the chapter to pinpoint what students may or may not know about the concepts in the chapter. Students will revisit this survey after they complete the chapter to see if their perceptions have changed.

Lesson ResourcesStudy Guide and Intervention These masters provide vocabulary, key concepts, additional worked-out examples and Guided Practice exercises to use as a reteaching activity. It can also be used in conjunction with the Student Edition as an instructional tool for students who have been absent.

Skills Practice This master focuses more on the computational nature of the lesson. Use as an additional practice option or as homework for second-day teaching of the lesson.

Practice This master closely follows the types of problems found in the Exercises section of the Student Edition and includes word problems. Use as an additional practice option or as homework for second-day teaching of the lesson.

Word Problem Practice This master includes additional practice in solving word problems that apply the concepts of the lesson. Use as an additional practice or as homework for second-day teaching of the lesson.

Enrichment These activities may extend the concepts of the lesson, offer a historical or multicultural look at the concepts, or widen students’ perspectives on the mathematics they are learning. They are written for use with all levels of students.

Graphing Calculator, TI-Nspire, or Spreadsheet Activities These activities present ways in which technology can be used with the concepts in some lessons of this chapter. Use as an alternative approach to some concepts or as an integral part of your lesson presentation.

The Chapter 6 Resource Masters includes the core materials needed for Chapter 6. These materials include worksheets, extensions, and assessment options. The answers for these pages appear at the back of this booklet.

All of the materials found in this booklet are included for viewing, printing, and editing at connectED.mcgraw-hill.com.

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Assessment OptionsThe assessment masters in the Chapter 6 Resource Masters offer a wide range of assessment tools for formative (monitoring) assessment and summative (final) assessment.

Student Recording Sheet This master cor-responds with the standardized test practice at the end of the chapter.

Extended Response Rubric This master provides information for teachers and students on how to assess performance on open-ended questions.

Quizzes Four free-response quizzes offer assessment at appropriate intervals in the chapter.

Mid-Chapter Test This 1-page test provides an option to assess the first half of the chapter. It parallels the timing of the Mid-Chapter Quiz in the Student Edition and includes both multiple-choice and free-response questions.

Vocabulary Test This test is suitable for all students. It includes a list of vocabulary words and 9 questions to assess students’ knowledge of those words. This can also be used in conjunction with one of the leveled chapter tests.

Leveled Chapter Tests• Form 1 contains multiple-choice

questions and is intended for use with below grade level students.

• Forms 2A and 2B contain multiple-choice questions aimed at on grade level students. These tests are similar in format to offer comparable testing situations.

• Forms 2C and 2D contain free-response questions aimed at on grade level students. These tests are similar in format to offer comparable testing situations.

• Form 3 is a free-response test for use with above grade level students.

All of the above mentioned tests include a free-response Bonus question.

Extended-Response Test Performance assessment tasks are suitable for all students. Sample answers and a scoring rubric are included for evaluation.

Standardized Test Practice These three pages are cumulative in nature. It includes three parts: multiple-choice questions with bubble-in answer format, griddable questions with answer grids, and short-answer free-response questions.

Answers• The answers for the Anticipation Guide

and Lesson Resources are provided as reduced pages.

• Full-size answer keys are provided for the assessment masters.

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Chapter 6 1 Glencoe Algebra 1

This is an alphabetical list of the key vocabulary terms you will learn in Chapter 6. As you study the chapter, complete each term’s definition or description. Remember to add the page number where you found the term. Add these pages to your Algebra Study Notebook to review vocabulary at the end of the chapter.

Student-Built Glossary6

Vocabulary TermFound

on PageDefi nition/Description/Example

consistentkuhn·SIHS·tuhnt

dependent

eliminationih·LIH·muh·NAY·shuhn

inconsistent

independent

substitutionSUHB·stuh·TOO·shuhn

system of equations

system of inequalities

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Before you begin Chapter 6

• Read each statement.

• Decide whether you Agree (A) or Disagree (D) with the statement.

• Write A or D in the first column OR if you are not sure whether you agree or disagree, write NS (Not Sure).

After you complete Chapter 6

• Reread each statement and complete the last column by entering an A or a D.

• Did any of your opinions about the statements change from the first column?

• For those statements that you mark with a D, use a piece of paper to write an example of why you disagree.

6 Anticipation GuideSolving Systems of Linear Equations

STEP 1A, D, or NS

StatementSTEP 2A or D

1. A solution of a system of equations is any ordered pair that satisfies one of the equations

2. A system of equations of parallel lines will have no solutions.

3. A system of equations of two perpendicular lines will have infinitely many solutions.

4. It is not possible to have exactly two solutions to a system of linear equations

5. The most accurate way to solve a system of equations is tograph the equations to see where they intersect.

6. To solve a system of equations, such as 2x - y = 21 and3y = 2x - 6, by substitution, solve one of the equations for one variable and substitute the result into the other equation.

7. When solving a system of equations, a result that is a true statement, such as -5 = -5, means the equations do not share a common solution.

8. Adding the equations 3x - 4y = 8 and 2x + 4y = 7 results in a 0 coefficient for y.

9. The equation 7x - 2y = 12 can be multiplied by 2 so that the coefficient of y is -4.

10. The result of multiplying -7x - 3y = 11 by -3 is -1x + 9y = 11.

Step 1

Step 2


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Capítulo 6 4 Álgebra 1 de Glencoe

6

Antes de comenzar el Capítulo 6

• Lee cada enunciado.

• Decide si estás de acuerdo (A) o en desacuerdo (D) con el enunciado.

• Escribe A o D en la primera columna O si no estás seguro(a) de la respuesta, escribe NS (No estoy seguro(a)).

Después de completar el Capítulo 6

• Vuelve a leer cada enunciado y completa la última columna con una A o una D.

• ¿Cambió cualquiera de tus opiniones sobre los enunciados de la primera columna?

• En una hoja de papel aparte, escribe un ejemplo de por qué estás en desacuerdo con los enunciados que marcaste con una D.

Ejercicios preparatoriosResuelve sistemas de ecuaciones lineales

Paso 1

PASO 1A, D o NS

EnunciadoPASO 2A o D

1. Una solución de un sistema de ecuaciones es cualquier par ordenado que satisface una de las ecuaciones

2. Un sistema de ecuaciones de rectas paralelas no tendrá soluciones.

3. Un sistema de ecuaciones de dos rectas perpendiculares tendrá un número infinito de soluciones.

4. No es posible tener exactamente dos soluciones para un sistema de ecuaciones lineales.

5. La forma más precisa de resolver un sistema de ecuaciones es graficar las ecuaciones y ver dónde se intersecan.

6. Para resolver un sistema de ecuaciones, como 2x - y = 21 y 3y = 2x - 6, por sustitución, despeja una variable en una de la ecuaciones y reemplaza el resultado en la otra ecuación.

7. Cuando se resuelve un sistema de ecuaciones, un resultado que es un enunciado verdadero, como -5 = -5, significa que las ecuaciones no comparten una solución.

8. El sumar las ecuaciones 3x - 4y = 8 y 2x + 4y = 7 resulta en un coeficiente de 0 para y.

9. La ecuación 7x - 2y = 12 se puede multiplicar por 2, de modo que el coeficiente de y es -4.

10. El resultado de multiplicar -7x - 3y = 11 por -3 es -1x + 9y = 11.

Paso 2


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Possible Number of Solutions Two or more linear equations involving the same variables form a system of equations. A solution of the system of equations is an ordered pair of numbers that satisfies both equations. The table below summarizes information about systems of linear equations.

Graph of a System intersecting lines same line parallel lines

Number of Solutions exactly one solution infi nitely many solutions no solution

Terminology

consistent and consistent and inconsistent

independent dependent

Use the graph at the right to determine whether each system is consistent or inconsistent and if it is independent or dependent.

a. y = -x + 2 y = x + 1

Since the graphs of y = -x + 2 and y = x + 1 intersect, there is one solution. Therefore, the system is consistent and independent.

b. y = -x + 2 3x + 3y = -3

Since the graphs of y = -x + 2 and 3x + 3y = -3 are parallel, there are no solutions. Therefore, the system is inconsistent.

c. 3x + 3y = -3 y = -x - 1

Since the graphs of 3x + 3y = -3 and y = -x - 1 coincide, there are infinitely many solutions. Therefore, the system is consistent and dependent.

ExercisesUse the graph at the right to determine whether each system is consistent or inconsistent and if it is independent or dependent.

1. y = -x - 3 2. 2x + 2y = -6y = x - 1 y = -x - 3

3. y = -x - 3 4. 2x + 2y = -62x + 2y = 4 3x + y = 3

6-1 Study Guide and InterventionGraphing Systems of Equations

x

y

O

y = -x - 3

3x + y = 3

2x + 2y = -6

2x + 2y = 4

y = x - 1

x

y

O

y = x + 1

y = -x - 1

3x + 3y = -3

y = -x + 2

Example

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Solve by Graphing One method of solving a system of equations is to graph the equations on the same coordinate plane.

Graph each system and determine the number of solutions that it has. If it has one solution, name it.

a. x + y = 2 x - y = 4

The graphs intersect. Therefore, there is one solution. The point (3, -1) seems to lie on both lines. Check this estimate by replacing x with 3 and y with -1 in each equation. x + y = 2 3 + (-1) = 2 � x - y = 4 3 - (-1) = 3 + 1 or 4 �The solution is (3, -1).

b. y = 2x + 1 2y = 4x + 2

The graphs coincide. Therefore there are infinitely many solutions.

ExercisesGraph each system and determine the number of solutions it has. If it has one solution, name it.

1. y = -2 2. x = 2 3. y = 1 −

2 x

3x - y = -1 2x + y = 1 x + y = 3

4. 2x + y = 6 5. 3x + 2y = 6 6. 2y = -4x + 4 2x - y = -2 3x + 2y = -4 y = -2x + 2

6-1 Study Guide and Intervention (continued)

Graphing Systems of Equations

x

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x

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x - y = 4

x + y = 2

Example

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1. y = x - 1 2. x - y = -4

y = -x + 1 y = x + 4

3. y = x + 4 4. y = 2x - 32x - 2y = 2 2x - 2y = 2


5. 2x - y = 1 6. x = 1 7. 3x + y = -3 y = -3 2x + y = 4 3x + y = 3

x

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8. y = x + 2 9. x + 3y = -3 10. y - x = -1x - y = -2 x - 3y = -3 x + y = 3

x

y

11. x - y = 3 12. x + 2y = 4 13. y = 2x + 3

x - 2y = 3 y = -

1 −

2 x + 2 3y = 6x - 6

x

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6-1 Skills PracticeGraphing Systems of Equations

x

y

y = -x + 1

x - y = -4 2x - 2y = 2

y = 2x - 3

y = x - 1

y = x + 4

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1. x + y = 3 2. 2x - y = -3x + y = -3 4x - 2y = -6

3. x + 3y = 3 4. x + 3y = 3x + y = -3 2x - y = -3


5. 3x - y = -2 6. y = 2x - 3 7. x + 2y = 3 3x - y = 0 4x = 2y + 6 3x - y = -5

x

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8. BUSINESS Nick plans to start a home-based business producing and selling gourmet dog treats. He figures it will cost $20 in operating costs per week plus $0.50 to produce each treat. He plans to sell each treat for $1.50.

a. Graph the system of equations y = 0.5x + 20 and y = 1.5x to represent the situation.

b. How many treats does Nick need to sell per week to break even?

9. SALES A used book store also started selling used CDs and videos. In the first week, the store sold 40 used CDs and videos, at $4.00 per CD and $6.00 per video. The sales for both CDs and videos totaled $180.00

a. Write a system of equations to represent the situation.

b. Graph the system of equations.

c. How many CDs and videos did the store sell in the first week?

6-1 PracticeGraphing Systems of Equations

x

y

x + y = -3

x + 3y = 3 2x - y = -3

4x - 2y = -6

x + y = 3

x

y

x

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O

Sales ($)

Dog Treats

Co

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$)

5 1510 20 25 30 35 40 450

40

35

30

25

20

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5

CD Sales ($)

CD and Video Sales

Vid

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($)

5 1510 20 25 30 35 40 450

40

35

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5


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1. BUSINESS The widget factory will sell a total of y widgets after x days according to the equation y = 200x + 300. The gadget factory will sell y gadgets after x days according to the equation y = 200x + 100. Look at the graph of the system of equations and determine whether it has no solution, one solution, or infinitely many solutions.

2. ARCHITECTURE An office building has two elevators. One elevator starts out on the 4th floor, 35 feet above the ground, and is descending at a rate of 2.2 feet per second. The other elevator starts out at ground level and is rising at a rate of 1.7 feet per second. Write a system of equations to represent the situation.

3. FITNESS Olivia and her brother William had a bicycle race. Olivia rode at a speed of 20 feet per second while William rode at a speed of 15 feet per second. To be fair, Olivia decided to give William a 150-foot head start. The race ended in a tie. How far away was the finish line from where Olivia started?

4. AVIATION Two planes are in flight near a local airport. One plane is at an altitude of 1000 meters and is ascending at a rate of 400 meters per minute. The second plane is at an altitude of 5900 meters and is descending at a rate of 300 meters per minute.

a. Write a system of equations that represents the progress of each plane

b. Make a graph that represents the progress of each plane.

6-1 Word Problem PracticeGraphing Systems of Equations

Days210 43

y

x5 6

Item

s so

ld

300

400

200

100

500

800

900

700

600

Gadgets

Widgetsy = 200x + 300

y = 200x + 100

Time (min.)0

y

x

Alt

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(m)


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Enrichment

Graphing a TripThe distance formula, d = rt, is used to solve many types of

t

d

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slope is 50

slope is –25

2

50

problems. If you graph an equation such as d = 50t, the graph is a model for a car going at 50 mph. The time the car travels is t; the distance in miles the car covers is d. The slope of the line is the speed.Suppose you drive to a nearby town and return. You average 50 mph on the trip out but only 25 mph on the trip home. The round trip takes 5 hours. How far away is the town?The graph at the right represents your trip. Notice that the return trip is shown with a negative slope because you are driving in the opposite direction.

Solve each problem.

1. Estimate the answer to the problem in the above example. About how far away is the town?

Graph each trip and solve the problem.

2. An airplane has enough fuel for 3 hours of safe flying. On the trip out the pilot averages 200 mph flying against a headwind. On the trip back, the pilot averages 250 mph. How long a trip out can the pilot make?

3. You drive to a town 100 miles away. 4. You drive at an average speed of 50 mph On the trip out you average 25 mph. to a discount shopping plaza, spend On the trip back you average 50 mph. 2 hours shopping, and then return at an How many hours do you spend driving? average speed of 25 mph. The entire trip takes 8 hours. How far away is the shopping plaza?

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Graphing Calculator ActivitySolution to a System of Linear Equations

A graphing calculator can be used to solve a system of linear equations graphically. The solution of a system of linear equations can be found by using the TRACE feature or by using the intersect command under the CALC menu.

ExercisesSolve each system of linear equations.

1. y = 2 2. y = -x + 3 3. x + y = -1 5x + 4y = 18 y = x + 1 2x - y = -8

4. -3x + y = 10 5. -4x + 3y = 10 6. 5x + 3y = 11 -x + 2y = 0 7x + y = 20 x - 5y = 5

7. 3x - 2y = -4 8. 3x + 2y = 4 9. 4x - 5y = 0 -4x + 3y = 5 -6x - 4y = -8 6x - 5y = 10

Solve each system of linear equations.

a. x + y = 0 x - y = -4

Using TRACE: Solve each equation for y and enter each equation into Y=. Then graph using Zoom 8: ZInteger. Use TRACE to find the solution.Keystrokes: Y= (–) ENTER + 4 ZOOM 6

ZOOM 8 ENTER TRACE .

The solution is (-2, 2).

b. 2x + y = 4 4x + 3y = 3

Using CALC: Solve each equation for y, enter each into the calculator, and graph. Use CALC to determine the solution.

Keystrokes: Y= (–) 2 + 4 ENTER ( (–) 4 ÷ 3 )

+ 1 ZOOM 6 2nd [CALC] 5 ENTER ENTER ENTER .

To change the x-value to a fraction, press 2nd [QUIT]

MATH ENTER ENTER .

The solution is (4.5, -5) or (

9 −

2 , -5

)

.

[-47, 47] scl:10 by [-31, 31]

scl:10

[-10, 10] scl:1 by [-10, 10]

scl:1

6-1

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Study Guide and InterventionSubstitution

Solve by Substitution One method of solving systems of equations is substitution.

Use substitution to solve the system of equations.y = 2x4x - y = -4

Substitute 2x for y in the second equation. 4x - y = -4 Second equation

4x - 2x = -4 y = 2x

2x = -4 Combine like terms.

x = -2 Divide each side by 2

and simplify.

Use y = 2x to find the value of y.y = 2x First equation

y = 2(-2) x = -2

y = -4 Simplify.

The solution is (-2, -4).

Solve for one variable, then substitute.x + 3y = 72x - 4y = -6

Solve the first equation for x since the coefficient of x is 1. x + 3y = 7 First equation

x + 3y - 3y = 7 - 3y Subtract 3y from each side.

x = 7 - 3y Simplify.

Find the value of y by substituting 7 - 3y for x in the second equation. 2x - 4y = -6 Second equation

2(7 - 3y) - 4y = -6 x = 7 - 3y

14 - 6y - 4y = -6 Distributive Property

14 - 10y = -6 Combine like terms.

14 - 10y - 14 = -6 - 14 Subtract 14 from each side.

-10y = -20 Simplify.

y = 2 Divide each side by -10

and simplify.

Use y = 2 to find the value of x.x = 7 - 3yx = 7 - 3(2)x = 1The solution is (1, 2).

ExercisesUse substitution to solve each system of equations.

1. y = 4x 2. x = 2y 3. x = 2y - 33x - y = 1 y = x - 2 x = 2y + 4

4. x - 2y = -1 5. x - 4y = 1 6. x + 2y = 03y = x + 4 2x - 8y = 2 3x + 4y = 4

7. 2b = 6a - 14 8. x + y = 16 9. y = -x + 3 3a - b = 7 2y = -2x + 2 2y + 2x = 4

10. x = 2y 11. x - 2y = -5 12. -0.2x + y = 0.50.25x + 0.5y = 10 x + 2y = -1 0.4x + y = 1.1

6-2

Example 1 Example 2


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Study Guide and Intervention (continued)

Substitution

Solve Real-World Problems Substitution can also be used to solve real-world problems involving systems of equations. It may be helpful to use tables, charts, diagrams, or graphs to help you organize data.

CHEMISTRY How much of a 10% saline solution should be mixed with a 20% saline solution to obtain 1000 milliliters of a 12% saline solution?Let s = the number of milliliters of 10% saline solution.Let t = the number of milliliters of 20% saline solution.Use a table to organize the information.

10% saline 20% saline 12% saline

Total milliliters s t 1000

Milliliters of saline 0.10 s 0.20 t 0.12(1000)

Write a system of equations.s + t = 10000.10s + 0.20t = 0.12(1000)Use substitution to solve this system. s + t = 1000 First equation

s = 1000 - t Solve for s.

0.10s + 0.20t = 0.12(1000) Second equation

0.10(1000 - t) + 0.20t = 0.12(1000) s = 1000 - t

100 - 0.10t + 0.20t = 0.12(1000) Distributive Property

100 + 0.10t = 0.12(1000) Combine like terms.

0.10t = 20 Simplify.

0.10t −

0.10 = 20 −

0.10 Divide each side by 0.10.

t = 200 Simplify.

s + t = 1000 First equation

s + 200 = 1000 t = 200

s = 800 Solve for s.

800 milliliters of 10% solution and 200 milliliters of 20% solution should be used.

Exercises 1. SPORTS At the end of the 2007–2008 football season, 38 Super Bowl games had been

played with the current two football leagues, the American Football Conference (AFC) and the National Football Conference (NFC). The NFC won two more games than the AFC. How many games did each conference win?

2. CHEMISTRY A lab needs to make 100 gallons of an 18% acid solution by mixing a 12% acid solution with a 20% solution. How many gallons of each solution are needed?

3. GEOMETRY The perimeter of a triangle is 24 inches. The longest side is 4 inches longer than the shortest side, and the shortest side is three-fourths the length of the middle side. Find the length of each side of the triangle.

6-2

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Use substitution to solve each system of equations.

1. y = 4x 2. y = 2xx + y = 5 x + 3y = -14

3. y = 3x 4. x = -4y2x + y = 15 3x + 2y = 20

5. y = x - 1 6. x = y - 7x + y = 3 x + 8y = 2

7. y = 4x - 1 8. y = 3x + 8y = 2x - 5 5x + 2y = 5

9. 2x - 3y = 21 10. y = 5x - 8y = 3 - x 4x + 3y = 33

11. x + 2y = 13 12. x + 5y = 43x - 5y = 6 3x + 15y = -1

13. 3x - y = 4 14. x + 4y = 82x - 3y = -9 2x - 5y = 29

15. x - 5y = 10 16. 5x - 2y = 142x - 10y = 20 2x - y = 5

17. 2x + 5y = 38 18. x - 4y = 27x - 3y = -3 3x + y = -23

19. 2x + 2y = 7 20. 2.5x + y = -2x - 2y = -1 3x + 2y = 0

6-2 Skills PracticeSubstitution


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Use substitution to solve each system of equations.

1. y = 6x 2. x = 3y 3. x = 2y + 72x + 3y = -20 3x - 5y = 12 x = y + 4

4. y = 2x - 2 5. y = 2x + 6 6. 3x + y = 12y = x + 2 2x - y = 2 y = -x - 2

7. x + 2y = 13 8. x - 2y = 3 9. x - 5y = 36 - 2x - 3y = -18 4x - 8y = 12 2x + y = -16

10. 2x - 3y = -24 11. x + 14y = 84 12. 0.3x - 0.2y = 0.5x + 6y = 18 2x - 7y = -7 x - 2y = -5

13. 0.5x + 4y = -1 14. 3x - 2y = 11 15. 1 −

2 x + 2y = 12

x + 2.5y = 3.5 x - 1 −

2 y = 4 x - 2y = 6

16. 1 −

3 x - y = 3 17. 4x - 5y = -7 18. x + 3y = -4

2x + y = 25 y = 5x 2x + 6y = 5

19. EMPLOYMENT Kenisha sells athletic shoes part-time at a department store. She can earn either $500 per month plus a 4% commission on her total sales, or $400 per month plus a 5% commission on total sales.


b. What is the total price of the athletic shoes Kenisha needs to sell to earn the same income from each pay scale?

c. Which is the better offer?

20. MOVIE TICKETS Tickets to a movie cost $7.25 for adults and $5.50 for students. A group of friends purchased 8 tickets for $52.75.


b. How many adult tickets and student tickets were purchased?

6-2 PracticeSubstitution


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1. BUSINESS Mr. Randolph finds that the supply and demand for gasoline at his station are generally given by the following equations.

x - y = -2x + y = 10

Use substitution to find the equilibrium point where the supply and demand lines intersect.

2. GEOMETRY The measures of complementary angles have a sum of 90 degrees. Angle A and angle B are complementary, and their measures have a difference of 20°. What are the measures of the angles?

AB

3. MONEY Harvey has some $1 bills and some $5 bills. In all, he has 6 bills worth $22. Let x be the number of $1 bills and let y be the number of $5 bills. Write a system of equations to represent the information and use substitution to determine how many bills of each denomination Harvey has.

4. POPULATION Sanjay is researching population trends in South America. He found that the population of Ecuador to increased by 1,000,000 and the population of Chile to increased by 600,000 from 2004 to 2009. The table displays the information he found.

Source: World Almanac

If the population growth for each country continues at the same rate, in what year are the populations of Ecuador and Chile predicted to be equal?

5. CHEMISTRY Shelby and Calvin are doing a chemistry experiment. They need 5 ounces of a solution that is 65% acid and 35% distilled water. There is no undiluted acid in the chemistry lab, but they do have two flasks of diluted acid: Flask A contains 70% acid and 30% distilled water. Flask B contains 20% acid and 80% distilled water.

a. Write a system of equations that Shelby and Calvin could use to determine how many ounces they need to pour from each flask to make their solution.

b. Solve your system of equations. How many ounces from each f lask do Shelby and Calvin need?

Word Problem PracticeSubstitution

6-2

Country2004

Population

5-Year

Population

Change

Ecuador 13,000,000 +1,000,000

Chile 16,000,000 +600,000

010_018_ALG1_A_CRM_C06_CR_660280.indd 16010_018_ALG1_A_CRM_C06_CR_660280.indd 16 12/21/10 11:48 PM12/21/10 11:48 PM

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Enrichment

Intersection of Two Parabolas

Substitution can be used to find the intersection of two parabolas. Replace the y-value in one of the equations with the y-value in terms of x from the other equation.

Find the intersection of the two parabolas. y = x2 + 5x + 6 y = x2 + 4x + 3

Graph the equations.

37.5

y

x-10 10-5 5

12.5

25.0

50.0

From the graph, notice that the two graphs intersect in one point.

Use substitution to solve for the point of intersection. x2 + 5x + 6 = x2 + 4x + 3 5x + 6 = 4x + 3 Subtract x2 from each side.

x + 6 = 3 Subtract 4x from each side.

x = -3 Subtract 6 from each side.

The graphs intersect at x = -3.

Replace x with -3 in either equation to find the y-value. y = x2 + 5x + 6 Original equation

y = (-3)2 + 5(-3) + 6 Replace x with -3.

y = 9 – 15 + 6 or 0 Simplify.

The point of intersection is (-3, 0).

ExercisesUse substitution to find the point of intersection of the graphs of each pair of equations.

1. y = x2 + 8x + 7 2. y = x2 + 6x + 8 3. y = x2 + 5x + 6y = x2 + 2x + 1 y = x2 + 4x + 4 y = x2 + 7x + 6

6-2

Example


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Study Guide and InterventionElimination Using Addition and Subtraction

Elimination Using Addition In systems of equations in which the coefficients of the x or y terms are additive inverses, solve the system by adding the equations. Because one of the variables is eliminated, this method is called elimination.

Use elimination to solve the system of equations.x - 3y = 73x + 3y = 9

Write the equations in column form and add to eliminate y. x - 3y = 7 (+) 3x + 3y = 9 4x = 16Solve for x. 4x −

4 = 16 −

4

x = 4Substitute 4 for x in either equation and solve for y. 4 - 3y = 7 4 - 3y - 4 = 7 - 4 -3y = 3

-3y

−

-3 = 3 −

-3

y = -1The solution is (4, -1).

The sum of two numbers is 70 and their difference is 24. Find the numbers.

Let x represent one number and y represent the other number. x + y = 70 (+) x - y = 24 2x = 94 2x −

2 = 94 −

2

x = 47Substitute 47 for x in either equation. 47 + y = 70 47 + y - 47 = 70 - 47 y = 23The numbers are 47 and 23.

ExercisesUse elimination to solve each system of equations.

1. x + y = -4 2. 2x - 3y = 14 3. 3x - y = -9x - y = 2 x + 3y = -11 -3x - 2y = 0

4. -3x - 4y = -1 5. 3x + y = 4 6. -2x + 2y = 93x - y = -4 2x - y = 6 2x - y = -6

7. 2x + 2y = -2 8. 4x - 2y = -1 9. x - y = 23x - 2y = 12 -4x + 4y = -2 x + y = -3

10. 2x - 3y = 12 11. -0.2x + y = 0.5 12. 0.1x + 0.3y = 0.9 4x + 3y = 24 0.2x + 2y = 1.6 0.1x - 0.3y = 0.2

13. Rema is older than Ken. The difference of their ages is 12 and the sum of their ages is 50. Find the age of each.

14. The sum of the digits of a two-digit number is 12. The difference of the digits is 2. Find the number if the units digit is larger than the tens digit.

6-3

Example 1 Example 2


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Elimination Using Addition and Subtraction

Elimination Using Subtraction In systems of equations where the coefficients of the x or y terms are the same, solve the system by subtracting the equations.

Use elimination to solve the system of equations.2x - 3y = 115x - 3y = 14

2x - 3y = 11 Write the equations in column form and subtract.

(-) 5x - 3y = 14 -3x = -3 Subtract the two equations. y is eliminated.

-3x −

-3 = -3 −

-3 Divide each side by -3.

x = 1 Simplify.

2(1) - 3y = 11 Substitute 1 for x in either equation.

2 - 3y = 11 Simplify.

2 - 3y - 2 = 11 - 2 Subtract 2 from each side.

-3y = 9 Simplify.

-3y

−

-3 = 9 −

-3 Divide each side by -3.

y = -3 Simplify.

The solution is (1, -3).


1. 6x + 5y = 4 2. 3m - 4n = -14 3. 3a + b = 16x - 7y = -20 3m + 2n = -2 a + b = 3

4. -3x - 4y = -23 5. x - 3y = 11 6. x - 2y = 6-3x + y = 2 2x - 3y = 16 x + y = 3

7. 2a - 3b = -13 8. 4x + 2y = 6 9. 5x - y = 62a + 2b = 7 4x + 4y = 10 5x + 2y = 3

10. 6x - 3y = 12 11. x + 2y = 3.5 12. 0.2x + y = 0.74x - 3y = 24 x - 3y = -9 0.2x + 2y = 1.2

13. The sum of two numbers is 70. One number is ten more than twice the other number. Find the numbers.

14. GEOMETRY Two angles are supplementary. The measure of one angle is 10° more than three times the other. Find the measure of each angle.

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Skills PracticeElimination Using Addition and Subtraction

Use elimination to solve each system of equations.

1. x - y = 1 2. -x + y = 1x + y = 3 x + y = 11

3. x + 4y = 11 4. -x + 3y = 6x - 6y = 11 x + 3y = 18

5. 3x + 4y = 19 6. x + 4y = -83x + 6y = 33 x - 4y = -8

7. 3x + 4y = 2 8. 3x - y = -14x - 4y = 12 -3x - y = 5

9. 2x - 3y = 9 10. x - y = 4-5x - 3y = 30 2x + y = -4

11. 3x - y = 26 12. 5x - y = -6-2x - y = -24 -x + y = 2

13. 6x - 2y = 32 14. 3x + 2y = -194x - 2y = 18 -3x - 5y = 25

15. 7x + 4y = 2 16. 2x - 5y = -287x + 2y = 8 4x + 5y = 4

17. The sum of two numbers is 28 and their difference is 4. What are the numbers?

18. Find the two numbers whose sum is 29 and whose difference is 15.


20. Find the two numbers whose sum is 54 and whose difference is 4.

21. Two times a number added to another number is 25. Three times the first number minus the other number is 20. Find the numbers.

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Practice Elimination Using Addition and Subtraction


1. x - y = 1 2. p + q = -2 3. 4x + y = 23x + y = -9 p - q = 8 3x - y = 12

4. 2x + 5y = -3 5. 3x + 2y = -1 6. 5x + 3y = 222x + 2y = 6 4x + 2y = -6 5x - 2y = 2

7. 5x + 2y = 7 8. 3x - 9y = -12 9. -4c - 2d = -2-2x + 2y = -14 3x - 15y = -6 2c - 2d = -14

10. 2x - 6y = 6 11. 7x + 2y = 2 12. 4.25x - 1.28y = -9.22x + 3y = 24 7x - 2y = -30 x + 1.28y = 17.6

13. 2x + 4y = 10 14. 2.5x + y = 10.7 15. 6m - 8n = 3x - 4y = -2.5 2.5x + 2y = 12.9 2m - 8n = -3

16. 4a + b = 2 17. -

1 −

3 x - 4 −

3 y = -2 18. 3 −

4 x - 1 −

2 y = 8

4a + 3b = 10 1 −

3 x - 2 −

3 y = 4 3 −

2 x + 1 −

2 y = 19


20. Four times one number added to another number is 36. Three times the first number minus the other number is 20. Find the numbers.

21. One number added to three times another number is 24. Five times the first number added to three times the other number is 36. Find the numbers.

22. LANGUAGES English is spoken as the first or primary language in 78 more countries than Farsi is spoken as the first language. Together, English and Farsi are spoken as a first language in 130 countries. In how many countries is English spoken as the first language? In how many countries is Farsi spoken as the first language?

23. DISCOUNTS At a sale on winter clothing, Cody bought two pairs of gloves and four hats for $43.00. Tori bought two pairs of gloves and two hats for $30.00. What were the prices for the gloves and hats?

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Word Problem PracticeElimination Using Addition and Subtraction

1. NUMBER FUN Ms. Simms, the sixth grade math teacher, gave her students this challenge problem.

Twice a number added to another number is 15. The sum of the two numbers is 11.

Lorenzo, an algebra student who was Ms. Simms aide, realized he could solve the problem by writing the following system of equations.

2x + y = 15x + y = 11

Use the elimination method to solve the system and find the two numbers.

2. GOVERNMENT The Texas State Legislature is comprised of state senators and state representatives. The sum of the number of senators and representatives is 181. There are 119 more representatives than senators. How many senators and how many representatives make up the Texas State Legislature?

3. RESEARCH Melissa wondered how much it would cost to send a letter by mail in 1990, so she asked her father. Rather than answer directly, Melissa’s father gave her the following information. It would have cost $3.70 to send 13 postcards and 7 letters, and it would have cost $2.65 to send 6 postcards and 7 letters. Use a system of equations and elimination to find how much it cost to send a letter in 1990.

4. SPORTS As of 2010, the New York Yankees had won more World Series Championships than any other team. In fact, the Yankees had won 3 fewer than 3 times the number of World Series championships won by the second most-winning team, the St. Louis Cardinals. The sum of the two teams’ World Series championships is 37. How many times has each team won the World Series?

5. BASKETBALL In 2005, the average ticket prices for Dallas Mavericks games and Boston Celtics games are shown in the table below. The change in price is from the 2004 season to the 2005 season.

Source: Team Marketing Report

a. Assume that tickets continue to change at the same rate each year after 2005. Let x be the number of years after 2005, and y be the price of an average ticket. Write a system of equations to represent the information in the table.

b. In how many years will the average ticket price for Dallas approximately equal that of Boston?

6-3

Team Average

Ticket Price

Change in

Price

Dallas $53.60 $0.53

Boston $55.93 -$1.08


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Systems of equations can involve more than 2 equations and 2 variables. It is possible to solve a system of 3 equations and 3 variables using elimination.

Solve the following system. x + y + z = 6 3x - y + z = 8 x - z = 2

Step 1: Use elimination to get rid of the y in the first two equations. x + y + z = 6 3x – y + z = 8

4x + 2z = 14

Step 2: Use the equation you found in step 1 and the third equation to eliminate the z. 4x + 2z = 14

x – z = 2

Multiply the second equation by 2 so that the z’s will eliminate. 4x + 2z = 14 2x – 2z = 4 6x = 18 So, x = 3.

Step 3: Replace x with 3 in the third original equation to determine z. 3 – z = 2, so z = 1.

Step 4: Replace x with 3 and z with 1 in either of the first two original equation to determine the value of y. 3 + y + 1 = 6 or 4 + y = 6. So, y = 2.So, the solution to the system of equations is (3, 2, 1).

ExercisesSolve each system of equations.

1. 3x + 2y + z = 42 2. x + y + z = -3 3. x + y + z = 7 2y + z + 12 = 3x 2x + 3y + 5z = -4 x + 2y + z = 10

x – 3y = 0 2y – z = 4 2y + z = 5

EnrichmentSolving Systems of Equations in Three Variables

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Study Guide and InterventionElimination Using Multiplication

Elimination Using Multiplication Some systems of equations cannot be solved simply by adding or subtracting the equations. In such cases, one or both equations must first be multiplied by a number before the system can be solved by elimination.

Use elimination to solve the system of equations.x + 10y = 34x + 5y = 5

If you multiply the second equation by -2, you can eliminate the y terms. x + 10y = 3 (+) -8x - 10y = -10 -7x = -7 -7x −

-7 = -7 −

-7

x = 1Substitute 1 for x in either equation. 1 + 10y = 3 1 + 10y - 1 = 3 - 1 10y = 2

10y

−

10 = 2 −

10

y =

1 −

5

The solution is (

1, 1 −

5 )

.

Use elimination to solve the system of equations.3x - 2y = -72x - 5y = 10

If you multiply the first equation by 2 and the second equation by -3, you can eliminate the x terms. 6x - 4y = -14(+) -6x + 15y = -30 11y = -44

11y

−

11 = -

44 −

11

y = -4Substitute -4 for y in either equation. 3x - 2(-4) = -7 3x + 8 = -7 3x + 8 -8 = -7 -8 3x = -15

3x −

3 = -

15 −

3

x = -5The solution is (-5, -4).


1. 2x + 3y = 6 2. 2m + 3n = 4 3. 3a - b = 2x + 2y = 5 -m + 2n = 5 a + 2b = 3

4. 4x + 5y = 6 5. 4x - 3y = 22 6. 3x - 4y = -46x - 7y = -20 2x - y = 10 x + 3y = -10

7. 4x - y = 9 8. 4a - 3b = -8 9. 2x + 2y = 55x + 2y = 8 2a + 2b = 3 4x - 4y = 10

10. 6x - 4y = -8 11. 4x + 2y = -5 12. 2x + y = 3.5 4x + 2y = -3 -2x - 4y = 1 -x + 2y = 2.5

13. GARDENING The length of Sally’s garden is 4 meters greater than 3 times the width. The perimeter of her garden is 72 meters. What are the dimensions of Sally’s garden?

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6-4 Study Guide and Intervention (continued)

Elimination Using Multiplication

Solve Real-World Problems Sometimes it is necessary to use multiplication before elimination in real-world problems.

CANOEING During a canoeing trip, it takes Raymond 4 hours to paddle 12 miles upstream. It takes him 3 hours to make the return trip paddling downstream. Find the speed of the canoe in still water.

Read You are asked to find the speed of the canoe in still water.

Solve Let c = the rate of the canoe in still water.Let w = the rate of the water current.

r t d r · t = d

Against the Current c – w 4 12 (c – w)4 = 12

With the Current c + w 3 12 (c + w)3 = 12

So, our two equations are 4c - 4w = 12 and 3c + 3w = 12.Use elimination with multiplication to solve the system. Since the problem asks for c, eliminate w.4c - 4w = 12 ⇒ Multiply by 3 ⇒ 12c - 12w = 363c + 3w = 12 ⇒ Multiply by 4 ⇒ (+) 12c + 12w = 48 24c = 84 w is eliminated.

24c−

24=

84−

24Divide each side by 24.

c = 3.5 Simplify.The rate of the canoe in still water is 3.5 miles per hour.

Exercises 1. FLIGHT An airplane traveling with the wind flies 450 miles in 2 hours. On the return

trip, the plane takes 3 hours to travel the same distance. Find the speed of the airplane if the wind is still.

2. FUNDRAISING Benji and Joel are raising money for their class trip by selling gift wrapping paper. Benji raises $39 by selling 5 rolls of red wrapping paper and 2 rolls of foil wrapping paper. Joel raises $57 by selling 3 rolls of red wrapping paper and 6 rolls of foil wrapping paper. For how much are Benji and Joel selling each roll of red and foil wrapping paper?

Example


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Skills PracticeElimination Using Multiplication


1. x + y = -9 2. 3x + 2y = -95x - 2y = 32 x - y = -13

3. 2x + 5y = 3 4. 2x + y = 3-x + 3y = -7 -4x - 4y = -8

5. 4x - 2y = -14 6. 2x + y = 03x - y = -8 5x + 3y = 2

7. 5x + 3y = -10 8. 2x + 3y = 143x + 5y = -6 3x - 4y = 4

9. 2x - 3y = 21 10. 3x + 2y = -265x - 2y = 25 4x - 5y = -4

11. 3x - 6y = -3 12. 5x + 2y = -32x + 4y = 30 3x + 3y = 9

13. Two times a number plus three times another number equals 13. The sum of the two numbers is 7. What are the numbers?

14. Four times a number minus twice another number is -16. The sum of the two numbers is -1. Find the numbers.

15. FUNDRAISING Trisha and Byron are washing and vacuuming cars to raise money for a class trip. Trisha raised $38 washing 5 cars and vacuuming 4 cars. Byron raised $28 by washing 4 cars and vacuuming 2 cars. Find the amount they charged to wash a car and vacuum a car.

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PracticeElimination Using Multiplication


1. 2x - y = -1 2. 5x - 2y = -10 3. 7x + 4y = -43x - 2y = 1 3x + 6y = 66 5x + 8y = 28

4. 2x - 4y = -22 5. 3x + 2y = -9 6. 4x - 2y = 323x + 3y = 30 5x - 3y = 4 -3x - 5y = -11

7. 3x + 4y = 27 8. 0.5x + 0.5y = -2 9. 2x - 3 −

4 y = -7

5x - 3y = 16 x - 0.25y = 6 x + 1 −

2 y = 0

10. 6x - 3y = 21 11. 3x + 2y = 11 12. -3x + 2y = -152x + 2y = 22 2x + 6y = -2 2x - 4y = 26

13. Eight times a number plus five times another number is -13. The sum of the two numbers is 1. What are the numbers?

14. Two times a number plus three times another number equals 4. Three times the first number plus four times the other number is 7. Find the numbers.

15. FINANCE Gunther invested $10,000 in two mutual funds. One of the funds rose 6% in one year, and the other rose 9% in one year. If Gunther’s investment rose a total of $684 in one year, how much did he invest in each mutual fund?

16. CANOEING Laura and Brent paddled a canoe 6 miles upstream in four hours. The return trip took three hours. Find the rate at which Laura and Brent paddled the canoe in still water.

17. NUMBER THEORY The sum of the digits of a two-digit number is 11. If the digits are reversed, the new number is 45 more than the original number. Find the number.

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1. SOCCER Suppose a youth soccer field has a perimeter of 320 yards and its length measures 40 yards more than its width. Ms. Hughey asks her players to determine the length and width of their field. She gives them the following system of equations to represent the situation. Use elimination to solve the system to find the length and width of the field.

2L + 2W = 320 L – W = 40

2. SPORTS The Fan Cost Index (FCI) tracks the average costs for attending sporting events, including tickets, drinks, food, parking, programs, and souvenirs. According to the FCI, a family of four would spend a total of $592.30 to attend two Major League Baseball (MLB) games and one National Basketball Association (NBA) game. The family would spend $691.31 to attend one MLB and two NBA games. Write and solve a system of equations to find the family’s costs for each kind of game according to the FCI.

3. ART Mr. Santos, the curator of the children’s museum, recently made two purchases of clay and wood for a visiting artist to sculpt. Use the table to find the cost of each product per kilogram.

Clay (kg) Wood (kg) Total Cost

5 4 $35.50

3.5 6 $50.45

4. TRAVEL Antonio flies from Houston to Philadelphia, a distance of about 1340 miles. His plane travels with the wind and takes 2 hours and 20 minutes. At the same time, Paul is on a plane from Philadelphia to Houston. Since his plane is heading against the wind, Paul’s flight takes 2 hours and 50 minutes. What was the speed of the wind in miles per hour?

5. BUSINESS Suppose you start a business assembling and selling motorized scooters. It costs you $1500 for tools and equipment to get started, and the materials cost $200 for each scooter. Your scooters sell for $300 each.

a. Write and solve a system of equations representing the total costs and revenue of your business.

b. Describe what the solution means in terms of the situation.

c. Give an example of a reasonable number of scooters you could assemble and sell in order to make a profit, and find the profit you would make for that number of scooters.

Word Problem PracticeElimination Using Multiplication

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Enrichment

George Washington Carver and Percy Julian In 1990, George Washington Carver and Percy Julian became the first African Americans elected to the National Inventors Hall of Fame. Carver (1864–1943) was an agricultural scientist known worldwide for developing hundreds of uses for the peanut and the sweet potato. His work revitalized the economy of the southern United States because it was no longer dependent solely upon cotton. Julian (1898–1975) was a research chemist who became famous for inventing a method of making a synthetic cortisone from soybeans. His discovery has had many medical applications, particularly in the treatment of arthritis.There are dozens of other African American inventors whose accomplishments are not as well known. Their inventions range from common household items like the ironing board to complex devices that have revolutionized manufacturing. The exercises that follow will help you identify just a few of these inventors and their inventions.

Match the inventors with their inventions by matching each system with its solution. (Not all the solutions will be used.)

1. Sara Boone x + y = 2 A. (1, 4) automatic traffic signal x - y = 10

2. Sarah Goode x = 2 - y B. (4, -2) eggbeater 2y + x = 9

3. Frederick M. y = 2x + 6 C. (-2, 3) fire extinguisher Jones y = -x - 3

4. J. L. Love 2x + 3y = 8 D. (-5, 7) folding cabinet bed 2x - y = -8

5. T. J. Marshall y - 3x = 9 E. (6, -4) ironing board 2y + x = 4

6. Jan Matzeliger y + 4 = 2x F. (-2, 4) pencil sharpener 6x - 3y = 12

7. Garrett A. 3x - 2y = -5 G. (-3, 0) portable x-ray machine Morgan 3y - 4x = 8

8. Norbert Rillieux 3x - y = 12 H. (2, -3) player piano y - 3x = 15

I. no solution evaporating pan for refining sugar

J. infinitely lasting (shaping) many machine for solutions manufacturing shoes

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Study Guide and Intervention Applying Systems of Linear Equations

Determine The Best Method You have learned five methods for solving systems of linear equations: graphing, substitution, elimination using addition, elimination using subtraction, and elimination using multiplication. For an exact solution, an algebraic method is best.

At a baseball game, Henry bought 3 hotdogs and a bag of chips for $14. Scott bought 2 hotdogs and a bag of chips for $10. Each of the boys paid the same price for their hotdogs, and the same price for their chips. The following system of equations can be used to represent the situation. Determine the best method to solve the system of equations. Then solve the system.

3x + y = 142x + y = 10

• Since neither the coefficients of x nor the coefficients of y are additive inverses, you cannot use elimination using addition.

• Since the coefficient of y in both equations is 1, you can use elimination using subtraction. You could also use the substitution method or elimination using multiplication

The following solution uses elimination by subtraction to solve this system. 3x + y = 14 Write the equations in column form and subtract. (-) 2x + (-) y = (-)10

x = 4 The variable y is eliminated. 3(4) + y = 14 Substitute the value for x back into the first equation.

y = 2 Solve for y.

This means that hot dogs cost $4 each and a bag of chips costs $2.

ExercisesDetermine the best method to solve each system of equations. Then solve the system.

1. 5x + 3y = 16 2. 3x - 5y = 73x - 5y = -4 2x + 5y = 13

3. y + 3x = 24 4. -11x - 10y = 175x - y = 8 5x – 7y = 50

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Example


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Apply Systems Of Linear Equations When applying systems of linear equations to problem situations, it is important to analyze each solution in the context of the situation.

BUSINESS A T-shirt printing company sells T-shirts for $15 each. The company has a fixed cost for the machine used to print the T-shirts and an additional cost per T-shirt. Use the table to estimate the number of T-shirts the company must sell in order for the income to equal expenses.

Understand You know the initial income and the initial expense and the rates of change of each quantity with each T-shirt sold.

Plan Write an equation to represent the income and the expenses. Then solve to find how many T-shirts need to be sold for both values to be equal.

Solve Let x = the number of T-shirts sold and let y = the total amount.

income y = 0 + 15x

expenses y = 3000 + 5x

You can use substitution to solve this system.

y = 15x The fi rst equation.

15x = 3000 + 5x Substitute the value for y into the second equation.

10x = 3000 Subtract 10x from each side and simplify.

x = 300 Divide each side by 10 and simplify.

This means that if 300 T-shirts are sold, the income and expenses of the T-shirt company are equal.

Check Does this solution make sense in the context of the problem? After selling 300 T-shirts, the income would be about 300 × $15 or $4500. The costs would be about $3000 + 300 × $5 or $4500.

ExercisesRefer to the example above. If the costs of the T-shirt company change to the given values and the selling price remains the same, determine the number of T-shirts the company must sell in order for income to equal expenses.

1. printing machine: $5000.00; 2. printing machine: $2100.00; T-shirt: $10.00 each T-shirt: $8.00 each

3. printing machine: $8800.00; 4. printing machine: $1200.00;T-shirt: $4.00 each T-shirt: $12.00 each


Applying Systems of Linear Equations

6-5

T-shirt Printing Cost

printing

machine $3000.00

blank

T-shirt $5.00

Example

total

amount

initial

amount

rate of change times

number of T-shirts sold


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Determine the best method to solve each system of equations. Then solve the system.

1. 5x + 3y = 16 2. 3x – 5y = 73x – 5y = -4 2x + 5y = 13

3. y = 3x - 24 4. -11x – 10y = 175x - y = 8 5x – 7y = 50

5. 4x + y = 24 6. 6x – y = -1455x - y = 12 x = 4 – 2y

7. VEGETABLE STAND A roadside vegetable stand sells pumpkins for $5 each and squashes for $3 each. One day they sold 6 more squash than pumpkins, and their sales totaled $98. Write and solve a system of equations to find how many pumpkins and squash they sold?

8. INCOME Ramiro earns $20 per hour during the week and $30 per hour for overtime on the weekends. One week Ramiro earned a total of $650. He worked 5 times as many hours during the week as he did on the weekend. Write and solve a system of equations to determine how many hours of overtime Ramiro worked on the weekend.

9. BASKETBALL Anya makes 14 baskets during her game. Some of these baskets were worth 2-points and others were worth 3-points. In total, she scored 30 points. Write and solve a system of equations to find how 2-points baskets she made.

6-5 Skills PracticeApplying Systems of Linear Equations


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1. 1.5x – 1.9y = -29 2. 1.2x – 0.8y = -6 3. 18x –16y = -312x – 0.9y = 4.5 4.8x + 2.4y = 60 78x –16y = 408

4. 14x + 7y = 217 5. x = 3.6y + 0.7 6. 5.3x – 4y = 43.514x + 3y = 189 2x + 0.2y = 38.4 x + 7y = 78

7. BOOKS A library contains 2000 books. There are 3 times as many non-fiction books as fiction books. Write and solve a system of equations to determine the number of non-fiction and fiction books.

8. SCHOOL CLUBS The chess club has 16 members and gains a new member every month. The film club has 4 members and gains 4 new members every month. Write and solve a system of equations to find when the number of members in both clubs will be equal.

9. Tia and Ken each sold snack bars and magazine subscriptions for a school fund-raiser, as shown in the table. Tia earned $132 and Ken earned $190.

a. Define variable and formulate a system of linear equation from this situation.

b. What was the price per snack bar? Determine the reasonableness of your solution.

PracticeApplying Systems of Linear Equations

6-5

ItemNumber Sold

Tia Ken

snack bars 16 20

magazine

subscriptions4 6


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Word Problem PracticeApplying Systems of Linear Equations

1. MONEY Veronica has been saving dimes and quarters. She has 94 coins in all, and the total value is $19.30. How many dimes and how many quarters does she have?

2. CHEMISTRY How many liters of 15% acid and 33% acid should be mixed to make 40 liters of 21% acid solution?

Concentration

of Solution

Amount of

Solution (L)

Amount

of Acid

15% x

33% y

21% 40

3. BUILDINGS The Sears Tower in Chicago is the tallest building in North America. The total height of the tower t and the antenna that stands on top of it a is 1729 feet. The difference in heights between the building and the antenna is 279 feet. How tall is the Sears Tower?

4. PRODUCE Roger and Trevor went shopping for produce on the same day. They each bought some apples and some potatoes. The amount they bought and the total price they paid are listed in the table below.

What was the price of apples and potatoes per pound?

5. SHOPPING Two stores are having a sale on T-shirts that normally sell for $20. Store S is advertising an s percent discount, and Store T is advertising at dollar discount. Rose spends $63 for three T-shirts from Store S and one from Store T. Manny spends $140 on five T-shirts from Store S and four from Store T. Find the discount at each store.

6. TRANSPORTATION A Speedy River barge bound for New Orleans leaves Baton Rouge, Louisiana, at 9:00 A.M. and travels at a speed of 10 miles per hour. A Rail Transport freight train also bound for New Orleans leaves Baton Rouge at 1:30 P.M. the same day. The train travels at 25 miles per hour, and the river barge travels at 10 miles per hour. Both the barge and the train will travel 100 miles to reach New Orleans.

a. How far will the train travel before catching up to the barge?

b. Which shipment will reach New Orleans first? At what time?

c. If both shipments take an hour to unload before heading back to Baton Rouge, what is the earliest time that either one of the companies can begin to load grain to ship in Baton Rouge?

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Apples

(Ib)

Potatoes

(Ib)

Total Cost

($)

Roger 8 7 18.85

Trevor 2 10 12.88


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Cramer’s RuleCramer’s Rule is a method for solving a system of equations. To use Cramer’s Rule, set up a matrix to represent the equations. A matrix is a way of organizing data.

Solve the following system of equations using Cramer’s Rule. 2x + 3y = 13 x + y = 5

Step 1: Set up a matrix representing the coefficients of x and y.

A = x y

Step 2: Find the determinant of matrix A.

If a matrix A = ⎪

a c b d ⎥

, then the determinant, det(A) = ad – bc.

det(A) = 2(1) – 1(3) = -1

Step 3: Replace the first column in A with 13 and 5 and find the determinant of the new matrix.

A1 = ⎪

13 5 3 1 ⎥

; det (A1) = 13(1) – 5(3) = -2

Step 4: To find the value of x in the solution to the system of equations,

determine the value of det(A1) −

det(A) .

det(A1) −

det(A) = -2 −

-1 or 2

Step 5: Repeat the process to find the value of y. This time, replace the second column with 13 and 5 and find the determinant.

A2 = ⎪

2 1 13

5 ⎥

; det (A2) = 2(5) – 1(13) = -3 and det(A2) −

det(A) = -3 −

-1 or 3.

So, the solution to the system of equations is (2, 3).

Exercises Use Cramer’s Rule to solve each system of equations.

1. 2x + y = 1 2. x + y = 4 3. x – y = 43x + 5y = 5 2x – 3y = -2 3x – 5y = 8

4. 4x – y = 3 5. 3x – 2y = 7 6. 6x – 5y = 1x + y = 7 2x + y = 14 3x + 2y = 5

Enrichment6-5

Example

⎪

2 1 3 1

⎥


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Study Guide and InterventionSystems of Inequalities

Systems of Inequalities The solution of a system of inequalities is the set of all ordered pairs that satisfy both inequalities. If you graph the inequalities in the same coordinate plane, the solution is the region where the graphs overlap.

Solve the system of inequalities by graphing.y > x + 2y ≤ -2x - 1

The solution includes the ordered pairs in the intersection of the graphs. This region is shaded at the right. The graphs of y = x + 2 and y = -2x - 1 are boundaries of this region. The graph of y = x + 2 is dashed and is not included in the graph of y > x + 2.

Solve the system of inequalities by graphing.x + y > 4x + y < -1

The graphs of x + y = 4 and x + y = -1 are parallel. Because the two regions have no points in common, the system of inequalities has no solution.

ExercisesSolve each system of inequalities by graphing.

1. y > -1 2. y > -2x + 2 3. y < x + 1x < 0 y ≤ x + 1 3x + 4y ≥ 12

x

y

O

x

y

O

x

y

O

4. 2x + y ≥ 1 5. y ≤ 2x + 3 6. 5x - 2y < 6x - y ≥ -2 y ≥ -1 + 2x y > -x + 1

x

y

O

x

y

O

x

y

O

x

y

O

x + y = 1

x + y = 4

x

y

O

y = x + 2

y = -2x - 1

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Example 1

Example 2


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Systems of Inequalities

Apply Systems of Inequalities In real-world problems, sometimes only whole numbers make sense for the solution, and often only positive values of x and y make sense.

BUSINESS AAA Gem Company produces necklaces and bracelets. In a 40-hour week, the company has 400 gems to use. A necklace requires 40 gems and a bracelet requires 10 gems. It takes 2 hours to produce a necklace and a bracelet requires one hour. How many of each type can be produced in a week?

Let n = the number of necklaces that will be produced and b = the number of bracelets that will be produced. Neither n or b can be a negative number, so the following system of inequalities represents the conditions of the problems.

n ≥ 0b ≥ 0b + 2n ≤ 4010b + 40n ≤ 400The solution is the set ordered pairs in the intersection of the graphs. This region is shaded at the right. Only whole-number solutions, such as (5, 20) make sense in this problem.

ExercisesFor each exercise, graph the solution set. List three possible solutions to the problem.

1. HEALTH Mr. Flowers is on a restricted 2. RECREATION Maria had $150 in gift diet that allows him to have between certificates to use at a record store. She 1600 and 2000 Calories per day. His bought fewer than 20 recordings. Each daily fat intake is restricted to between tape cost $5.95 and each CD cost $8.95. 45 and 55 grams. What daily Calorie How many of each type of recording might and fat intakes are acceptable? she have bought?

60

50

40

30

20

10

Calories

Fat

Gra

ms

1000 2000 30000

30

25

20

15

10

5

Tapes

Co

mp

act

Dis

cs

5 10 15 20 25 300

Necklaces

Bra

cele

ts

10 20 30 40 500

50

40

30

20

10

10b + 40n = 400

b + 2n = 40

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Skills PracticeSystems of Inequalities

Solve each system of inequalities by graphing.

1. x > -1 2. y > 2 3. y > x + 3y ≤ -3 x < -2 y ≤ -1

x

y

O

x

y

O

x

y

O

4. x < 2 5. x + y ≤ -1 6. y - x > 4y - x ≤ 2 x + y ≥ 3 x + y > 2

x

y

O

y

xO

x

y

O

7. y > x + 1 8. y ≥ -x + 2 9. y < 2x + 4y ≥ -x + 1 y < 2x - 2 y ≥ x + 1

x

y

O

x

y

O

x

y

O

Write a system of inequalities for each graph.

10.

x

y

O

11.

x

y

O

12.

x

y

O

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PracticeSystems of Inequalities

$4 Stones

Turquoise Stones

$6 S

ton

es

1 32 4 5 6 70

6

5

4

3

2

1

Gym (hours)

Diego’s RoutineW

alki

ng

(m

iles)

1 32 4 5 6 7 80

16

14

12

10

8

6

4

2

6-6

Solve each system of inequalities by graphing.

1. y > x - 2 2. y ≥ x + 2 3. x + y ≥ 1y ≤ x y > 2x + 3 x + 2y > 1

x

y

O

x

y

O

x

y

O

4. y < 2x - 1 5. y > x - 4 6. 2x - y ≥ 2y > 2 - x 2x + y ≤ 2 x - 2y ≥ 2

x

y

O

x

y

O

x

y

O

7. FITNESS Diego started an exercise program in which each week he works out at the gym between 4.5 and 6 hours and walks between 9 and 12 miles.

a. Make a graph to show the number of hours Diego works out at the gym and the number of miles he walks per week.

b. List three possible combinations of working out and walking that meet Diego’s goals.

8. SOUVENIRS Emily wants to buy turquoise stones on her trip to New Mexico to give to at least 4 of her friends. The gift shop sells stones for either $4 or $6 per stone. Emily has no more than $30 to spend.

a. Make a graph showing the numbers of each price of stone Emily can purchase.

b. List three possible solutions.


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Word Problem PracticeSystems of Inequalities

1. PETS Renée’s Pet Store never has more than a combined total of 20 cats and dogs and never more than 8 cats. This is represented by the inequalities x ≤ 8 and x + y ≤ 20. Solve the system of inequalities by graphing.

2. WAGES The minimum wage for one group of workers in Texas is $7.25 per hour effective Sept. 1, 2008. The graph below shows the possible weekly wages for a person who makes at least minimum wages and works at most 40 hours. Write the system of inequalities for the graph.

3. FUND RAISING The Camp Courage Club plans to sell tins of popcorn and peanuts as a fundraiser. The Club members have $900 to spend on products to sell and want to order up to 200 tins in all. They also want to order at least as many tins of popcorn as tins of peanuts. Each tin of popcorn costs $3 and each tin of peanuts costs $4. Write a system of equations to represent the conditions of this problem.

4. BUSINESS For maximum efficiency, a factory must have at least 100 workers, but no more than 200 workers on a shift. The factory also must manufacture at least 30 units per worker.

a. Let x be the number of workers and let y be the number of units. Write four inequalities expressing the conditions in the problem given above.

b. Graph the systems of inequalities.

c. List at least three possible solutions.

O

Do

gs

Cats

4

6

2

8

10

12

14

16

18

20y

642 10 14 16 18 208 12 x

x

y

25

50

75

100

Wag

es (

$)

125

150

175

200

225

250

275

300

Hours5 10 15 20 25 30 35 40 45 50

O

Un

its

(100

0)

2

3

1

4

5

6

7

Workers

y

150100

50 250 350200 300 400

x

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Enrichment

Describing Regions

The shaded region inside the triangle can be described with a system of three inequalities. y < -x + 1 y > 1 −

3 x - 3

y > -9x - 31

Write systems of inequalities to describe each region. You may first need to divide a region into triangles or quadrilaterals.

1.

x

y

O

2.

x

y

O

3.

x

y

O

6-6

x

y

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Graphing Calculator ActivityShading Absolute Value Inequalities

Absolute value inequalities of the form y ≤ a|x - h| + k or y ≥ a|x - h| + k can be graphed using the SHADE command or by using the shading feature in the Y= screen.

Graph y ≤ 2|x - 3| + 2.

Method 1 SHADE command

Enter the boundary equation y = 2|x - 3| + 2 into Y1. From the home screen, use SHADE to shade the region under the boundary equation because the inequality uses <. Use Ymin as the lower boundary and the boundary equation, Y1, as the upper boundary.Keystrokes: 2nd [DRAW] 7 VARS ENTER 4 , VARS ENTER ENTER ) ENTER .

Method 2 Y=

On the Y= screen, move the cursor to the far left and repeatedly press ENTER to toggle through the graph choices. Be sure to select the option to shade below . Then enter the boundary equation and graph.

The same techniques can be used to solve a system of absolute value inequalities.

ExercisesSolve each system of inequalities by graphing.

1. y ≤ 2|x - 4| + 1 2. y < 1 −

2 |x - 3| + 2 3. y ≤ 4|x - 4| + 3 4. y ≤

1 −

4 |x + 1| + 1

y > |x + 3| + 6 y ≤ |x + 1| + 4 y ≥ 3|x - 4| - 2 y ≥ 2|x + 3| + 5

Solve the system y ≥ |x + 5| - 4 and y ≤ 3|x - 6| + 1.

Enter y = |x + 5| - 4 in Y1 and set the graph to shade above. Then enter y = 3|x - 6| + 1 in Y2 and set the graph to shade below. Graph the inequalities. Notice that a different pattern is used for the second inequality. The region where the two patterns overlap is the solution to the system.

[-10, 10] scl:1 by [-10, 10] scl:1

[-10, 10] scl:1 by [-10, 10] scl:1

[-20, 20] scl:2 by [-20, 20] scl:2

6-6

Example 1

Example 2


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6-6 Spreadsheet ActivitySystems of Inequalities

ExercisesUse the spreadsheet to determine whether TopSport can make the following combinations of shoes.

1. 1000 Runners, 1100 Flyers 2. 1200 Runners, 1000 Flyers



6. If TopSport can either buy another cutting machine or hire more assemblers, which would make more combinations of shoe production possible? Explain.

TopSport Shoe Company has a total of 9600 minutes of machine time each week to cut the materials for the two types of athletic shoes they make. There are a total of 28,000 minutes of worker time per week for assembly. It takes 3 minutes to cut and 12 minutes to assemble a pair of Runners and 2 minutes to cut and 10 minutes to assemble a pair of Flyers. Is it possible for the company to make 1200 pairs of Runners and 1400 pairs of Flyers in a week?

Step 1 Represent the situation using a system of inequalities. Let r represent the number of Runners and f represent the number of Flyers.

3r + 2f ≤ 9600 12r + 10f ≤ 28,000

Step 2 Columns A and B contain the values of r and f. Columns C and D contain the formulas for the inequalities. The formulas will return TRUE or FALSE. If both inequalities are true for an ordered pair, then the ordered pair is a solution to the system of inequalities.

Since one of the inequalities is false for (1200, 1400), the ordered pair is not part of the solution set. The company cannot make 1200 pairs of Runners and 1400 pairs of Flyers in a week.The formula in cell D2 is

12*A2 + 10*B2 <= 28,000.

+ ≤ + ≤

Example



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Student Recording SheetUse this recording sheet with pages 386–387 of the Student Edition.

6

Read each question. Then fill in the correct answer.

1. A B C D

2. F G H J

3. A B C D

4. F G H J

5. A B C D

6. F G H J

7. A B C D

8. F G H J

Multiple Choice

Extended Response

Record your answers for Question 18 on the back of this paper.

9.

Record your answer in the blank.

For gridded response questions, also enter your answer in the grid by writing each number or symbol in a box. Then fill in the corresponding circle for that number or symbol.

9. (grid in)

10.

11. (grid in)

12.

13.

14. (grid in)

15.

16.

17.

Short Response/Gridded Response

9

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0

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General Scoring Guidelines• If a student gives only a correct numerical answer to a problem but does not show how he

or she arrived at the answer, the student will be awarded only 1 credit. All extended-response questions require the student to show work.

• A fully correct answer for a multiple-part question requires correct responses for all parts of the question. For example, if a question has three parts, the correct response to one or two parts of the question that required work to be shown is not considered a fully correct response.

• Students who use trial and error to solve a problem must show their method. Merely showing that the answer checks or is correct is not considered a complete response for full credit.

Exercise 18 Rubric

6 Rubric for Scoring Extended Response

Score Specifi c Criteria

4 The equation and calculations written in part a correctly represent the information provided. They show that 250 canned goods were collected on Day 1 and that after 5 days, 5(250) or 1250 cans will be collected. In part b, students should note that the total number of cans collected each day will vary and that the total after Day 5 is highly unlikely to be exactly 1250.

3 A generally correct solution, but may contain minor flaws in reasoning or computation.

2 A partially correct interpretation and/or solution to the problem.

1 A correct solution with no evidence or explanation.

0 An incorrect solution indicating no mathematical understanding of the concept or task, or no solution is given.


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Chapter 6 Quiz 2(Lessons 6-3 and 6-4)

CCCh tt QQ iiNAME DATE PERIOD

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SCORE

Graph each system of equations. Then determine whether the system has no solution, one solution, or infinitely many solutions. If the system has one solution, name it.

1. y =

3 −

2 x 2. x - 2y =

-2 y = -x + 5 x - 2y = 3

For Questions 3 and 4, use substitution to solve each system of equations. If the system does not have exactly one solution, state whether it has no solutions or infinitely many solutions.

3. 3x - 2y = -7 4. -6x - 2y = -20y = x + 4 y = -3x + 10

5. MULTIPLE CHOICE In order for José and Marty to compete against each other during the wrestling season next year they need to be in the same weight category. José weighs 180 pounds and plans to gain 2 pounds per week. Marty weighs 249 pounds and plans to lose 1 pound per week. In how manyweeks will they weigh the same?

A 23 B 34.5 C 69 D 226

For Questions 1–4, use elimination to solve each system of equations.

1. x + y = 4 2. -2x + y = 5x - y = 7 2x + 3y = 3

3. 4x + 6y = -10 4. 2x + 3y = 18x - 3y = 25 5x - 4y = 14

5. MULTIPLE CHOICE If 5x - 3y = 7 and -3x - 5y = 23, what is the value of x?

A (-1, -4) B (-4, -1) C -4 D -1

1.

2.

3.

4.

5.

y

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y

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2.

3.

4.

5.

6 Chapter 6 Quiz 1(Lessons 6-1 and 6-2)

SCORE


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SCORE

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Chapter 6 Quiz 4(Lessons 6-6)

6


1. 2y + 1 = x 2. 7x + 5y = 293x + y = 17 21x - 25y = -33

3. -2x + 5y = 14 4. 4x + 2y = 1 -2x - 3y = -2 -4x + 4y = 5

5. MULTIPLE CHOICE Timothy and his brothers are selling lemonade and cookies to raise funds for a camping trip. Paul bought two cups of lemonade and two cookies for $3 and Nancy bought three cookies and two cups of lemonade for $4.00. What is the cost of a cup of lemonade?

A $2 B $1 C $0.75 D $0.50

6 Chapter 6 Quiz 3(Lessons 6-5)

1.

2.

3.

4.

5.

1. Solve the system of inequalities by graphing. y < -3x + 2

y - 5x ≤ 3

2. Solve the system of inequalities by graphing. y < 1 −

2 x + 2

y ≥ -3x -1

3. MULTIPLE CHOICE LaShawn designs websites for local businesses. He charges $25 an hour to build a website, and charges $15 an hour to update websites once he builds them. He wants to earn at least $100 every week, but he does not want to work more than 6 hours each week. What is a possible solution to describe how many hours LaShawn can spend building a website (x) and updating a website (y) in a week?

A (1, 4) B (1, 6) C (2, 3) D (3, 3)

1.

2.

3.

x

y

y

x

SCORE


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Part I Write the letter for the correct answer in the blank at the right of each question.

Use the graph for Questions 1 and 2.For Questions 1 and 2, determine how many solutions exist for each system of equations.

A no solutionB one solution

C infinitely many solutionsD cannot be determined

1. x + y = 3 2. x + y = 3 x - y = 3 x + y = -2

3. The solution to which system of equations has a positive y value? F x + y = 3 G x + y = 3 H x - 3y = -2 J x + y = 3

x - 3y = -2 x + y = -2 x + y = -2 x - y = 3

4. If y = 5x - 3 and 3x - y = -1, what is the value of y? A 2 B -1 C 7 D -8

5. Use elimination to solve the system x - 5y = -6of equations for y. x + 2y = 8

F 2−

3G 2 H 4 J 4 2−

3

6. How many solutions does the system 2y = 10x - 14 and 5x - y = 7 have? A one B two C none D infinitely many

Part II

7. Graph the system of equations. Then determine whether it has no solution, one solution, or infinitely many solutions. If the system has one solution name it.

x - 3y = -3 x + 3y = 9

8. Use substitution to solve the 4x + y = 0system of equations. 2y + x = -7

For Questions 9–11, use elimination to solve each system of equations.

9. x - 4y = 11 10. 2r + 3t = 9 11. 9x + 2y = -17 5x - 7y = -10 3r + 2t = 12 -11x + 2y = 3

12. At the end of a recent WNBA regular season, the Phoenix Mercury had 12 more victories than losses. The number of victories they had was one more than two times the number of losses. How many regular season games did the Phoenix Mercury play during that season?

y

x

x + y = 3

x - 3y = -2

x - y = 3x + y = -2

y = x + 13

23

7.

8.

9.

10.

11.

12.

y

x

Chapter 6 Mid-Chapter Test(Lessons 6-1 through 6-4)

6

y

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6.


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SCORE 6 Chapter 6 Vocabulary Test

Choose from the terms above to complete each sentence.

1. Two equations, such as y = 3x - 6 and y = 12 + 4x, together are

called a(n) .

2. If the graphs of the equations of a system intersect or coincide,

the system of equations is .

3. If a system of equations has an infinite number of solutions,

the system is .

4. Sometimes adding two equations of a system will give an equation with only one variable. This is helpful when you are

solving the system by .

Define the terms in your own words.

5. substitution

6. independent

consistentdependent

eliminationinconsistent

independentsubstitution

system of equationssystem of inequalities

?

?

?

?

1.

2.

3.

4.


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SCORE

Write the letter for the correct answer in the blank at the right of each question.

Use the graph for Questions 1–4.

1. The ordered pair (3, 2) is the solution of which system of equations?

A y = -

1 −

3 x + 3 C y = -

1 −

3 x + 3

y = -3x + 2 y = 2x -4

B y = -

1 −

3 x + 3 D y = -3x + 2

y = 2x y = 2x - 4

For Questions 2–4, find how many solutions exist for each system of equations. F no solution H one solution G infinitely many solutions J cannot be determined 2. y = 2x 3. y = 2x 4. y = -3x + 2

y = 2x - 4 y - 2x = 0 y = 2x

5. When solving the system of equations, r = 4 - twhich expression could be substituted 3r + 2t = 15for r in the second equation?

A 4 - t B 4 - r C t - 4 D 4 − t

6. If x = 2 and 3x + y = 5, what is the value of y? F 0 G -1 H 11 J 10

7. Use substitution to solve the system n = 3m -11of equations. 2m + 3n = 0

A (-2, 3) B (-3, 2) C (3, -2) D (2, -3)

8. Use elimination to solve the system x - y = 5of equations. x + y = 3

F (4, 1) G (4, -1) H (-4, 1) J (-4, -1)

9. Use elimination to solve the system x + 6y = 10of equations. x + 5y = 9

A (1, 4) B (4, 1) C (-1, -4) D (-4, -1)

10. Use elimination to find the value of x in 2x + 2y = 10the solution for the system of equations. 2x - 3y = 5

F 1 G 10 H 4 J -2

11. To eliminate the variable y in the system of equations, 6x + 4y = 22multiply the second equation by which number? 2x - y = 1

A 3 B 9 C 22 D 4

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

Chapter 6 Test, Form 1

y

xO

y - 2x = 0

y = 2x - 4

y =-3x + 2

y = 2x

y = - x + 313

6


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6 Chapter 6 Test, Form 1 (continued)

For Questions 12 and 13, determine the best method to solve the system of equations. A substitution C elimination using subtraction B elimination using addition D elimination using multiplication

12. 5x - 2y = 4 13. y = 3x + 122x + 2y = 8 2x + y = 16

14. The length of a rectangle is three times the width. The sum of the length and the width is 24 inches. What is the length of the rectangle?F 3 inches G 6 inches H 9 inches J 18 inches

15. An airport shuttle company owns sedans that have a maximum capacity of 3 passengers and vans that have a maximum capacity of 8 passengers. Their 12 vehicles have a combined maximum capacity of 61 passengers. How many vans does the company own?

A 5 B 8 C 12 D 7

For Questions 16 and 17, use the graph. 16. What is a solution to the system of inequalities?

A (0, 2) C (1, 1)

B (0, 0) D (2, 4)

17. What system of inequalities is represented? F y < x + 2 G y > x + 2 H y > x + 2 J y < x + 2 y ≤ -x + 1 y ≤ -x + 1 y ≥ -x + 1 y ≥ -x + 1

Bonus Determine the best method to solve x - 3y = 0 the system of equations. Then solve 2x - 7y = 0the system.

12. 13.

14.

15.

16.

17.

B:

x

y


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SCORE


Use the graph for Questions 1–4.For Questions 1 and 2, determine how many solutions exist for each system of equations.

A no solutionB one solutionC infinitely many solutionsD cannot be determined

1. y = 3x + 3 2. x + 2y = -13x - y = 2 2x + 3y = 0

3. The solution to which system of equations has an x value of 3? F x + 2y = -1 G 3x - y = 2 H y = 3x + 3 J 2x + 3y = 0

y = 3x + 3 x + 2y = -1 2x + 3y = 0 x + 2y = -1

4. The solution to which system of equations has a y value of 0? A x + 2y = -1 B 3x - y = 2 C y = 3x + 3 D 2x + 3y = 0

y = 3x + 3 x + 2y = -1 2x + 3y = 0 x + 2y = -1

5. When solving the system of equations, x + 2y = 15which expression could be substituted 5x + y = 21for x in the second equation?

F 15 - 2y G 21 - 5x H 15 - x −

2 J

21- 2y −

5

6. If x = 2y + 3 and 4x - 5y = 9, what is the value of y? A 2 B 1 C -1 D -2

7. Use elimination to solve the system x + 7y = 16 and 3x - 7y = 4 for x. F 3 G 4 H 5 J -6

8. Use elimination to solve the system x - 5y = 20 and x + 3y = -4 for x. A 5 B -3 C 10 D -40

9. Use elimination to solve the system 8x - 7y = 5 and 3x - 5y = 9 for y. F -2 G 8 H -3 J -1

10. Use elimination to solve the system 4x + 6y = 10 and 2x + 5y = 1 for x.

A 11 B 5 1 −

2 C -2 D -

1 −

2

11. The substitution method should be used to solve which system of equations?

F 4x + 3y = 6 G 2x + 5y = 1 H 6x + 2y = 1 J y = 3x + 15x - 3y = 2 2x - 3y = 4 3x + 4y = 7 2x + 4y = 5

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

y

xO

2x + 3y = 0

x + 2y = -1

y = 3x + 3

y = 3x - 2

3x - y = 2

6 Chapter 6 Test, Form 2A

053_060_ALG1_A_CRM_C06_AS_660280.indd 53053_060_ALG1_A_CRM_C06_AS_660280.indd 53 12/22/10 12:04 AM12/22/10 12:04 AM

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6 Chapter 6 Test, Form 2A (continued)

12.

13.

14.

15.

16.

17.

12. Use substitution to solve the system x + 2y = 1 and 2x + 5y = 3.A (-1, 1) B (1, -1) C (-5, 3) D (-1, -1)

For Questions 13 and 14, solve the system and find values of y. 13. 3x - 5y = -35 F 4 G 4 −

5 H -4 J -

4

−

5

2x - 5y = -30

14. 3x + 4y = -30 A -6 B 6 C -12 D 12 2x - 5y = 72

15. Coffee Cafe makes 90 pounds of coffee that costs $6 per pound. The types of coffee used to make this mixture cost $7 per pound and $4 per pound. How many pounds of the $7-per-pound coffee should be used in this mixture?

F 30 lb G 40 lb H 50 lb J 60 lb

16. Your teacher is giving a test that has 5 more four-point questions than six-point questions. The test is worth 120 points. Which system represents this information? A x + 5 = y B x + y = 5 C x - y = 5 D x - y = 5

4x + 6y = 120 6x + 4y =120 6x + 4y = 120 4x + 6y = 120

17. What system of inequalities is represented in the graph?

F y < -2x + 1 H y < -2x + 1 y ≤ 1 −

5 x - 1 y ≥ 1 −

5 x - 1

G y > -2x + 1 J y > -2x + 1 y ≤ 1 −

5 x - 1 y ≥ 1 −

5 x - 1

Bonus Where on the graph of 2x - 6y = 7 is the x-coordinate twice the y-coordinate? B:

x

y


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SCORE


Use the graph for Questions 1–4. For Questions 1 and 2, determine how many solutions exist for each system of equations.

A no solutionB one solutionC infinitely many solutionsD cannot be determined

1. 3x - 3y = -6 2. x + y = 0y = x + 2 3x - y = 2

3. The solution to which system of equations has an x value of -1?F x + y = 0 G 3x - y = 2 H x + y = 0 J y = -x - 2

3x - y = 2 y = -x - 2 3x - 3y = -6 x + y = 0

4. The solution to which system of equations has a y value of -2? A x + y = 0 B 3x - y = 2 C x + y = 0 D y = -x - 2 3x - y = 2 y = -x - 2 3x - 3y = -6 x + y = 0

5. When solving the system of equations, 3x + y = 14which expression could be substituted x + 4y = 3for y in the second equation?

F 3 - 4y G 3 - x −

4 H

14 - y −

3 J 14 - 3x

6. If x = 5y - 1 and 2x + 5y = -32, what is the value of y? A -2 B 2 C 1 D -1

7. Use elimination to solve the system 3x + 5y = 16 and 8x - 5y = 28 for x. F -6 G 5 H 4 J 4 −

5

8. Use elimination to solve the system x - 4y = 1 and x + 2y = 19 for x. A -11 B 3 C 25 D 13

9. Use elimination to solve the system 4x + 7y = -14 and 8x + 5y = 8 for x. F 3

1 −

2 G -1

1 −

2 H 8 J -4

10. Use elimination to solve the system 5x + 4y = -10 and 3x + 6y = -6 for y. A -2 B -5 C 0 D 2

11. The substitution method should be used to solve which system of equations? F 5x - 7y = 16 G 4x + 3y = -5 H x = 3y + 1 J 2x + 6y = 3 2x - 7y = 12 6x - 3y = 2 2x + y = 7 3x + 2y = -1

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

y

xO

y = x + 2

y = -x - 2

x + y = 0

3x - y = 2

3x - 3y = - 6

6 Chapter 6 Test, Form 2B


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6 Chapter 6 Test, Form 2B (continued)

12. Use substitution to solve the system x - 2y = 1 and 6x - 5y = 20.A (2, 5) B (-5, -2) C (5, 2) D (-2, -5)

For Questions 13 and 14, solve the system and find the value of y. 13. 2x + 3y = 1 5x - 4y = -32 F -8 3 −

7 G 8 3 −

7 H -3 J 3

14. 6x + 3y = 12 5x + 3y = 0

A -20 B -1 9 −

11 C 20 D 1 9 −

11

15. Colortime Bakers wants to make 30 pounds of a berry mix that costs $3 per pound to use in their pancake mix. They are using blueberries that cost $2 per pound and blackberries that cost $3.50 per pound. How many pounds of blackberries should be used in this mixture?

F 15 lb G 20 lb H 10 lb J 30 lb

16. Your teacher is giving a test that has 12 more three-point questions than five-point questions. The test is worth 100 points. Which system represents this information? A x + y = 12 B x + y = 12 C x - y = 12 D x - y = 12

3x + 5y = 100 5x + 3y = 100 3x + 5y = 100 5x + 3y = 100

17. What system of inequalities is represented in the graph?

F y ≥ - 3x G y ≥ - 3x H y ≤ - 3x J y ≤ - 3x

y < - 1 −

2 x - 2 y > - 1 −

2 x - 2 y < - 1 −

2 x - 2 y > - 1 −

2 x - 2

x

y

Bonus Manuel is 8 years older than his sister. Three years ago he was 3 times older than his sister. How old is each now?

12.

13.

14.

15.

16.

17.

B:


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y

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1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

Use the graph at the right to determine whether each system has no solution, one solution, or infinitely many solutions.

1. y = -x x + y = 3

2. x - 2y = -34x + y = 6


3. y = -x + 4 4. 2x - y = -3y = x - 4 6x - 3y = -9

Use substitution to solve each system of equations. If the system does not have exactly one solution, state whether it has no solution or infinitely many solutions.

5. y = 3x 6. 5x - y = 10x + y = 4 7x - 2y = 11

Use elimination to solve each system of equations. 7. x + 4y = -8 8. 2x + 5y = 3

x - 4y = -8 -x + 3y = -7

9. 2x - 5y = -16 10. 2x + 5y = 9-2x + 3y = 12 2x + y = 13


11. x = 2y - 1 12. 5x - y = 173x + y = 11 3x - y = 13

13. The sum of two numbers is 17 and their difference is 29. What are the two numbers?

14. Adult tickets for the school musical sold for $3.50 and student tickets sold for $2.50. Three hundred twenty-one tickets were sold altogether for $937.50. How many of each kind of ticket were sold?

15. Ayana has $2.35 in nickels and dimes. If she has 33 coins in all, find the number of nickels and dimes.

y

xO

y = -x

x + y = 0

x + y = 3

x - 2y = -3

4x + y = 6

y

xO

6 Chapter 6 Test, Form 2C


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For Questions 16 and 17, solve the system of inequalities by graphing.

16. y > x + 2

y ≤ -2x - 1

17. y ≤ 1 −

4 x + 2

y ≥ 2x

18. To qualify for a certain need-based scholarship, a student must get a score of at least 75 on a qualification test. In addition, the student’s family must make less than $40,000 a year. Define the variables, write a system of inequalities to represent this situation, and name one possible solution.

Bonus Mavis is 5 years older than her brother. Five years ago she was 2 times older than her brother. How old is each now?

16.

x

y

17.

x

y

18.

6 Chapter 6 Test, Form 2C (continued)

B:


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SCORE

Use the graph at the right to determine whether each system has no solution, one solution, or infinitely many solutions.

1. x + y = 3y = -x + 3

2. 3x - y = -32x - 3y = -2


3. y = -x + 3 4. 2x - y = 5y = x - 3 4x - 2y = 10


5. y = 2x 6. 2x - y = 32x + y = 8 5x + 7y = 17


7. 2x + 3y = 19 8. 6x + 4y = 202x - 3y = 1 4x - 2y = 4

9. 2x + 2y = 6 10. 7x + 3y = 13x - 2y = -11 9x + 3y = -3


11. y = 3x + 1 12. 5x - 15y = -20x - 2y = 8 5x - 4y = -9

13. The sum of two numbers is 16 and their difference is 20. What are the two numbers?

14. Kyle just started a new job that pays $7 per hour. He had been making $5 per hour at his old job. Kyle worked a total of 54 hours last month and made $338 before deductions. How many hours did he work at his new job?

15. Brent has $3.35 in quarters and dimes. If he has 23 coins in all, find the number of quarters and dimes.

1.

2. 3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

y

x

y

x

y

xO

2x - 3y = -2x + y = 3

y = -x + 3

x + y = -1

3x - y = -3

6 Chapter 6 Test, Form 2D


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16. y < 1 −

3 x + 1

y ≤ 2x - 3

17. y ≤ x + 3

y > - 1 −

2 x - 2

18. To qualify for a certain car loan, a customer must have a credit score of at least 600. In addition, the cost of the car must be at least $5000. Define the variables, write a system of inequalities to represent this situation, and name one possible solution.

Bonus Find the point on the graph of 3x - 4y = 9 where the y-coordinate is 3 times the x-coordinate.

16.

x

y

17.

x

y

18.

6 Chapter 6 Test, Form 2D (continued)

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SCORE

Graph each system of equations. Determine whether the system has no solution, one solution, or infinitely many solutions. If the system has one solution, name it.

1. 1 −

3 y = x 2. x + 3y = 3

y + x + 4 = 0 3y = -x + 9


3. y = 2x - 7 4. 4y - 3x = 5

3x - 4y = 8 3 −

4 x = y - 4

5. x - 2y = -3 6. y = -x + 3y = 3x - 1 x + y = -1


7. 6x - 7y = 21 8. 0.2x + 0.5y = 0.73x + 7y = 6 -0.2x - 0.6y = -1.4

9. 2x + 2 −

3 y = -8 10. 1 −

2 x + 2 −

5 y = -10

1 −

2 x - 1 −

3 y = 1 3x + 6y = -6


11. x + y = 147 12. 7y = 2 1 −

2 - 2x

25x + 10y = 2415 5x = 3y - 4

13. Three times one number added to five times a second number is 68. Three times the second number minus four times the first number is 6. What are the two numbers?

14. The difference of two numbers is 5. Five times the lesser number minus the greater number is 9. What are the two numbers?

15. A trail mix that costs $2.45 per pound is mixed with a trail mix that costs $2.30 per pound. How much of each type of trail mix must be used to have 30 pounds of a trail mix that costs $2.35 per pound?

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

y

x

6 Chapter 6 Test, Form 3

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16. A boat travels 60 miles downstream in the same time it takes to go 36 miles upstream. The speed of the boat in still water is 15 mph greater than the speed of the current. Find the speed of the current.


17. y < - 1 −

4 x + 2

y ≤ x - 1

18. -2x - y > 1

1 −

2 x - y ≤ 1

19. Mr. Nordmann gets a commission of $2.50 on each pair of women’s shoes he sells, and a commission of $3 on each pair of men’s shoes he sells. To meet his sales targets, he must sell at least 10 pairs of women’s shoes and at least 5 pairs of men’s shoes. He also wants to make at least $60 a week in commissions. Define the variables, write a system of inequalities to represent this situation, and name one possible solution.

Bonus Find the area of the polygon formed by the system of equations y = 0, y = 2x + 4 and -3y - 4x = -12.

16.

17.

x

y

18.

x

y

19.

B:

6 Chapter 6 Test, Form 3 (continued)


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SCORE

Demonstrate your knowledge by giving a clear, concise solution to each problem. Be sure to include all relevant drawings and justify your answers. You may show your solution in more than one way or investigate beyond the requirements of the problem.

1. Ruth invests $10,000 in two accounts. One account has an annual interest rate of 7%, and the other account has an annual interest rate of 5%. Let I represent the total interest earned in one year. Then Ruth’s investments can be modeled by the system of equations x + y = 10,000 and 0.07x + 0.05y = I.

a. Determine a solution for this system of equations that represents a possible investment, and find the value of I that corresponds to your solution.

b. Find the solution for the system if I = 800. Explain why this solution does not represent a possible investment.

2. A car rental company wants to charge a rate of $A per day plus $B per mile to rent a compact car. Their leading competitor charges $15 per day plus $0.25 per mile to rent a compact car.

a. Explain how the system of equations y = A + Bx and y = 15 + 0.25x compares the total cost of renting a compact car for one day from this company and from their leading competitor.

b. Describe how the value of A and the value of B affects the comparison of the total cost of any one-day rental from these two companies.

3. A bookstore makes a profit of $2.50 on each book they sell, and $0.75 on each magazine they sell. Each week the store sells x books and y magazines. Let $P be the weekly profit, and $S be the weekly sales for the bookstore.

a. Write a system of equations that models the possible weekly sales and weekly profit for the book store. Then describe possible values for P and S.

b. Choose a value for P and a value for S, and substitute the values into the system of equations from part a. Make a graph of this system of equations, and describe what the graph represents.

4. A bird observatory is keeping a tally on the number of adult and baby condors nesting on their grounds. There are 21 birds in total. The number of adult condors is 3 fewer than 5 times the number of babies.

a. Write a system of linear equations to model the situation. b. How many adult and baby condors are there?

6 Chapter 6 Extended-Response Test


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SCORE 6 Standardized Test Practice(Chapters 1–6)

1. Evaluate [1 + 4(5)] + [3(9) - 7]. (Lesson 1-2)

A 45 B 27 C 41 D 31

2. Rewrite 3(4a - 2b + c) using the Distributive Property. (Lesson 1-4)

F 12a - 2b + c G 12a -2b + 3c H 12a - 6b + c J 12a - 6b + 3c

3. How many liters of a 10% saline solution must be added to 4 liters of a 40% saline solution to obtain a 15% saline solution? (Lesson 2-9)

A 20 L B 4 L C 2 L D 48 L

4. When Nick was traveling in Montreal, the currency exchange rate between the U.S. and Canada could be modeled by d = 1.02c whered represents the number of U.S. dollars and c represents the numberof Canadian dollars. Solve the equation for Canadian dollar amountsof $1, $2, $5, and $20. (Lesson 3-1)

F {(1, 1.02), (2, 2.04), (5, 5.10), (20, 20.40)} G {(1, 1), (2, 2), (5, 5), (20, 20)} H {(1, 0.98), (2, 1.96), (5, 4.9), (20, 19.61)} J {(1, 2.02), (2, 3.02), (5, 6.02), (20, 21.02)}

5. If a line passes through (0, -6) and has a slope of -3, what is the equation of the line? (Lesson 4-2)

A y = -6x - 3 B x = -6y - 3 C y = -3x - 6 D x = -3y - 6

6. If r is the slope of a line, and m is the slope of a line perpendicular to that line, what is the relationship between r and m? (Lesson 4-4)

F There is no relationship. H r = m G r = -m J r = - 1

−

m

7. Which inequality does not have the solution {t | t > 4}? (Lesson 5-2)

A -t < -4 B 3t > 12 C t −

2 > 2 D - t −

8 > -

1 −

2

8. Solve 4 - 2r ≥ 3(5 - r) + 7(r + 1). (Lesson 5-3)

F {

r | r ≤ - 3 −

2 }

G {r | r ≤ -3} H {r | r ≤ -2} J {

r | r ≤ -

9 −

4 }

9. Write a compound inequality for the graph. (Lesson 5-4)

A x < -1 and x ≥ 2 C x ≤ -1 or x > 2 B x < -1 or x ≥ 2 D x ≤ -1 and x > 2

10. Evaluate x2 + 5(y - 3) when x = -3 and y = 14. (Lesson 1-2)

F 64 G 58 H 61 J 29

Part 1: Multiple Choice

Instructions: Fill in the appropriate circle for the best answer.

-1-2-3-4 0 1 2 43

1. A B C D

2. F G H J

3. A B C D

4. F G H J

5. A B C D

6. F G H J

7. A B C D

8.

9. A B C D

10. F G H J

F G H J


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6 Standardized Test Practice (continued)

11. Solve 8x - 5 = 19. (Lesson 2-3)

A 7 −

2 B -3 C 3 D 6

12. Solve -

4 − x = 6 −

9 . (Lesson 2-3)

F -6 G 6 H 12 J -12

13. Use the graph to determine how many solutions exist for the system -4x + 3y = 12 and x + y = 2. (Lesson 6-1)

A 0 C 2 B 1 D infinitely many

14. Use elimination to solve the system 2x + y = -8 and -2x + 3y = -8 for x. (Lesson 6-3)

F -2 G 2 H -4 J 4

15. The substitution method could be used to solve which system of equations? (Lesson 6-2)

A 5x - 7y = 8 C 3x - 3y = 5 2x + 6y = 6 -2x + 2y = -6

B 3x + 2y = 8 D x = 4y + 6 4x + 3y = 5 3x - 2y = 3

16. Which graph represents the solution of � n + 5 � > 1? (Lesson 5-5)

F H

G J

For Questions 17 and 18, determine the value that is missing. 17. The solution set 17. 18.

is {n � n ≥ 15} for the inequality n - 7 ≥ ___. (Lesson 5-1)

18. If � a - 8 � = 17, then a = ___ or a = -9. (Lesson 2-5)

-4-3-2-1 0 2 3 4 5 6 71-1 0 1 2 3 4 5 6 7 8 9 10

-9-8-7-6-5-4-3-2-1 0 21-1-2-3-5-6 -4 0 1 2 3 4 5

y

xO

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8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

11. A B C D

12. F G H J

13. A B C D

14. F G H J

15. A B C D

16. F G H J

Part 2: Gridded ResponsePart 2: Gridded Response

Instructions: Enter your answer by writing each digit of the answer in a column

box and then shading in the appropriate circle that corresponds to that entry.

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0


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19.

20.

21.

22.

23.

24.

25.

26.

27.

28a.

28b.

Part 3: Short Response

Instructions: Write your answers in the space provided.

19. State whether the percent of change is a percent of increase or a percent of decrease. Then find the percent of change. original: 76; new: 57 (Lesson 2-7)

20. Express the relation {(-2, 1), (3, -1), (2, -2), (-2, 0)} as a mapping. (Lesson 1-6)

21. Determine whether -6, -3, 0, 3 … is an arithmetic sequence. If it is, state the common difference. (Lesson 3-4)

22. Find the slope of the line that passes through (-2, 0) and (5, -8). (Lesson 3-3)

23. The Lopez family drove 165 miles in 3 hours. Write a direct variation equation for the distance driven in any time. How far can the Lopez family drive in 5 hours? (Lesson 3-4)

24. Write an equation of a line that passes through (-2, -1) with slope 3. (Lesson 4-3)

25. Solve the system of inequalities by graphing. (Lesson 6-6) 2x - y ≥ 4 x - 2y < 4

26. Solve 12 + r < 15. Then graph the solution. (Lesson 5-1)

27. Define a variable, write a compound inequality, and solve the problem. (Lesson 5-4)

Seven less than twice a number is greater than 13 or less than or equal to -5.

28. Three times a first number minus a second number equals negative forty. The first number plus twice the second number equals negative four. (Lesson 6-5)

a. Define variables and formulate a system of linear equations from this situation.

b. What are the numbers?

y

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-1-2-3-4 0 1 2 43

6 Standardized Test Practice (continued)


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1. Graph 3x - y = 1.

2. Solve 4x + 9 = 4x + 13.

3. Find the value of r so that the line through (2, -3) and (-4, r) has a slope of - 1 −

2 .

4. A giraffe can travel 800 feet in 20 seconds. Write a direct variation equation for the distance traveled in any time.

5. Find the 25th term of the arithmetic sequence with first term 7 and common difference -2.

6. Write an equation of the line whose slope is 2 and whose y-intercept is 9.

7. Write an equation of the line that passes through (-1, -7) and (1, 3).

8. Write y - 4 = - 3 −

2 (x + 6) in standard form.

9. Write the slope-intercept form of an equation of the linethat passes through (-2, 0) and is parallel to the graph of y = -3x - 2.

10. The table below shows the distance driven during four different trips and the duration of each trip. Draw a scatter plot and determine what relationship exists, if any, in the data. Write an equation for a line of fit for the data.

Time (hours) 1 2 2.5 4

Distance (miles) 50 85 120 180

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

y

xO

6 Unit 2 Test (Chapters 3–6)

40

0

80

120

160

1 2 3 4

Dis

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mile

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Time (hours)


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11. The table below shows the cost to ride the New York City subway in various years.

Year 1985 1987 1990 1994 2000 2007

Subway Fare $0.90 $1.00 $1.15 $1.25 $1.50 $2.00

Source: Metropolitan Transportation Authority (MTA)

Use a regression line to estimate the cost of a subway ride in 2014.

Solve each inequality. 12. 4x - 5 < 7x + 10 13. 2(5a - 4) - 3(6 + 2a) ≤ 6

Solve each compound inequality. 14. 5 < 2t + 7 < 11 15. 13 < 4 - 3v or 2v - 14 > 8

For Questions 16 and 17, solve each open sentence. Then graph the solution set.

16. |3b - 5| ≤ 7 17. |w + 5| > 1

18. Use a graph to determine whether the system x - y = 4 and y = x has no solution, one solution, or infinitely many solutions.

For Questions 19–22, determine the best method to solve each system of equations. Then solve the system.

19. x + y = 2 20. -x - 5y = 7y = 2x -1 x + y = 1

21. 3x + y = 26 22. 4x - 8y = 523x + 3y = 18 7x + 4y = 1

23. Solve the system of inequalities by graphing. 2x - y > 3x + 2y ≤ 4

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23. y

xO

-3-2-1 0 1 2 4 53

y

xO

6 Unit 2 Test (continued)

-7-6-5-4-8 -3-2-1 0


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Chapter 6 A1 Glencoe Algebra 1

Chapter Resources

Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.

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sibl

e to

hav

e ex

actl

y tw

o so

luti

ons

to a

sys

tem

of

lin

ear

equ

atio

ns

5.

Th

e m

ost

accu

rate

way

to

solv

e a

syst

em o

f eq

uat

ion

s is

to

grap

h t

he

equ

atio

ns

to s

ee w

her

e th

ey i

nte

rsec

t.

6.

To

solv

e a

syst

em o

f eq

uat

ion

s, s

uch

as

2x -

y =

21

and

3y =

2x

- 6

, by

subs

titu

tion

, sol

ve o

ne

of t

he

equ

atio

ns

for

one

vari

able

an

d su

bsti

tute

th

e re

sult

in

to t

he

oth

er e

quat

ion

.

7.

Wh

en s

olvi

ng

a sy

stem

of

equ

atio

ns,

a r

esu

lt t

hat

is

a tr

ue

stat

emen

t, s

uch

as

-5

= -

5, m

ean

s th

e eq

uat

ion

s do

not

sh

are

a co

mm

on s

olu

tion

.

8.

Add

ing

the

equ

atio

ns

3x -

4y

= 8

an

d 2x

+ 4

y =

7 r

esu

lts

in a

0 c

oeff

icie

nt

for

y.

9.

Th

e eq

uat

ion

7x

- 2

y =

12

can

be

mu

ltip

lied

by

2 so

th

at t

he

coef

fici

ent

of y

is

-4.

10.

Th

e re

sult

of

mu

ltip

lyin

g -

7x -

3y

= 1

1 by

-3

is

-1x

+ 9

y =

11.

Step

1

Step

2

D A D A D A D A A D

001_

009_

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AM


NA

ME

DAT

E

P

ER

IOD

Lesson 6-1

Cha

pte

r 6

5 G

lenc

oe A

lgeb

ra 1

Poss

ible

Nu

mb

er o

f So

luti

on

s T

wo

or m

ore

lin

ear

equ

atio

ns

invo

lvin

g th

e sa

me

vari

able

s fo

rm a

sys

tem

of

equ

atio

ns.

A s

olu

tion

of

the

syst

em o

f eq

uat

ion

s is

an

ord

ered

pa

ir o

f n

um

bers

th

at s

atis

fies

bot

h e

quat

ion

s. T

he

tabl

e be

low

su

mm

ariz

es i

nfo

rmat

ion

ab

out

syst

ems

of l

inea

r eq

uat

ion

s.

Gra

ph

of

a S

yste

m

inte

rsec

ting

lines

sa

me

line

para

llel l

ines

N

um

ber

of

So

luti

on

s ex

actly

one

sol

utio

n in

fi nite

ly m

any

solu

tions

no

sol

utio

n

Te

rmin

olo

gy

cons

iste

nt a

nd

cons

iste

nt a

nd

inco

nsis

tent

inde

pend

ent

depe

nden

t

U

se t

he

grap

h a

t th

e ri

ght

to d

eter

min

e

wh

eth

er e

ach

sys

tem

is

con

sist

ent

or i

nco

nsi

sten

t an

d

if i

t is

in

dep

end

ent

or d

epen

den

t.

a. y

= -

x +

2

y =

x +

1S

ince

th

e gr

aph

s of

y =

-x

+ 2

an

d y

= x

+ 1

in

ters

ect,

th

ere

is o

ne

solu

tion

. Th

eref

ore,

th

e sy

stem

is

con

sist

ent

and

inde

pen

den

t.b

. y

= -

x +

2

3x +

3y

= -

3S

ince

th

e gr

aph

s of

y =

-x

+ 2

an

d 3x

+ 3

y =

-3

are

para

llel

, th

ere

are

no

solu

tion

s. T

her

efor

e, t

he

syst

em i

s in

con

sist

ent.

c. 3

x +

3y

= -

3

y =

-x

- 1

Sin

ce t

he

grap

hs

of 3

x +

3y

= -

3 an

d y

= -

x -

1 c

oin

cide

, th

ere

are

infi

nit

ely

man

y so

luti

ons.

Th

eref

ore,

th

e sy

stem

is

con

sist

ent

and

depe

nde

nt.

Exer

cise

sU

se t

he

grap

h a

t th

e ri

ght

to d

eter

min

e w

het

her

eac

h

syst

em i

s co

nsi

sten

t or

in

con

sist

ent

and

if

it i

s in

dep

end

ent

or d

epen

den

t.

1. y

= -

x -

3

2. 2

x +

2y

= -

6y

= x

- 1

y

= -

x -

3

3. y

= -

x -

3

4. 2

x +

2y

= -

62x

+ 2

y =

4

3x

+ y

= 3

6-1

Stud

y G

uide

and

Inte

rven

tion

Gra

ph

ing

Syste

ms o

f E

qu

ati

on

s

x

y

O

y =

-x

- 33x

+ y

= 3

2x +

2y

= -

6

2x +

2y

= 4

y =

x -

1

x

y O

y =

x +

1

y =

-x

- 1

3x +

3y

= -

3

y =

-x

+ 2

Exam

ple

x

y

Ox

y

Ox

y

O

con

sist

ent;

ind

epen

den

tco

nsi

sten

t; d

epen

den

t

inco

nsi

sten

tco

nsi

sten

t; in

dep

end

ent

001_

009_

ALG

1_A

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M_C

06_C

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6028

0.in

dd

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/21/

10

8:43

AM

Answers (Anticipation Guide and Lesson 6-1)

A01_A21_ALG1_A_CRM_C06_AN_660280.indd A1A01_A21_ALG1_A_CRM_C06_AN_660280.indd A1 12/21/10 8:49 AM12/21/10 8:49 AM

Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 6

6 G

lenc

oe A

lgeb

ra 1

Solv

e b

y G

rap

hin

g O

ne

met

hod

of

solv

ing

a sy

stem

of

equ

atio

ns

is t

o gr

aph

th

e eq

uat

ion

s on

th

e sa

me

coor

din

ate

plan

e.

G

rap

h e

ach

sys

tem

an

d d

eter

min

e th

e n

um

ber

of

solu

tion

s th

at i

t h

as. I

f it

has

on

e so

luti

on, n

ame

it.

a. x

+ y

= 2

x -

y =

4T

he

grap

hs

inte

rsec

t. T

her

efor

e, t

her

e is

on

e so

luti

on. T

he

po

int

(3, -

1) s

eem

s to

lie

on

bot

h l

ines

. Ch

eck

this

est

imat

e by

rep

laci

ng

x w

ith

3 a

nd

y w

ith

-1

in e

ach

equ

atio

n.

x

+ y

= 2

3 +

(-

1) =

2 �

x

- y

= 4

3 -

(-

1) =

3 +

1 o

r 4

�

Th

e so

luti

on i

s (3

, -1)

.b

. y

= 2

x +

1

2y =

4x

+ 2

Th

e gr

aph

s co

inci

de. T

her

efor

e th

ere

are

infi

nit

ely

man

y so

luti

ons.

Exer

cise

sG

rap

h e

ach

sys

tem

an

d d

eter

min

e th

e n

um

ber

of

solu

tion

s it

has

. If

it h

as o

ne

solu

tion

, nam

e it

.

1. y

= -

2 o

ne;

(-

1, -

2)

2. x

= 2

on

e; (

2, -

3)

3. y

= 1 −

2 x

on

e; (

2, 1

)

3

x -

y =

-1

2

x +

y =

1

x +

y =

3

4. 2

x +

y =

6 o

ne;

(1,

4)

5. 3

x +

2y

= 6

no

so

luti

on

6.

2y

= -

4x +

4 i

nfi

nit

ely

2x -

y =

-2

3x

+ 2

y =

-4

y =

-2x

+ 2

m

any

6-1

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Gra

ph

ing

Syste

ms o

f E

qu

ati

on

s

x

y O

y =

2x

+ 1

2y =

4x

+ 2x

y

O( 3

, –1) x -

y =

4

x +

y =

2

Exam

ple

x

y

O

3x -

y =

-1

( –1,

–2)

y =

-2

x

y

O

2x +

y =

6

2x -

y =

-2

( 1, 4

)

x

y

O

x =

2

2x +

y =

1( 2

, –3)

x

y

O

3x +

2y

= 6

3x +

2y

= -

4

x

y

O

x +

y =

3

y =

1 2x( 2

, 1)

x

y

O

2y =

-4x

+ 4

y =

-2x

+ 2

001_

009_

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10

8:43

AM


NA

ME

DAT

E

P

ER

IOD

Lesson 6-1

Cha

pte

r 6

7 G

lenc

oe A

lgeb

ra 1

Use

th

e gr

aph

at

the

righ

t to

det

erm

ine

wh

eth

er

ea

ch s

yste

m i

s co

nsi

sten

t or

in

con

sist

ent

and

if

it i

s

ind

epen

den

t or

dep

end

ent.

1. y

= x

- 1

co

nsi

sten

t;

2. x

- y

= -

4 co

nsi

sten

t;

y =

-x

+ 1

in

dep

end

ent

y =

x +

4 d

epen

den

t

3. y

= x

+ 4

4.

y =

2x

- 3

2x -

2y

= 2

2

x -

2y

= 2

i

nco

nsi

sten

t c

on

sist

ent;

in

dep

end

ent

Gra

ph

eac

h s

yste

m a

nd

det

erm

ine

the

nu

mb

er o

f so

luti

ons

that

it

has

. If

it

has

on

e so

luti

on, n

ame

it.

5. 2

x -

y =

1

6. x

= 1

7.

3x

+ y

= -

3 n

o

y =

-3

on

e; (

-1,

-3)

2

x +

y =

4 o

ne;

(1,

2)

3x

+ y

= 3

so

luti

on

x

y

O

x =

1 2x +

y =

4

( 1, 2

)

8. y

= x

+ 2

in

fi n

itel

y

9. x

+ 3

y =

-3

on

e;

10. y

- x

= -

1x

- y

= -

2 m

any

x -

3y

= -

3 (-

3, 0

) x

+ y

= 3

on

e; (

2, 1

)

x

y

y =

x +

2

x -

y =

-2

11

. x -

y =

3

12. x

+ 2

y =

4 i

nfi

nit

ely

13

. y =

2x

+ 3

no

so

luti

on

x

- 2

y =

3 o

ne;

(3,

0)

y =

- 1 −

2 x

+ 2

man

y 3

y =

6x

- 6

x

y

O

y =

2x

+ 3

3y =

6x

- 6

6-1

Skill

s Pr

acti

ceG

rap

hin

g S

yste

ms o

f E

qu

ati

on

s

x

y

y =

-x

+ 1

x -

y =

-4

2x -

2y

= 2

y =

2x

- 3

y =

x -

1

y =

x +

4

x

y

O

2x -

y =

1

y =

-3

( -1,

-3)

x

y

O

3x +

y =

3

3x +

y =

-3

x

y

O

x -

3y

= -

3

x +

3y

= -

3

( -3,

0)

x

y

x +

y =

3

( 2, 1

)

y -

x =

-1

x

y

x -

2y

= 3

x -

y =

3

( 3, 0

)

x

y

O

y =

- 1 – 2

x +

2

x +

2y

= 4

001_

009_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

712

/21/

10

8:43

AM

Answers (Lesson 6-1)


An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF 2nd



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 6

8 G

lenc

oe A

lgeb

ra 1

Use

th

e gr

aph

at

the

righ

t to

det

erm

ine

wh

eth

er

ea

ch s

yste

m i

s co

nsi

sten

t or

in

con

sist

ent

and

if

it

is i

nd

epen

den

t or

dep

end

ent.

1. x

+ y

= 3

2.

2x

- y

= -

3x

+ y

= -

3 4

x -

2y

= -

6in

con

sist

ent

co

nsi

sten

t an

d

3. x

+ 3

y =

3

4. x

+ 3

y =

3x

+ y

= -

3 2

x -

y =

-3

con

sist

ent

and

c

on

sist

ent

and

ind

epen

den

t

Gra

ph

eac

h s

yste

m a

nd

det

erm

ine

the

nu

mb

er o

f so

luti

ons

that

it

has

. If

it h

as

one

solu

tion

, nam

e it

.

5. 3

x -

y =

-2

no

6.

y =

2x

- 3

in

fi n

itel

y

7. x

+ 2

y =

3

on

e;

3x -

y =

0

solu

tio

n

4x

= 2

y +

6

man

y 3

x -

y =

-5

(-1,

2)

x

y

O

3x -

y =

-2

3x -

y =

0

8. B

USI

NES

S N

ick

plan

s to

sta

rt a

hom

e-ba

sed

busi

nes

s pr

odu

cin

g an

d se

llin

g go

urm

et d

og t

reat

s. H

e fi

gure

s it

w

ill

cost

$20

in

ope

rati

ng

cost

s pe

r w

eek

plu

s $0

.50

to

prod

uce

eac

h t

reat

. He

plan

s to

sel

l ea

ch t

reat

for

$1.

50.

a.

Gra

ph t

he

syst

em o

f eq

uat

ion

s y

= 0

.5x

+ 2

0 an

d y

= 1

.5x

to r

epre

sen

t th

e si

tuat

ion

.

b.

How

man

y tr

eats

doe

s N

ick

nee

d to

sel

l pe

r w

eek

to

brea

k ev

en?

20

9. S

ALE

S A

use

d bo

ok s

tore

als

o st

arte

d se

llin

g u

sed

CD

s an

d vi

deos

. In

th

e fi

rst

wee

k, t

he

stor

e so

ld 4

0 u

sed

CD

s an

d vi

deos

, at

$4.0

0 pe

r C

D a

nd

$6.0

0 pe

r vi

deo.

Th

e sa

les

for

both

CD

s an

d vi

deos

tot

aled

$18

0.00

a.

Wri

te a

sys

tem

of

equ

atio

ns

to

re

pres

ent

the

situ

atio

n.

b.

Gra

ph t

he

syst

em o

f eq

uat

ion

s.

c.

How

man

y C

Ds

and

vide

os d

id t

he

stor

e se

ll i

n t

he

firs

t w

eek?

6-1

Prac

tice

Gra

ph

ing

Syste

ms o

f E

qu

ati

on

s

x

y

x +

y =

-3

x +

3y

= 3

2x -

y =

-3

4x -

2y

= -

6

x +

y =

3

x

y

3x -

y =

-5

x +

2y

= 3

( –1,

2)

x

y

O

y =

2x

- 3

4x =

2y

+ 6

Sale

s ($

)

Do

g T

reats

Cost ($)

515

1020

2530

3540

450

40 35 30 25 20 15 10 5

y =

0.5

x +

20

y =

1.5

x

( 20,

30)

CD

Sal

es (

$)

CD

an

d V

ideo

Sale

s

Video Sales ($)

515

1020

2530

3540

450

40 35 30 25 20 15 10 5

4c +

6v

= 1

80( 3

0, 1

0)

c +

v =

40

c +

v =

40,

4c +

6v =

180

30 C

Ds

and

10

vid

eos

ind

epen

den

t

dep

end

ent

001_

009_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

812

/21/

10

8:43

AM


NA

ME

DAT

E

P

ER

IOD

Lesson 6-1

Cha

pte

r 6

9 G

lenc

oe A

lgeb

ra 1

1. B

USI

NES

S T

he

wid

get

fact

ory

wil

l se

ll a

tot

al o

f y

wid

gets

aft

er x

day

s ac

cord

ing

to t

he

equ

atio

n y

= 2

00x

+ 3

00.

Th

e ga

dget

fac

tory

wil

l se

ll y

gad

gets

af

ter

x da

ys a

ccor

din

g to

th

e eq

uat

ion

y

= 2

00x

+ 1

00. L

ook

at t

he

grap

h o

f th

e sy

stem

of

equ

atio

ns

and

dete

rmin

e w

het

her

it

has

no

solu

tion

, on

e so

luti

on,

or i

nfi

nit

ely

man

y so

luti

ons.

n

o s

olu

tio

n

2. A

RC

HIT

ECTU

RE

An

off

ice

buil

din

g h

as

two

elev

ator

s. O

ne

elev

ator

sta

rts

out

on

the

4th

flo

or, 3

5 fe

et a

bove

th

e gr

oun

d,

and

is d

esce

ndi

ng

at a

rat

e of

2.2

fee

t pe

r se

con

d. T

he

oth

er e

leva

tor

star

ts o

ut

at

grou

nd

leve

l an

d is

ris

ing

at a

rat

e of

1.

7 fe

et p

er s

econ

d. W

rite

a s

yste

m o

f eq

uat

ion

s to

rep

rese

nt

the

situ

atio

n.

Sam

ple

an

swer

: y =

35

- 2

.2x;

y =

1.7

x

3. F

ITN

ESS

Oli

via

and

her

bro

ther

Wil

liam

h

ad a

bic

ycle

rac

e. O

livi

a ro

de a

t a

spee

d of

20

feet

per

sec

ond

wh

ile

Wil

liam

rod

e at

a s

peed

of

15 f

eet

per

seco

nd.

To

be

fair

, Oli

via

deci

ded

to g

ive

Wil

liam

a

150-

foot

hea

d st

art.

Th

e ra

ce e

nde

d in

a

tie.

How

far

aw

ay w

as t

he

fin

ish

lin

e fr

om w

her

e O

livi

a st

arte

d?

600

ft

4. A

VIA

TIO

N T

wo

plan

es a

re i

n f

ligh

t n

ear

a lo

cal

airp

ort.

On

e pl

ane

is a

t an

al

titu

de o

f 10

00 m

eter

s an

d is

asc

endi

ng

at a

rat

e of

400

met

ers

per

min

ute

. Th

e se

con

d pl

ane

is a

t an

alt

itu

de o

f 59

00 m

eter

s an

d is

des

cen

din

g at

a r

ate

of 3

00 m

eter

s pe

r m

inu

te.

a. W

rite

a s

yste

m o

f eq

uat

ion

s th

at

repr

esen

ts t

he

prog

ress

of

each

pla

ne

y =

40

0x +

10

00;

y =

590

0 -

30

0x

b.

Mak

e a

grap

h t

hat

rep

rese

nts

th

e pr

ogre

ss o

f ea

ch p

lan

e.

6-1

Wor

d Pr

oble

m P

ract

ice

Gra

ph

ing

Syste

ms o

f E

qu

ati

on

s

Day

s2

10

43

y

x5

6

Items sold

300

400

200

100

500

800

900

700

600

Gad

get

s

Wid

get

sy

= 2

00x

+ 3

00

y =

200

x +

100

Tim

e (m

in.)

0

y

x

Altitude (m)

Firs

t Pla

ne

Seco

nd P

lane

21

43

56

78

910

2000

1000

4000

5000

6000

3000

001_

009_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

912

/24/

10

11:4

1 A

M



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF 2nd



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 6

10

G

lenc

oe A

lgeb

ra 1

Enri

chm

ent

Gra

ph

ing

a T

rip

Th

e di

stan

ce f

orm

ula

, d =

rt,

is u

sed

to s

olve

man

y ty

pes

of

t

d O

slop

e is

50

slop

e is

–25

2

50

prob

lem

s. I

f yo

u g

raph

an

equ

atio

n s

uch

as

d =

50t

, th

e gr

aph

is

a m

odel

for

a c

ar g

oin

g at

50

mph

. Th

e ti

me

the

car

trav

els

is t

; th

e di

stan

ce i

n m

iles

th

e ca

r co

vers

is

d. T

he

slop

e of

th

e li

ne

is

the

spee

d.S

upp

ose

you

dri

ve t

o a

nea

rby

tow

n a

nd

retu

rn. Y

ou a

vera

ge

50 m

ph o

n t

he

trip

ou

t bu

t on

ly 2

5 m

ph o

n t

he

trip

hom

e. T

he

rou

nd

trip

tak

es 5

hou

rs. H

ow f

ar a

way

is

the

tow

n?

Th

e gr

aph

at

the

righ

t re

pres

ents

you

r tr

ip. N

otic

e th

at t

he

retu

rn

trip

is

show

n w

ith

a n

egat

ive

slop

e be

cau

se y

ou a

re d

rivi

ng

in t

he

oppo

site

dir

ecti

on.

Sol

ve e

ach

pro

ble

m.

1. E

stim

ate

the

answ

er t

o th

e pr

oble

m i

n t

he

abov

e ex

ampl

e. A

bou

t h

ow f

ar a

way

is

the

tow

n?

a

bo

ut

80 m

iles

Gra

ph

eac

h t

rip

an

d s

olve

th

e p

rob

lem

.

2. A

n a

irpl

ane

has

en

ough

fu

el f

or 3

hou

rs o

f sa

fe f

lyin

g. O

n t

he

trip

ou

t th

e pi

lot

aver

ages

20

0 m

ph f

lyin

g ag

ain

st a

hea

dwin

d. O

n t

he

trip

bac

k, t

he

pilo

t av

erag

es 2

50 m

ph. H

ow

lon

g a

trip

ou

t ca

n t

he

pilo

t m

ake?

a

bo

ut

1 2

−

3 h

ou

rs a

nd

330 m

iles

3. Y

ou d

rive

to

a to

wn

100

mil

es a

way

. 4.

You

dri

ve a

t an

ave

rage

spe

ed o

f 50

mph

O

n t

he

trip

ou

t yo

u a

vera

ge 2

5 m

ph.

to

a di

scou

nt

shop

pin

g pl

aza,

spe

nd

On

th

e tr

ip b

ack

you

ave

rage

50

mph

. 2

hou

rs s

hop

pin

g, a

nd

then

ret

urn

at

an

How

man

y h

ours

do

you

spe

nd

driv

ing?

a

vera

ge s

peed

of

25 m

ph. T

he

enti

re t

rip

t

akes

8 h

ours

. How

far

aw

ay i

s th

e

sh

oppi

ng

plaz

a?

6

ho

urs

1

00 m

iles

t

d

O2

50

t

d

O50

2

6-1

t

d

O1

100

010_

018_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

1012

/21/

10

8:43

AM


NA

ME

DAT

E

P

ER

IOD

Lesson 6-1

Cha

pte

r 6

11

Gle

ncoe

Alg

ebra

1

Gra

phin

g Ca

lcul

ator

Act

ivit

yS

olu

tio

n t

o a

Syste

m o

f Lin

ear

Eq

uati

on

s

A g

raph

ing

calc

ula

tor

can

be

use

d to

sol

ve a

sys

tem

of

lin

ear

equ

atio

ns

grap

hic

ally

. Th

e so

luti

on o

f a

syst

em o

f li

nea

r eq

uat

ion

s ca

n b

e fo

un

d by

usi

ng

the

TR

AC

E f

eatu

re o

r by

u

sin

g th

e in

ters

ect

com

man

d u

nde

r th

e C

AL

C m

enu

.

Exer

cise

sS

olve

eac

h s

yste

m o

f li

nea

r eq

uat

ion

s.

1.

y =

2

2. y

= -

x +

3

3. x

+ y

= -

1

5x

+ 4

y =

18

y =

x +

1

2x

- y

= -

8

(2

, 2)

(1, 2)

(-

3, 2)

4. -

3x +

y =

10

5. -

4x +

3y

= 1

0 6.

5x

+ 3

y =

11

-

x +

2y

= 0

7

x +

y =

20

x -

5y

= 5

(-

4,

-2)

(2, 6)

(2.5

, -

0.5

)

7. 3

x -

2y

= -

4 8.

3x

+ 2

y =

4

9. 4

x -

5y

= 0

-

4x +

3y

= 5

-

6x -

4y

= -

8 6

x -

5y

= 1

0

(-

2,

-1)

in

fi n

ite s

olu

tio

ns

(5, 4)

S

olve

eac

h s

yste

m o

f li

nea

r eq

uat

ion

s.

a. x

+ y

= 0

x

- y

= -

4

Usi

ng T

RA

CE:

Sol

ve e

ach

equ

atio

n f

or y

an

d en

ter

each

equ

atio

n

into

Y=.

Th

en g

raph

usi

ng

Zoo

m 8

: ZIn

tege

r. U

se T

RA

CE

to

fin

d th

e so

luti

on.

Key

stro

kes:

Y=

(–)

EN

TE

R

+

4

ZO

OM

6

ZO

OM

8 E

NT

ER

T

RA

CE

.

Th

e so

luti

on i

s (-

2, 2

).

b.

2x +

y =

4

4x +

3y

= 3

Usi

ng

CA

LC:

Sol

ve e

ach

equ

atio

n f

or y

, en

ter

each

in

to t

he

calc

ula

tor,

and

grap

h. U

se C

AL

C t

o de

term

ine

the

solu

tion

.

Key

stro

kes:

Y=

(–) 2

+ 4

EN

TE

R

( (

–)

4 ÷

3

)

+

1

ZO

OM

6

2nd

[C

AL

C]

5 E

NT

ER

E

NT

ER

E

NT

ER

.

To

chan

ge t

he

x-va

lue

to a

fra

ctio

n, p

ress

2nd

[Q

UIT

]

MA

TH

EN

TE

R E

NT

ER

.

Th

e so

luti

on i

s (4

.5, -

5) o

r ( 9 −

2 , -

5 ) .

[-47, 47]

scl:

10 b

y [

-31, 31]

scl:

10

[-10, 10]

scl:

1 b

y [

-10, 10]

scl:

1

6-1

Exam

ple

010_

018_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

1112

/24/

10

7:07

AM



An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 6

12

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

Su

bsti

tuti

on

Solv

e b

y Su

bst

itu

tio

n O

ne

met

hod

of

solv

ing

syst

ems

of e

quat

ion

s is

su

bst

itu

tion

.

U

se s

ub

stit

uti

on t

o so

lve

the

syst

em o

f eq

uat

ion

s.y

= 2

x4x

- y

= -

4

Su

bsti

tute

2x

for

y in

th

e se

con

d eq

uat

ion

.

4x -

y =

-4

Sec

ond

equa

tion

4x

- 2

x =

-4

y =

2x

2x

= -

4 C

ombi

ne li

ke te

rms.

x

= -

2 Di

vide

eac

h si

de b

y 2

and

sim

plify

.

Use

y =

2x

to f

ind

the

valu

e of

y.

y =

2x

Firs

t eq

uatio

n

y =

2(-

2)

x =

-2

y =

-4

Sim

plify

.

Th

e so

luti

on i

s (-

2, -

4).

S

olve

for

on

e va

riab

le,

then

su

bst

itu

te.

x +

3y

= 7

2x -

4y

= -

6

Sol

ve t

he

firs

t eq

uat

ion

for

x s

ince

th

e co

effi

cien

t of

x i

s 1.

x

+ 3

y =

7

F

irst

equa

tion

x +

3y

- 3

y =

7 -

3y

Sub

trac

t 3y

fro

m e

ach

side

.

x

= 7

- 3

y S

impl

ify.

Fin

d th

e va

lue

of y

by

subs

titu

tin

g 7

- 3

y fo

r x

in t

he

seco

nd

equ

atio

n.

2x

- 4

y =

-6

Sec

ond

equa

tion

2(7

- 3

y) -

4y

= -

6 x

= 7

- 3

y

14

- 6

y -

4y

= -

6 D

istr

ibut

ive

Pro

pert

y

14

- 1

0y =

-6

Com

bine

like

term

s.

14 -

10y

- 1

4 =

-6

- 1

4 S

ubtr

act

14 f

rom

eac

h si

de.

-10

y =

-20

S

impl

ify.

y =

2

Divi

de e

ach

side

by

-10

and

sim

plify

.

Use

y =

2 t

o fi

nd

the

valu

e of

x.

x =

7 -

3y

x =

7 -

3(2

)x

= 1

Th

e so

luti

on i

s (1

, 2).

Exer

cise

sU

se s

ub

stit

uti

on t

o so

lve

each

sys

tem

of

equ

atio

ns.

1. y

= 4

x 2.

x =

2y

3. x

= 2

y -

33x

- y

= 1

(-

1, -

4)

y =

x -

2 (

4, 2

) x

= 2

y +

4 n

o s

olu

tio

n

4. x

- 2

y =

-1

5. x

- 4

y =

1

infi

nit

ely

6.

x +

2y

= 0

3y =

x +

4 (

5, 3

) 2

x -

8y

= 2

man

y 3

x +

4y

= 4

(4,

-2)

7. 2

b =

6a

- 1

4 in

fi n

itel

y

8. x

+ y

= 1

6 9.

y =

-x

+ 3

n

o3a

- b

= 7

m

any

2y

= -

2x +

2 n

o s

olu

tio

n

2y

+ 2

x =

4 s

olu

tio

n

10. x

= 2

y (2

0, 1

0)

11. x

- 2

y =

-5

12. -

0.2x

+ y

= 0

.50.

25x

+ 0

.5y

= 1

0

x +

2y

= -

1 (-

3, 1

) 0

.4x

+ y

= 1

.1 (

1, 0

.7)

6-2

Exam

ple

1Ex

amp

le 2

010_

018_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

1212

/21/

10

8:43

AM


NA

ME

DAT

E

P

ER

IOD

Lesson 6-2

Cha

pte

r 6

13

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Su

bsti

tuti

on

Solv

e R

eal-

Wo

rld

Pro

ble

ms

Su

bsti

tuti

on c

an a

lso

be u

sed

to s

olve

rea

l-w

orld

pr

oble

ms

invo

lvin

g sy

stem

s of

equ

atio

ns.

It

may

be

hel

pfu

l to

use

tab

les,

ch

arts

, dia

gram

s,

or g

raph

s to

hel

p yo

u o

rgan

ize

data

.

C

HEM

ISTR

Y H

ow m

uch

of

a 10

% s

alin

e so

luti

on s

hou

ld b

e m

ixed

w

ith

a 2

0% s

alin

e so

luti

on t

o ob

tain

100

0 m

illi

lite

rs o

f a

12%

sal

ine

solu

tion

?L

et s

= t

he

nu

mbe

r of

mil

lili

ters

of

10%

sal

ine

solu

tion

.L

et t

= t

he

nu

mbe

r of

mil

lili

ters

of

20%

sal

ine

solu

tion

.U

se a

tab

le t

o or

gan

ize

the

info

rmat

ion

.

10%

sal

ine

20%

sal

ine

12%

sal

ine

Tota

l mill

ilite

rss

t10

00

Mill

ilite

rs o

f sa

line

0.10

s0.

20 t

0.12

(100

0)

Wri

te a

sys

tem

of

equ

atio

ns.

s +

t =

100

00.

10s

+ 0

.20t

= 0

.12(

1000

)U

se s

ubs

titu

tion

to

solv

e th

is s

yste

m.

s

+ t

= 1

000

Firs

t eq

uatio

n

s

= 1

000

- t

S

olve

for

s.

0.

10s

+ 0

.20t

= 0

.12(

1000

) S

econ

d eq

uatio

n

0.10

(100

0 -

t)

+ 0

.20t

= 0

.12(

1000

) s

= 1

000

- t

10

0 -

0.1

0t +

0.2

0t =

0.1

2(10

00)

Dis

trib

utiv

e P

rope

rty

10

0 +

0.1

0t =

0.1

2(10

00)

Com

bine

like

term

s.

0.

10t

= 2

0 S

impl

ify.

0.

10t

−

0.10

=

20

−

0.10

D

ivid

e ea

ch s

ide

by 0

.10.

t

= 2

00

Sim

plify

.

s

+ t

= 1

000

Firs

t eq

uatio

n

s +

200

= 1

000

t =

200

s

= 8

00

Sol

ve fo

r s.

800

mil

lili

ters

of

10%

sol

uti

on a

nd

200

mil

lili

ters

of

20%

sol

uti

on s

hou

ld b

e u

sed.

Exer

cise

s 1

. SPO

RTS

At

the

end

of t

he

2007

–200

8 fo

otba

ll s

easo

n, 3

8 S

upe

r B

owl

gam

es h

ad b

een

pl

ayed

wit

h t

he

curr

ent

two

foot

ball

lea

gues

, th

e A

mer

ican

Foo

tbal

l C

onfe

ren

ce (

AF

C)

and

the

Nat

ion

al F

ootb

all

Con

fere

nce

(N

FC

). T

he

NF

C w

on t

wo

mor

e ga

mes

th

an t

he

AF

C. H

ow m

any

gam

es d

id e

ach

con

fere

nce

win

? A

FC

18;

NF

C 2

0

2. C

HEM

ISTR

Y A

lab

nee

ds t

o m

ake

100

gall

ons

of a

n 1

8% a

cid

solu

tion

by

mix

ing

a 12

%

acid

sol

uti

on w

ith

a 2

0% s

olu

tion

. How

man

y ga

llon

s of

eac

h s

olu

tion

are

nee

ded?

3. G

EOM

ETRY

Th

e pe

rim

eter

of

a tr

ian

gle

is 2

4 in

ches

. Th

e lo

nge

st s

ide

is 4

in

ches

lon

ger

than

th

e sh

orte

st s

ide,

an

d th

e sh

orte

st s

ide

is t

hre

e-fo

urt

hs

the

len

gth

of

the

mid

dle

side

. Fin

d th

e le

ngt

h o

f ea

ch s

ide

of t

he

tria

ngl

e. 6

in.,

8 in

., 10

in.

6-2

Exam

ple

25 g

al o

f 12

% s

olu

tio

n a

nd

75

gal

of

20%

so

luti

on

010_

018_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

1312

/21/

10

8:43

AM



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 6

14

Gle

ncoe

Alg

ebra

1

Use

su

bst

itu

tion

to

solv

e ea

ch s

yste

m o

f eq

uat

ion

s.

1. y

= 4

x 2.

y =

2x

x +

y =

5 (

1, 4

) x

+ 3

y =

-14

(-

2, -

4)

3. y

= 3

x 4.

x =

-4y

2x +

y =

15

(3, 9

) 3

x +

2y

= 2

0 (8

, -2)

5. y

= x

- 1

6.

x =

y -

7x

+ y

= 3

(2,

1)

x +

8y

= 2

(-

6, 1

)

7. y

= 4

x -

1

8. y

= 3

x +

8y

= 2

x -

5 (

-2,

-9)

5

x +

2y

= 5

(-

1, 5

)

9. 2

x -

3y

= 2

1 10

. y =

5x

- 8

y =

3 -

x (

6, -

3)

4x

+ 3

y =

33

(3, 7

)

11. x

+ 2

y =

13

12. x

+ 5

y =

43x

- 5

y =

6 (

7, 3

) 3

x +

15y

= -

1 n

o s

olu

tio

n

13. 3

x -

y =

4

14. x

+ 4

y =

82x

- 3

y =

-9

(3, 5

) 2

x -

5y

= 2

9 (1

2, -

1)

15. x

- 5

y =

10

16. 5

x -

2y

= 1

42x

- 1

0y =

20

infi

nit

ely

man

y 2

x -

y =

5 (

4, 3

)

17. 2

x +

5y

= 3

8 18

. x -

4y

= 2

7x

- 3

y =

-3

(9, 4

) 3

x +

y =

-23

(-

5, -

8)

19. 2

x +

2y

= 7

20

. 2.5

x +

y =

-2

x -

2y

= -

1 (2

, 3 −

2 )

3x

+ 2

y =

0 (

-2,

3)

6-2

Skill

s Pr

acti

ceS

ub

sti

tuti

on

010_

018_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

1412

/21/

10

8:43

AM


NA

ME

DAT

E

P

ER

IOD

Lesson 6-2

Cha

pte

r 6

15

Gle

ncoe

Alg

ebra

1

Use

su

bst

itu

tion

to

solv

e ea

ch s

yste

m o

f eq

uat

ion

s.

1. y

= 6

x 2.

x =

3y

3. x

= 2

y +

72x

+ 3

y =

-20

(-

1, -

6)

3x

- 5

y =

12

(9, 3

) x

= y

+ 4

(1,

-3)

4. y

= 2

x -

2

5. y

= 2

x +

6

6. 3

x +

y =

12

y =

x +

2 (

4, 6

) 2

x -

y =

2 n

o s

olu

tio

n

y =

-x

- 2

(7,

-9)

7. x

+ 2

y =

13

(-3,

8)

8. x

- 2

y =

3 i

nfi

nit

ely

9. x

- 5

y =

36

(-4,

-8)

- 2

x -

3y

= -

18

4x

- 8

y =

12

man

y 2

x +

y =

-16

10. 2

x -

3y

= -

24

11. x

+ 1

4y =

84

12. 0

.3x

- 0

.2y

= 0

.5x

+ 6

y =

18

(-6,

4)

2x

- 7

y =

-7

(14,

5)

x -

2y

= -

5 (5

, 5)

13. 0

.5x

+ 4

y =

-1

14. 3

x -

2y

= 1

1 15

. 1 −

2 x

+ 2

y =

12

x

+ 2

.5y

= 3

.5 (

6, -

1)

x -

1 −

2 y

= 4

(5,

2)

x -

2y

= 6

(12

, 3)

16. 1

−

3 x -

y =

3

17. 4

x -

5y

= -

7 18

. x +

3y

= -

4

2

x +

y =

25

y =

5x

2x

+ 6

y =

5

(

12, 1

) (

1 −

3 ,

1 2 −

3 )

no

so

luti

on

19. E

MPL

OY

MEN

T K

enis

ha

sell

s at

hle

tic

shoe

s pa

rt-t

ime

at a

dep

artm

ent

stor

e. S

he

can

ea

rn e

ith

er $

500

per

mon

th p

lus

a 4%

com

mis

sion

on

her

tot

al s

ales

, or

$400

per

mon

th

plu

s a

5% c

omm

issi

on o

n t

otal

sal

es.

a. W

rite

a s

yste

m o

f eq

uat

ion

s to

rep

rese

nt

the

situ

atio

n.

y =

0.0

4x +

50

0 an

d y

= 0

.05x

+ 4

00

b.

Wh

at i

s th

e to

tal

pric

e of

th

e at

hle

tic

shoe

s K

enis

ha

nee

ds t

o se

ll t

o ea

rn t

he

sam

e in

com

e fr

om e

ach

pay

sca

le?

$10,

00

0

c. W

hic

h i

s th

e be

tter

off

er?

the

fi rs

t o

ffer

if s

he

exp

ects

to

sel

l les

s th

an

$10,

00

0 in

sh

oes

, an

d t

he

seco

nd

off

er if

sh

e ex

pec

ts t

o s

ell m

ore

th

an

$10,

00

0 in

sh

oes

20. M

OV

IE T

ICK

ETS

Tic

kets

to

a m

ovie

cos

t $7

.25

for

adu

lts

and

$5.5

0 fo

r st

ude

nts

. A

grou

p of

fri

ends

pu

rch

ased

8 t

icke

ts f

or $

52.7

5.

a.

Wri

te a

sys

tem

of

equ

atio

ns

to r

epre

sen

t th

e si

tuat

ion

.

b.

How

man

y ad

ult

tic

kets

an

d st

ude

nt

tick

ets

wer

e pu

rch

ased

? 5

adu

lt a

nd

3 s

tud

ent

6-2

Prac

tice

Su

bsti

tuti

on

x +

y =

8 a

nd

7.2

5x +

5.5

y =

52.

75

010_

018_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

1512

/21/

10

8:43

AM



An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 6

16

Gle

ncoe

Alg

ebra

1

1. B

USI

NES

S M

r. R

ando

lph

fin

ds t

hat

th

e su

pply

an

d de

man

d fo

r ga

soli

ne

at h

is

stat

ion

are

gen

eral

ly g

iven

by

the

foll

owin

g eq

uat

ion

s.x

-y

=-

2x

+y

= 1

0

Use

su

bsti

tuti

on t

o fi

nd

the

equ

ilib

riu

m

poin

t w

her

e th

e su

pply

an

d de

man

d li

nes

in

ters

ect.

2. G

EOM

ETRY

Th

e m

easu

res

of

com

plem

enta

ry a

ngl

es h

ave

a su

m o

f 90

deg

rees

. An

gle

A a

nd

angl

e B

are

co

mpl

emen

tary

, an

d th

eir

mea

sure

s h

ave

a di

ffer

ence

of

20°.

Wh

at a

re t

he

mea

sure

s of

th

e an

gles

?

AB

3. M

ON

EY H

arve

y h

as s

ome

$1 b

ills

an

d so

me

$5 b

ills

. In

all

, he

has

6 b

ills

wor

th

$22.

Let

x b

e th

e n

um

ber

of $

1 bi

lls

and

let

y be

th

e n

um

ber

of $

5 bi

lls.

Wri

te a

sy

stem

of

equ

atio

ns

to r

epre

sen

t th

e in

form

atio

n a

nd

use

su

bsti

tuti

on t

o de

term

ine

how

man

y bi

lls

of e

ach

de

nom

inat

ion

Har

vey

has

.

x +

y =

6, x

+ 5

y =

22;

he

has

fo

ur

$5 b

ills

and

tw

o $

1 b

ills.

4. P

OPU

LATI

ON

San

jay

is r

esea

rch

ing

popu

lati

on t

ren

ds i

n S

outh

Am

eric

a. H

e fo

un

d th

at t

he

popu

lati

on o

f E

cuad

or t

o in

crea

sed

by 1

,000

,000

an

d th

e po

pula

tion

of

Ch

ile

to i

ncr

ease

d by

60

0,00

0 fr

om 2

004

to 2

009.

Th

e ta

ble

disp

lays

th

e in

form

atio

n h

e fo

un

d.

Sour

ce:W

orld

Alm

anac

If t

he

popu

lati

on g

row

th f

or e

ach

cou

ntr

y co

nti

nu

es a

t th

e sa

me

rate

, in

wh

at y

ear

are

the

popu

lati

ons

of E

cuad

or a

nd

Ch

ile

pred

icte

d to

be

equ

al?

5. C

HEM

ISTR

Y S

hel

by a

nd

Cal

vin

are

do

ing

a ch

emis

try

expe

rim

ent.

Th

ey n

eed

5 ou

nce

s of

a s

olu

tion

th

at i

s 65

% a

cid

and

35%

dis

till

ed w

ater

. Th

ere

is n

o u

ndi

lute

d ac

id i

n t

he

chem

istr

y la

b, b

ut

they

do

hav

e tw

o fl

asks

of

dilu

ted

acid

: F

lask

A c

onta

ins

70%

aci

d an

d 30

%

dist

ille

d w

ater

. Fla

sk B

con

tain

s 20

%

acid

an

d 80

% d

isti

lled

wat

er.

a. W

rite

a s

yste

m o

f eq

uat

ion

s th

at

Sh

elby

an

d C

alvi

n c

ould

use

to

dete

rmin

e h

ow m

any

oun

ces

they

n

eed

to p

our

from

eac

h f

lask

to

mak

e th

eir

solu

tion

.S

amp

le a

nsw

er:

a +

b =

5;

0.7a

+ 0

.2b

= 0

.65(

5)

b.

Sol

ve y

our

syst

em o

f eq

uat

ion

s. H

ow

man

y ou

nce

s fr

om e

ach

fla

sk d

o S

hel

by a

nd

Cal

vin

nee

d? 4

.5 o

z fr

om

Fla

sk A

an

d 0

.5 o

z fr

om

F

lask

B.

Wor

d Pr

oble

m P

ract

ice

Su

bsti

tuti

on

6-2

Co

un

try

2004

P

op

ula

tio

n

5-Ye

ar

Po

pu

lati

on

C

han

ge

Ecu

ador

13,0

00,0

00+

1,00

0,00

0

Chi

le16

,000

,000

+60

0,00

0

(4, 6

)

35°

and

55°

2041

010_

018_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

1612

/21/

10

8:43

AM


NA

ME

DAT

E

P

ER

IOD

Lesson 6-2

Cha

pte

r 6

17

Gle

ncoe

Alg

ebra

1

Enri

chm

ent

Inte

rsecti

on

of Tw

o P

ara

bo

las

Su

bsti

tuti

on c

an b

e u

sed

to f

ind

the

inte

rsec

tion

of

two

para

bola

s. R

epla

ce t

he

y-va

lue

in o

ne

of t

he

equ

atio

ns

wit

h t

he

y-va

lue

in t

erm

s of

x f

rom

th

e ot

her

equ

atio

n.

F

ind

th

e in

ters

ecti

on o

f th

e tw

o p

arab

olas

.

y =

x2 +

5x

+ 6

y

= x

2 + 4

x +

3

Gra

ph

th

e eq

uat

ion

s.

37.5

y

x-

1010

-5

5

12.5

25.0

50.0

Fro

m t

he

grap

h, n

otic

e th

at t

he

two

grap

hs

inte

rsec

t in

on

e po

int.

Use

su

bsti

tuti

on t

o so

lve

for

the

poin

t of

in

ters

ecti

on.

x2 +

5x

+ 6

= x

2 + 4

x +

3

5x +

6 =

4x

+ 3

S

ubtr

act

x2 fr

om e

ach

side

.

x

+ 6

= 3

S

ubtr

act

4x f

rom

eac

h si

de.

x

= -

3 S

ubtr

act

6 fr

om e

ach

side

.

Th

e gr

aph

s in

ters

ect

at x

= -

3.

Rep

lace

x w

ith

-3

in e

ith

er e

quat

ion

to

fin

d th

e y-

valu

e.

y =

x2 +

5x

+ 6

O

rigin

al e

quat

ion

y

= (-

3)2 +

5(-

3) +

6

Rep

lace

x w

ith -

3.

y

= 9

– 1

5 +

6 o

r 0

Sim

plify

.

Th

e po

int

of i

nte

rsec

tion

is

(-3,

0).

Exer

cise

sU

se s

ub

stit

uti

on t

o fi

nd

th

e p

oin

t of

in

ters

ecti

on o

f th

e gr

aph

s of

eac

h p

air

of e

qu

atio

ns.

1. y

= x

2 +

8x

+ 7

2.

y =

x2 +

6x

+ 8

3.

y =

x2 +

5x

+ 6

y =

x2 +

2x

+ 1

y

= x

2 + 4

x +

4

y =

x2 +

7x

+ 6

(-1,

0)

(-

2, 0

) (

0, 6

)

6-2

Exam

ple

010_

018_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

1712

/21/

10

8:43

AM



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 6

18

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

Elim

inati

on

Usin

g A

dd

itio

n a

nd

Su

btr

acti

on

Elim

inat

ion

Usi

ng

Ad

dit

ion

In

sys

tem

s of

equ

atio

ns

in w

hic

h t

he

coef

fici

ents

of

the

x or

y t

erm

s ar

e ad

diti

ve i

nve

rses

, sol

ve t

he

syst

em b

y ad

din

g th

e eq

uat

ion

s. B

ecau

se o

ne

of

the

vari

able

s is

eli

min

ated

, th

is m

eth

od i

s ca

lled

eli

min

atio

n.

U

se e

lim

inat

ion

to

solv

e th

e sy

stem

of

equ

atio

ns.

x -

3y

= 7

3x +

3y

= 9

Wri

te t

he

equ

atio

ns

in c

olu

mn

for

m a

nd

add

to e

lim

inat

e y.

x

- 3

y =

7

(+)

3x +

3y

=

9

4x

= 1

6S

olve

for

x.

4x

−

4 =

16

−

4

x

= 4

Su

bsti

tute

4 f

or x

in

eit

her

equ

atio

n a

nd

solv

e fo

r y.

4

- 3

y =

7 4

- 3

y -

4 =

7 -

4

-3y

= 3

-

3y

−

-3

=

3 −

-

3

y =

-1

Th

e so

luti

on i

s (4

, -1)

.

T

he

sum

of

two

nu

mb

ers

is 7

0 an

d t

hei

r d

iffe

ren

ce i

s 24

. Fin

d

the

nu

mb

ers.

Let

x r

epre

sen

t on

e n

um

ber

and

y re

pres

ent

the

oth

er n

um

ber.

x

+ y

= 7

0 (+

) x

- y

= 2

4

2x

= 9

4

2x

−

2 =

94

−

2

x

= 4

7S

ubs

titu

te 4

7 fo

r x

in e

ith

er e

quat

ion

.

47 +

y =

70

47 +

y -

47

= 7

0 -

47

y

= 2

3T

he

nu

mbe

rs a

re 4

7 an

d 23

.

Exer

cise

sU

se e

lim

inat

ion

to

solv

e ea

ch s

yste

m o

f eq

uat

ion

s.

1. x

+ y

= -

4 2.

2x

- 3

y =

14

3. 3

x -

y =

-9

x -

y =

2 (

-1,

-3)

x

+ 3

y =

-11

(1,

-4)

-

3x -

2y

= 0

(-

2, 3

)

4. -

3x -

4y

= -

1 5.

3x

+ y

= 4

6.

-2x

+ 2

y =

93x

- y

= -

4 (-

1, 1

) 2

x -

y =

6 (

2, -

2)

2x

- y

= -

6 (-

3 −

2 ,

3 )

7. 2

x +

2y

= -

2 8.

4x

- 2

y =

-1

9. x

- y

= 2

3x -

2y

= 1

2 (2

, -3)

-

4x +

4y

= -

2 (-

1, -

3 −

2 )

x

+ y

= -

3 (-

1 −

2 ,

- 5 −

2 )

10. 2

x -

3y

= 1

2 11

. -0.

2x +

y =

0.5

12

. 0.1

x +

0.3

y =

0.9

4

x +

3y

= 2

4 (6

, 0)

0.2

x +

2y

= 1

.6 (

1, 0

.7)

0.1

x -

0.3

y =

0.2

13. R

ema

is o

lder

th

an K

en. T

he

diff

eren

ce o

f th

eir

ages

is

12 a

nd

the

sum

of

thei

r ag

es i

s 50

. Fin

d th

e ag

e of

eac

h.

Rem

a is

31

and

Ken

is 1

9.

14. T

he

sum

of

the

digi

ts o

f a

two-

digi

t n

um

ber

is 1

2. T

he

diff

eren

ce o

f th

e di

gits

is

2. F

ind

the

nu

mbe

r if

th

e u

nit

s di

git

is l

arge

r th

an t

he

ten

s di

git.

57

6-3

Exam

ple

1Ex

amp

le 2

( 11

−

2 , 7 −

6 )

010_

018_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

1812

/21/

10

8:43

AM


NA

ME

DAT

E

P

ER

IOD

Lesson 6-3

Cha

pte

r 6

19

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Elim

inati

on

Usin

g A

dd

itio

n a

nd

Su

btr

acti

on

Elim

inat

ion

Usi

ng

Su

btr

acti

on

In

sys

tem

s of

equ

atio

ns

wh

ere

the

coef

fici

ents

of

the

x or

y t

erm

s ar

e th

e sa

me,

sol

ve t

he

syst

em b

y su

btra

ctin

g th

e eq

uat

ion

s.

U

se e

lim

inat

ion

to

solv

e th

e sy

stem

of

equ

atio

ns.

2x -

3y

= 1

15x

- 3

y =

14

2x

- 3

y =

11

Writ

e th

e eq

uatio

ns in

col

umn

form

and

sub

trac

t.

(-)

5x -

3y

= 1

4

-3x

=

-3

S

ubtr

act

the

two

equa

tions

. y is

elim

inat

ed.

-

3x

−

-3

= -

3 −

-

3 D

ivid

e ea

ch s

ide

by -

3.

x

= 1

S

impl

ify.

2(

1) -

3y

= 1

1 S

ubst

itute

1 fo

r x

in e

ither

equ

atio

n.

2

- 3

y =

11

Sim

plify

.

2 -

3y

- 2

= 1

1 -

2

Sub

trac

t 2

from

eac

h si

de.

-

3y =

9

Sim

plify

.

-

3y

−

-3

=

9 −

-

3 D

ivid

e ea

ch s

ide

by -

3.

y

= -

3 S

impl

ify.

Th

e so

luti

on i

s (1

, -3)

.

Exer

cise

sU

se e

lim

inat

ion

to

solv

e ea

ch s

yste

m o

f eq

uat

ion

s.

1. 6

x +

5y

= 4

2.

3m

- 4

n =

-14

3.

3a

+ b

= 1

6x -

7y

= -

20

3m

+ 2

n =

-2

a +

b =

3

(-

1, 2

) (

-2,

2)

(-

1, 4

)

4. -

3x -

4y

= -

23

5. x

- 3

y =

11

6. x

- 2

y =

6-

3x +

y =

2

2x

- 3

y =

16

x +

y =

3

(1,

5)

(5,

-2)

(

4, -

1)

7. 2

a -

3b

= -

13

8. 4

x +

2y

= 6

9.

5x

- y

= 6

2a +

2b

= 7

4

x +

4y

= 1

0 5

x +

2y

= 3

(

- 1 −

2 ,

4 )

( 1 −

2 ,

2 )

(1,

-1)

10. 6

x -

3y

= 1

2 11

. x +

2y

= 3

.5

12. 0

.2x

+ y

= 0

.74x

- 3

y =

24

x -

3y

= -

9 0

.2x

+ 2

y =

1.2

(

-6,

-16

) (

-1.

5, 2

.5)

(1,

0.5

)

13. T

he

sum

of

two

nu

mbe

rs i

s 70

. On

e n

um

ber

is t

en m

ore

than

tw

ice

the

oth

er n

um

ber.

Fin

d th

e n

um

bers

. 50

; 20

14. G

EOM

ETRY

Tw

o an

gles

are

su

pple

men

tary

. Th

e m

easu

re o

f on

e an

gle

is 1

0° m

ore

than

th

ree

tim

es t

he

oth

er. F

ind

the

mea

sure

of

each

an

gle.

42.

5°;

137.

5°

6-3

Exam

ple

019_

027_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

1912

/21/

10

8:43

AM



An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 6

20

Gle

ncoe

Alg

ebra

1

Skill

s Pr

acti

ceE

lim

inati

on

Usin

g A

dd

itio

n a

nd

Su

btr

acti

on

Use

eli

min

atio

n t

o so

lve

each

sys

tem

of

equ

atio

ns.

1. x

- y

= 1

2.

-x

+ y

= 1

x +

y =

3 (

2, 1

) x

+ y

= 1

1 (5

, 6)

3. x

+ 4

y =

11

4. -

x +

3y

= 6

x -

6y

= 1

1 (1

1, 0

) x

+ 3

y =

18

(6, 4

)

5. 3

x +

4y

= 1

9 6.

x +

4y

= -

83x

+ 6

y =

33

(-3,

7)

x -

4y

= -

8 (-

8, 0

)

7. 3

x +

4y

= 2

8.

3x

- y

= -

14x

- 4

y =

12

(2, -

1)

-3x

- y

= 5

(-

1, -

2)

9. 2

x -

3y

= 9

10

. x -

y =

4-

5x -

3y

= 3

0 (-

3, -

5)

2x

+ y

= -

4 (0

, -4)

11. 3

x -

y =

26

12. 5

x -

y =

-6

-2x

- y

= -

24 (

10, 4

) -

x +

y =

2 (

-1,

1)

13. 6

x -

2y

= 3

2 14

. 3x

+ 2

y =

-19

4x -

2y

= 1

8 (7

, 5)

-3x

- 5

y =

25

(-5,

-2)

15. 7

x +

4y

= 2

16

. 2x

- 5

y =

-28

7x +

2y

= 8

(2,

-3)

4

x +

5y

= 4

(-

4, 4

)

17. T

he

sum

of

two

nu

mbe

rs i

s 28

an

d th

eir

diff

eren

ce i

s 4.

Wh

at a

re t

he

nu

mbe

rs?

12, 1

6

18. F

ind

the

two

nu

mbe

rs w

hos

e su

m i

s 29

an

d w

hos

e di

ffer

ence

is

15.

7, 2

2

19. T

he

sum

of

two

nu

mbe

rs i

s 24

an

d th

eir

diff

eren

ce i

s 2.

Wh

at a

re t

he

nu

mbe

rs?

13, 1

1

20. F

ind

the

two

nu

mbe

rs w

hos

e su

m i

s 54

an

d w

hos

e di

ffer

ence

is

4. 2

5, 2

9

21. T

wo

tim

es a

nu

mbe

r ad

ded

to a

not

her

nu

mbe

r is

25.

Th

ree

tim

es t

he

firs

t n

um

ber

min

us

the

oth

er n

um

ber

is 2

0. F

ind

the

nu

mbe

rs.

9, 7

6-3

019_

027_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

2012

/21/

10

8:43

AM


NA

ME

DAT

E

P

ER

IOD

Lesson 6-3

Cha

pte

r 6

21

Gle

ncoe

Alg

ebra

1

Prac

tice

E

lim

inati

on

Usin

g A

dd

itio

n a

nd

Su

btr

acti

on

Use

eli

min

atio

n t

o so

lve

each

sys

tem

of

equ

atio

ns.

1. x

- y

= 1

2.

p +

q =

-2

3. 4

x +

y =

23

x +

y =

-9

p -

q =

8

3x

- y

= 1

2

(

-4,

-5)

(

3, -

5)

(5,

3)

4. 2

x +

5y

= -

3 5.

3x

+ 2

y =

-1

6. 5

x +

3y

= 2

22x

+ 2

y =

6

4x

+ 2

y =

-6

5x

- 2

y =

2

(

6, -

3)

(-

5, 7

) (

2, 4

)

7. 5

x +

2y

= 7

8.

3x

- 9

y =

-12

9.

-4c

- 2

d =

-2

-2x

+ 2

y =

-14

3

x -

15y

= -

6 2

c -

2d

= -

14

(

3, -

4)

(-

7, -

1)

(-

2, 5

)

10. 2

x -

6y

= 6

11

. 7x

+ 2

y =

2

12. 4

.25x

- 1

.28y

= -

9.2

2x +

3y

= 2

4 7

x -

2y

= -

30

x +

1.2

8y =

17.

6

(

9, 2

) (

-2,

8)

(1.

6, 1

2.5)

13. 2

x +

4y

= 1

0 14

. 2.5

x +

y =

10.

7 15

. 6m

- 8

n =

3x

- 4

y =

-2.

5 2

.5x

+ 2

y =

12.

9 2

m -

8n

= -

3

(

2.5,

1.2

5)

(3.

4, 2

.2)

(1

1 −

2 ,

3 −

4 )

16. 4

a +

b =

2

17. -

1 −

3 x

- 4 −

3 y

= -

2 18

. 3 −

4 x

- 1 −

2 y

= 8

4

a +

3b

= 1

0

1 −

3 x

- 2 −

3 y

= 4

3 −

2 x

+ 1 −

2 y

= 1

9

(-

1 −

2 ,

4 )

(

10, -

1)

(12

, 2)

19. T

he

sum

of

two

nu

mbe

rs i

s 41

an

d th

eir

diff

eren

ce i

s 5.

Wh

at a

re t

he

nu

mbe

rs?

18, 2

3

20. F

our

tim

es o

ne

nu

mbe

r ad

ded

to a

not

her

nu

mbe

r is

36.

Th

ree

tim

es t

he

firs

t n

um

ber

min

us

the

oth

er n

um

ber

is 2

0. F

ind

the

nu

mbe

rs.

8, 4

21. O

ne

nu

mbe

r ad

ded

to t

hre

e ti

mes

an

oth

er n

um

ber

is 2

4. F

ive

tim

es t

he

firs

t n

um

ber

adde

d to

th

ree

tim

es t

he

oth

er n

um

ber

is 3

6. F

ind

the

nu

mbe

rs.

3, 7

22. L

AN

GU

AG

ES E

ngl

ish

is

spok

en a

s th

e fi

rst

or p

rim

ary

lan

guag

e in

78

mor

e co

un

trie

s th

an F

arsi

is

spok

en a

s th

e fi

rst

lan

guag

e. T

oget

her

, En

glis

h a

nd

Fars

i ar

e sp

oken

as

a fi

rst

lan

guag

e in

130

cou

ntr

ies.

In

how

man

y co

un

trie

s is

En

glis

h s

poke

n a

s th

e fi

rst

lan

guag

e? I

n h

ow m

any

cou

ntr

ies

is F

arsi

spo

ken

as

the

firs

t la

ngu

age?

E

ng

lish

: 10

4 co

un

trie

s, F

arsi

: 26

co

un

trie

s

23. D

ISC

OU

NTS

At

a sa

le o

n w

inte

r cl

oth

ing,

Cod

y bo

ugh

t tw

o pa

irs

of g

love

s an

d fo

ur

hat

s fo

r $4

3.00

. Tor

i bo

ugh

t tw

o pa

irs

of g

love

s an

d tw

o h

ats

for

$30.

00. W

hat

wer

e th

e pr

ices

fo

r th

e gl

oves

an

d h

ats?

6-3

glo

ves:

$8.

50, h

ats:

$6.

50.

019_

027_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

2112

/21/

10

8:43

AM



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 6

22

Gle

ncoe

Alg

ebra

1

Wor

d Pr

oble

m P

ract

ice

Elim

inati

on

Usin

g A

dd

itio

n a

nd

Su

btr

acti

on

1. N

UM

BER

FU

N M

s. S

imm

s, t

he

sixt

h

grad

e m

ath

tea

cher

, gav

e h

er s

tude

nts

th

is c

hal

len

ge p

robl

em.

Tw

ice

a n

um

ber

adde

d to

an

oth

er n

um

ber

is 1

5. T

he

sum

of

the

two

nu

mbe

rs i

s 11

.L

oren

zo, a

n a

lgeb

ra s

tude

nt

wh

o w

as

Ms.

Sim

ms

aide

, rea

lize

d h

e co

uld

sol

ve

the

prob

lem

by

wri

tin

g th

e fo

llow

ing

syst

em o

f eq

uat

ion

s. 2

x+

y=

15

x+

y=

11

Use

th

e el

imin

atio

n m

eth

od t

o so

lve

the

syst

em a

nd

fin

d th

e tw

o n

um

bers

.

2. G

OV

ERN

MEN

T T

he

Tex

as S

tate

L

egis

latu

re i

s co

mpr

ised

of

stat

e se

nat

ors

and

stat

e re

pres

enta

tive

s. T

he

sum

of

the

nu

mbe

r of

sen

ator

s an

d re

pres

enta

tive

s is

181

. Th

ere

are

119

mor

e re

pres

enta

tive

s th

an s

enat

ors.

How

m

any

sen

ator

s an

d h

ow m

any

repr

esen

tati

ves

mak

e u

p th

e T

exas

Sta

te

Leg

isla

ture

?

3. R

ESEA

RC

H M

elis

sa w

onde

red

how

m

uch

it

wou

ld c

ost

to s

end

a le

tter

by

mai

l in

199

0, s

o sh

e as

ked

her

fat

her

. R

ath

er t

han

an

swer

dir

ectl

y, M

elis

sa’s

fa

ther

gav

e h

er t

he

foll

owin

g in

form

atio

n. I

t w

ould

hav

e co

st $

3.70

to

sen

d 13

pos

tcar

ds a

nd

7 le

tter

s, a

nd

it

wou

ld h

ave

cost

$2.

65 t

o se

nd

6 po

stca

rds

and

7 le

tter

s. U

se a

sys

tem

of

equ

atio

ns

and

elim

inat

ion

to

fin

d h

ow

mu

ch i

t co

st t

o se

nd

a le

tter

in

199

0.

4. S

POR

TS A

s of

201

0, t

he

New

Yor

k Ya

nke

es h

ad w

on m

ore

Wor

ld S

erie

s C

ham

pion

ship

s th

an a

ny

oth

er t

eam

. In

fa

ct, t

he

Yan

kees

had

won

3 f

ewer

th

an

3 ti

mes

th

e n

um

ber

of W

orld

Ser

ies

cham

pion

ship

s w

on b

y th

e se

con

d m

ost-

win

nin

g te

am, t

he

St.

Lou

is C

ardi

nal

s.

Th

e su

m o

f th

e tw

o te

ams’

Wor

ld S

erie

s ch

ampi

onsh

ips

is 3

7. H

ow m

any

tim

es

has

eac

h t

eam

won

th

e W

orld

Ser

ies?

T

he

Car

din

als

hav

e 10

win

s an

d

the

Yan

kees

hav

e 27

win

s.

5. B

ASK

ETB

ALL

In

200

5, t

he

aver

age

tick

et p

rice

s fo

r D

alla

s M

aver

icks

gam

es

and

Bos

ton

Cel

tics

gam

es a

re s

how

n i

n

the

tabl

e be

low

. Th

e ch

ange

in

pri

ce i

s fr

om t

he

2004

sea

son

to

the

2005

sea

son

.

Sour

ce: T

eam

Mar

ketin

g Re

port

a. A

ssu

me

that

tic

kets

con

tin

ue

to

chan

ge a

t th

e sa

me

rate

eac

h y

ear

afte

r 20

05. L

et x

be

the

nu

mbe

r of

ye

ars

afte

r 20

05, a

nd

y be

th

e pr

ice

of

an a

vera

ge t

icke

t. W

rite

a s

yste

m o

f eq

uat

ion

s to

rep

rese

nt

the

info

rmat

ion

in

th

e ta

ble.

y =

0.5

3x +

53.

60y =

-1.

08x +

55.

93

b.

In h

ow m

any

year

s w

ill

the

aver

age

tick

et p

rice

for

Dal

las

appr

oxim

atel

y eq

ual

th

at o

f B

osto

n?

x =

1.4

5; 1

or

2 yr

6-3

Team

Ave

rag

e T

icke

t P

rice

Ch

ang

e in

P

rice

Dal

las

$53.

60$0

.53

Bos

ton

$55.

93-

$1.0

8

(4, 7

)

31

sta

te s

enat

ors

an

d

150

stat

e re

pre

sen

tati

ves

$0.2

5

019_

027_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

2212

/21/

10

8:43

AM


NA

ME

DAT

E

P

ER

IOD

Lesson 6-3

Cha

pte

r 6

23

Gle

ncoe

Alg

ebra

1

Sys

tem

s of

equ

atio

ns

can

in

volv

e m

ore

than

2 e

quat

ion

s an

d 2

vari

able

s. I

t is

pos

sibl

e to

so

lve

a sy

stem

of

3 eq

uat

ion

s an

d 3

vari

able

s u

sin

g el

imin

atio

n.

S

olve

th

e fo

llow

ing

syst

em.

x

+ y

+ z

= 6

3x

- y

+ z

= 8

x

- z

= 2

Ste

p 1

: U

se e

lim

inat

ion

to

get

rid

of t

he

y in

th

e fi

rst

two

equ

atio

ns.

x

+ y

+ z

= 6

3x

– y

+ z

= 8

4x

+ 2

z =

14

Ste

p 2

: U

se t

he

equ

atio

n y

ou f

oun

d in

ste

p 1

and

the

thir

d eq

uat

ion

to

elim

inat

e th

e z.

4x

+ 2

z =

14

x

– z

= 2

M

ult

iply

th

e se

con

d eq

uat

ion

by

2 so

th

at t

he

z’s

wil

l el

imin

ate.

4x

+ 2

z =

14

2x

– 2

z =

4

6x =

18

S

o, x

= 3

.

Ste

p 3

: R

epla

ce x

wit

h 3

in

th

e th

ird

orig

inal

equ

atio

n t

o de

term

ine

z.

3 –

z =

2, s

o z

= 1

.

Ste

p 4

: R

epla

ce x

wit

h 3

an

d z

wit

h 1

in

eit

her

of

the

firs

t tw

o or

igin

al

equ

atio

n t

o de

term

ine

the

valu

e of

y.

3

+ y

+ 1

= 6

or

4 +

y =

6.

S

o, y

= 2

.S

o, t

he

solu

tion

to

the

syst

em o

f eq

uat

ion

s is

(3,

2, 1

).

Exer

cise

sS

olve

eac

h s

yste

m o

f eq

uat

ion

s.

1. 3

x +

2y

+ z

= 4

2 2.

x

+ y

+ z

= -

3 3.

x

+ y

+ z

= 7

2

y +

z +

12

= 3

x

2x +

3y

+ 5

z =

-4

x

+ 2

y +

z =

10

x –

3y =

0

2y

– z

= 4

2 y +

z =

5

x =

9

x =

-5

x =

5

y =

3

y =

2

y =

3

z =

9

z =

0

z =

-1

Enri

chm

ent

So

lvin

g S

yste

ms o

f E

qu

ati

on

s i

n T

hre

e V

ari

ab

les

6-3

Exam

ple

019_

027_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

2312

/21/

10

8:43

AM



An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 6

24

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

Elim

inati

on

Usin

g M

ult

iplicati

on

Elim

inat

ion

Usi

ng

Mu

ltip

licat

ion

Som

e sy

stem

s of

equ

atio

ns

can

not

be

solv

ed

sim

ply

by a

ddin

g or

su

btra

ctin

g th

e eq

uat

ion

s. I

n s

uch

cas

es, o

ne

or b

oth

equ

atio

ns

mu

st

firs

t be

mu

ltip

lied

by

a n

um

ber

befo

re t

he

syst

em c

an b

e so

lved

by

elim

inat

ion

.

U

se e

lim

inat

ion

to

solv

e th

e sy

stem

of

equ

atio

ns.

x +

10y

= 3

4x +

5y

= 5

If y

ou m

ult

iply

th

e se

con

d eq

uat

ion

by

-2,

yo

u c

an e

lim

inat

e th

e y

term

s.

x +

10y

= 3

(+

) -

8x -

10y

= -

10

-7x

=

-7

-

7x

−

-7

= -

7 −

-

7

x

= 1

Su

bsti

tute

1 f

or x

in

eit

her

equ

atio

n.

1

+ 1

0y =

3 1

+ 1

0y -

1 =

3 -

1

10y

= 2

10

y −

10

=

2 −

10

y

= 1 −

5

Th

e so

luti

on i

s (1

, 1 −

5 ) .

U

se e

lim

inat

ion

to

solv

e th

e sy

stem

of

equ

atio

ns.

3x -

2y

= -

72x

- 5

y =

10

If y

ou m

ult

iply

th

e fi

rst

equ

atio

n b

y 2

and

the

seco

nd

equ

atio

n b

y -

3, y

ou c

an

elim

inat

e th

e x

term

s.

6x -

4y

= -

14(+

) -

6x +

15y

=

-30

11

y =

-44

11

y −

11

=

- 44

−

11

y

= -

4S

ubs

titu

te -

4 fo

r y

in e

ith

er e

quat

ion

. 3

x -

2(-

4) =

-7

3x

+ 8

= -

7

3x +

8 -

8 =

-7

-8

3x

= -

15

3x

−

3

= -

15

−

3

x

= -

5T

he

solu

tion

is

(-5,

-4)

.

Exer

cise

sU

se e

lim

inat

ion

to

solv

e ea

ch s

yste

m o

f eq

uat

ion

s.

1. 2

x +

3y

= 6

2.

2m

+ 3

n =

4

3. 3

a -

b =

2x

+ 2

y =

5 (

-3,

4)

-m

+ 2

n =

5 (

-1,

2)

a +

2b

= 3

(1,

1)

4. 4

x +

5y

= 6

5.

4x

- 3

y =

22

6. 3

x -

4y

= -

46x

- 7

y =

-20

(-

1, 2

) 2

x -

y =

10

(4, -

2)

x +

3y

= -

10 (

-4,

-2)

7. 4

x -

y =

9

8. 4

a -

3b

= -

8 9.

2x

+ 2

y =

55x

+ 2

y =

8 (

2, -

1)

2a

+ 2

b =

3

(- 1 −

2 ,

2 )

4x

- 4

y =

10

10. 6

x -

4y

= -

8 11

. 4x

+ 2

y =

-5

12. 2

x +

y =

3.5

(0.9

, 1.7

)4x

+ 2

y =

-3

-

2x -

4y

= 1

-x

+ 2

y =

2.5

13. G

AR

DEN

ING

Th

e le

ngt

h o

f S

ally

’s g

arde

n i

s 4

met

ers

grea

ter

than

3 t

imes

th

e w

idth

. T

he

peri

met

er o

f h

er g

arde

n i

s 72

met

ers.

Wh

at a

re t

he

dim

ensi

ons

of S

ally

’s

gard

en?

28 m

by

8 m

6-4

Exam

ple

1Ex

amp

le 2

(- 3 −

2 ,

1 −

2 )

(-1,

1 −

2 )

( 5 −

2 ,

0 )

019_

027_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

2412

/21/

10

8:43

AM

Lesson 6-4


NA

ME

DAT

E

P

ER

IOD

Lesson 6-4

Cha

pte

r 6

25

Gle

ncoe

Alg

ebra

1

6-4

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Elim

inati

on

Usin

g M

ult

iplicati

on

Solv

e R

eal-

Wo

rld

Pro

ble

ms

Som

etim

es i

t is

nec

essa

ry t

o u

se m

ult

ipli

cati

on b

efor

e el

imin

atio

n i

n r

eal-

wor

ld p

robl

ems.

C

AN

OEI

NG

Du

rin

g a

can

oein

g tr

ip, i

t ta

kes

Ray

mon

d 4

hou

rs t

o p

add

le 1

2 m

iles

up

stre

am. I

t ta

kes

him

3 h

ours

to

mak

e th

e re

turn

tri

p p

add

lin

g d

own

stre

am. F

ind

th

e sp

eed

of

the

can

oe i

n s

till

wat

er.

Rea

d

You

are

ask

ed t

o fi

nd

the

spee

d of

th

e ca

noe

in

sti

ll w

ater

.

Sol

ve

Let

c =

th

e ra

te o

f th

e ca

noe

in

sti

ll w

ater

.L

et w

= t

he

rate

of

the

wat

er c

urr

ent.

rt

dr

· t

= d

Ag

ain

st t

he

Cu

rren

tc

– w

412

(c –

w)4

= 1

2

Wit

h t

he

Cu

rren

tc

+ w

312

(c +

w)3

= 1

2

So,

ou

r tw

o eq

uat

ion

s ar

e 4c

- 4

w=

12

and

3c+

3w

= 1

2.U

se e

lim

inat

ion

wit

h m

ult

ipli

cati

on t

o so

lve

the

syst

em. S

ince

th

e pr

oble

m a

sks

for

c,

elim

inat

e w

.4c

- 4

w=

12

⇒ M

ult

iply

by

3 ⇒

12

c-

12w

= 3

63c

+ 3

w=

12

⇒ M

ult

iply

by

4 ⇒

(+

) 1

2c+

12w

= 4

8

24c

= 8

4 w

is e

limin

ated

.

24c

− 24=

84 − 24D

ivid

e ea

ch s

ide

by 2

4.

c

= 3

.5

Sim

plify

.T

he

rate

of

the

can

oe i

n s

till

wat

er i

s 3.

5 m

iles

per

hou

r.

Exer

cise

s 1

. FLI

GH

T A

n a

irpl

ane

trav

elin

g w

ith

th

e w

ind

flie

s 45

0 m

iles

in

2 h

ours

. On

th

e re

turn

tr

ip, t

he

plan

e ta

kes

3 h

ours

to

trav

el t

he

sam

e di

stan

ce. F

ind

the

spee

d of

th

e ai

rpla

ne

if t

he

win

d is

sti

ll.

187.

5 m

iles

per

ho

ur

2. F

UN

DR

AIS

ING

Ben

ji a

nd

Joel

are

rai

sin

g m

oney

for

th

eir

clas

s tr

ip b

y se

llin

g gi

ft

wra

ppin

g pa

per.

Ben

ji r

aise

s $3

9 by

sel

lin

g 5

roll

s of

red

wra

ppin

g pa

per

and

2 ro

lls

of

foil

wra

ppin

g pa

per.

Joel

rai

ses

$57

by s

elli

ng

3 ro

lls

of r

ed w

rapp

ing

pape

r an

d 6

roll

s of

fo

il w

rapp

ing

pape

r. F

or h

ow m

uch

are

Ben

ji a

nd

Joel

sel

lin

g ea

ch r

oll

of r

ed a

nd

foil

w

rapp

ing

pape

r? r

ed:

$5;

foil:

$7

Exam

ple

019_

027_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

2512

/21/

10

8:43

AM



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 6

26

Gle

ncoe

Alg

ebra

1

Skill

s Pr

acti

ceE

lim

inati

on

Usin

g M

ult

iplicati

on

Use

eli

min

atio

n t

o so

lve

each

sys

tem

of

equ

atio

ns.

1. x

+ y

= -

9 2.

3x

+ 2

y =

-9

5x -

2y

= 3

2 (2

, -11

) x

- y

= -

13 (

-7,

6)

3. 2

x +

5y

= 3

4.

2x

+ y

= 3

-x

+ 3

y =

-7

(4, -

1)

-4x

- 4

y =

-8

(1, 1

)

5. 4

x -

2y

= -

14

6. 2

x +

y =

03x

- y

= -

8 (-

1, 5

) 5

x +

3y

= 2

(-

2, 4

)

7. 5

x +

3y

= -

10

8. 2

x +

3y

= 1

43x

+ 5

y =

-6

(-2,

0)

3x

- 4

y =

4 (

4, 2

)

9. 2

x -

3y

= 2

1 10

. 3x

+ 2

y =

-26

5x -

2y

= 2

5 (3

, -5)

4

x -

5y

= -

4 (-

6, -

4)

11. 3

x -

6y

= -

3 12

. 5x

+ 2

y =

-3

2x +

4y

= 3

0 (7

, 4)

3x

+ 3

y =

9 (

-3,

6)

13. T

wo

tim

es a

nu

mbe

r pl

us

thre

e ti

mes

an

oth

er n

um

ber

equ

als

13. T

he

sum

of

the

two

nu

mbe

rs i

s 7.

Wh

at a

re t

he

nu

mbe

rs?

8, -

1

14. F

our

tim

es a

nu

mbe

r m

inu

s tw

ice

anot

her

nu

mbe

r is

-16

. Th

e su

m o

f th

e tw

o n

um

bers

is

-1.

Fin

d th

e n

um

bers

. -

3, 2

15. F

UN

DR

AIS

ING

Tri

sha

and

Byr

on a

re w

ash

ing

and

vacu

um

ing

cars

to

rais

e m

oney

for

a

clas

s tr

ip. T

rish

a ra

ised

$38

was

hin

g 5

cars

an

d va

cuu

min

g 4

cars

. Byr

on r

aise

d $2

8 by

w

ash

ing

4 ca

rs a

nd

vacu

um

ing

2 ca

rs. F

ind

the

amou

nt

they

ch

arge

d to

was

h a

car

an

d va

cuu

m a

car

.

6-4 was

h:

$6, v

acu

um

: $2

019_

027_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

2612

/21/

10

8:43

AM


NA

ME

DAT

E

P

ER

IOD

Lesson 6-4

Cha

pte

r 6

27

Gle

ncoe

Alg

ebra

1

Prac

tice

Elim

inati

on

Usin

g M

ult

iplicati

on

Use

eli

min

atio

n t

o so

lve

each

sys

tem

of

equ

atio

ns.

1. 2

x -

y =

-1

2. 5

x -

2y

= -

10

3. 7

x +

4y

= -

43x

- 2

y =

1

3x

+ 6

y =

66

5x

+ 8

y =

28

(

-3,

-5)

(

2, 1

0)

(-

4, 6

)

4. 2

x -

4y

= -

22

5. 3

x +

2y

= -

9 6.

4x

- 2

y =

32

3x +

3y

= 3

0 5

x -

3y

= 4

-

3x -

5y

= -

11

(

3, 7

) (

-1,

-3)

(

7, -

2)

7. 3

x +

4y

= 2

7 8.

0.5

x +

0.5

y =

-2

9. 2

x -

3 −

4 y

= -

7

5

x -

3y

= 1

6 x

- 0

.25y

= 6

x

+ 1 −

2 y

= 0

(

5, 3

) (

4, -

8)

(-

2, 4

)

10. 6

x -

3y

= 2

1 11

. 3x

+ 2

y =

11

12. -

3x +

2y

= -

152x

+ 2

y =

22

2x

+ 6

y =

-2

2x

- 4

y =

26

(

6, 5

) (

5, -

2)

(1,

-6)

13. E

igh

t ti

mes

a n

um

ber

plu

s fi

ve t

imes

an

oth

er n

um

ber

is -

13. T

he

sum

of

the

two

nu

mbe

rs i

s 1.

Wh

at a

re t

he

nu

mbe

rs?

-6,

7

14. T

wo

tim

es a

nu

mbe

r pl

us

thre

e ti

mes

an

oth

er n

um

ber

equ

als

4. T

hre

e ti

mes

th

e fi

rst

nu

mbe

r pl

us

fou

r ti

mes

th

e ot

her

nu

mbe

r is

7. F

ind

the

nu

mbe

rs.

5, -

2

15. F

INA

NC

E G

un

ther

in

vest

ed $

10,0

00 i

n t

wo

mu

tual

fu

nds

. On

e of

th

e fu

nds

ros

e 6%

in

on

e ye

ar, a

nd

the

oth

er r

ose

9% i

n o

ne

year

. If

Gu

nth

er’s

in

vest

men

t ro

se a

tot

al o

f $6

84

in o

ne

year

, how

mu

ch d

id h

e in

vest

in

eac

h m

utu

al f

un

d?$7

200

in t

he

6% f

un

d a

nd

$28

00

in t

he

9% f

un

d

16. C

AN

OEI

NG

Lau

ra a

nd

Bre

nt

padd

led

a ca

noe

6 m

iles

ups

trea

m i

n f

our

hou

rs. T

he

retu

rn t

rip

took

th

ree

hou

rs. F

ind

the

rate

at

wh

ich

Lau

ra a

nd

Bre

nt

padd

led

the

can

oe

in s

till

wat

er.

1.75

mi/h

17. N

UM

BER

TH

EORY

Th

e su

m o

f th

e di

gits

of

a tw

o-di

git

nu

mbe

r is

11.

If

the

digi

ts a

re

reve

rsed

, th

e n

ew n

um

ber

is 4

5 m

ore

than

th

e or

igin

al n

um

ber.

Fin

d th

e n

um

ber.

38

6-4

019_

027_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

2712

/21/

10

8:43

AM



An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 6

28

Gle

ncoe

Alg

ebra

1

1. S

OC

CER

Su

ppos

e a

you

th s

occe

r fi

eld

has

a p

erim

eter

of

320

yard

s an

d it

s le

ngt

h m

easu

res

40 y

ards

mor

e th

an i

ts

wid

th. M

s. H

ugh

ey a

sks

her

pla

yers

to

dete

rmin

e th

e le

ngt

h a

nd

wid

th o

f th

eir

fiel

d. S

he

give

s th

em t

he

foll

owin

g sy

stem

of

equ

atio

ns

to r

epre

sen

t th

e si

tuat

ion

. Use

eli

min

atio

n t

o so

lve

the

syst

em t

o fi

nd

the

len

gth

an

d w

idth

of

the

fiel

d.

2L +

2W

= 3

20 L

– W

= 4

0

w

idth

= 6

0 yd

; le

ng

th =

10

0 yd

2. S

POR

TS T

he

Fan

Cos

t In

dex

(FC

I)

trac

ks t

he

aver

age

cost

s fo

r at

ten

din

g sp

orti

ng

even

ts, i

ncl

udi

ng

tick

ets,

dri

nks

, fo

od, p

arki

ng,

pro

gram

s, a

nd

sou

ven

irs.

A

ccor

din

g to

th

e F

CI,

a f

amil

y of

fou

r w

ould

spe

nd

a to

tal

of $

592.

30 t

o at

ten

d tw

o M

ajor

Lea

gue

Bas

ebal

l (M

LB

) ga

mes

an

d on

e N

atio

nal

Bas

ketb

all A

ssoc

iati

on

(NB

A)

gam

e. T

he

fam

ily

wou

ld s

pen

d $6

91.3

1 to

att

end

one

ML

B a

nd

two

NB

A

gam

es. W

rite

an

d so

lve

a sy

stem

of

equ

atio

ns

to f

ind

the

fam

ily’

s co

sts

for

each

kin

d of

gam

e ac

cord

ing

to t

he

FC

I.N

BA

: $2

63.4

4; M

LB

: $1

64.4

3

3. A

RT

Mr.

San

tos,

th

e cu

rato

r of

th

e ch

ildr

en’s

mu

seu

m, r

ecen

tly

mad

e tw

o pu

rch

ases

of

clay

an

d w

ood

for

a vi

siti

ng

arti

st t

o sc

ulp

t. U

se t

he

tabl

e to

fin

d th

e co

st o

f ea

ch p

rodu

ct p

er k

ilog

ram

.

Cla

y (k

g)

Wo

od

(kg

)To

tal C

ost

54

$35.

50

3.5

6$5

0.45

C

lay

was

$0.

70 p

er k

ilog

ram

an

d

wo

od

was

$8.

00

per

kilo

gra

m.

4. T

RA

VEL

An

ton

io f

lies

fro

m H

oust

on

to P

hil

adel

phia

, a d

ista

nce

of

abou

t 13

40 m

iles

. His

pla

ne

trav

els

wit

h t

he

win

d an

d ta

kes

2 h

ours

an

d 20

min

ute

s.

At

the

sam

e ti

me,

Pau

l is

on

a p

lan

e fr

om

Ph

ilad

elph

ia t

o H

oust

on. S

ince

his

pla

ne

is h

eadi

ng

agai

nst

th

e w

ind,

Pau

l’s f

ligh

t ta

kes

2 h

ours

an

d 50

min

ute

s. W

hat

was

th

e sp

eed

of t

he

win

d in

mil

es p

er h

our?

abo

ut

50.6

7 m

ph

5. B

USI

NES

S S

upp

ose

you

sta

rt a

bu

sin

ess

asse

mbl

ing

and

sell

ing

mot

oriz

ed

scoo

ters

. It

cost

s yo

u $

1500

for

too

ls a

nd

equ

ipm

ent

to g

et s

tart

ed, a

nd

the

mat

eria

ls c

ost

$200

for

eac

h s

coot

er. Y

our

scoo

ters

sel

l fo

r $3

00 e

ach

.

a. W

rite

an

d so

lve

a sy

stem

of

equ

atio

ns

repr

esen

tin

g th

e to

tal

cost

s an

d re

ven

ue

of y

our

busi

nes

s.

y =

20

0x +

150

0 (

cost

);

y =

30

0x

(rev

enu

e);

solu

tio

n:

(15,

450

0)

b.

Des

crib

e w

hat

th

e so

luti

on m

ean

s in

te

rms

of t

he

situ

atio

n. T

he

solu

tio

n

rep

rese

nts

th

e p

oin

t at

wh

ich

co

st e

qu

als

reve

nu

e. T

his

is t

he

bre

ak-e

ven

po

int

for

the

busi

nes

s o

wn

er.

c. G

ive

an e

xam

ple

of a

rea

son

able

n

um

ber

of s

coot

ers

you

cou

ld a

ssem

ble

and

sell

in

ord

er t

o m

ake

a pr

ofit

, an

d fi

nd

the

prof

it y

ou w

ould

mak

e fo

r th

at n

um

ber

of s

coot

ers.

Sam

ple

an

swer

: 20

sco

ote

rs:

It c

ost

s $5

500

to a

ssem

ble

th

ese

sco

ote

rs, w

hic

h s

ell f

or

$60

00,

le

avin

g a

$50

0 p

rofi t

.

Wor

d Pr

oble

m P

ract

ice

Elim

inati

on

Usin

g M

ult

iplicati

on

6-4

028_

036_

ALG

1_A

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M_C

06_C

R_6

6028

0.in

dd

2812

/21/

10

8:43

AM


NA

ME

DAT

E

P

ER

IOD

Lesson 6-4

Cha

pte

r 6

29

Gle

ncoe

Alg

ebra

1

Enri

chm

ent

Geo

rge W

ash

ing

ton

Carv

er

an

d P

erc

y J

ulian

In

199

0, G

eorg

e W

ash

ingt

on C

arve

r an

d P

ercy

Ju

lian

bec

ame

the

firs

t A

fric

an A

mer

ican

s el

ecte

d to

th

e N

atio

nal

In

ven

tors

Hal

l of

Fam

e. C

arve

r (1

864–

1943

) w

as a

n a

gric

ult

ura

l sc

ien

tist

kn

own

wor

ldw

ide

for

deve

lopi

ng

hu

ndr

eds

of u

ses

for

the

pean

ut

and

the

swee

t po

tato

. His

wor

k re

vita

lize

d th

e ec

onom

y of

th

e so

uth

ern

Un

ited

Sta

tes

beca

use

it

was

no

lon

ger

depe

nde

nt

sole

ly u

pon

cot

ton

. Ju

lian

(18

98–1

975)

was

a r

esea

rch

ch

emis

t w

ho

beca

me

fam

ous

for

inve

nti

ng

a m

eth

od o

f m

akin

g a

syn

thet

ic c

orti

son

e fr

om s

oybe

ans.

His

di

scov

ery

has

had

man

y m

edic

al a

ppli

cati

ons,

par

ticu

larl

y in

th

e tr

eatm

ent

of a

rth

riti

s.T

her

e ar

e do

zen

s of

oth

er A

fric

an A

mer

ican

in

ven

tors

wh

ose

acco

mpl

ish

men

ts a

re n

ot a

s w

ell

know

n. T

hei

r in

ven

tion

s ra

nge

fro

m c

omm

on h

ouse

hol

d it

ems

like

th

e ir

onin

g bo

ard

to

com

plex

dev

ices

th

at h

ave

revo

luti

oniz

ed m

anu

fact

uri

ng.

Th

e ex

erci

ses

that

fol

low

wil

l h

elp

you

ide

nti

fy ju

st a

few

of

thes

e in

ven

tors

an

d th

eir

inve

nti

ons.

Mat

ch t

he

inve

nto

rs w

ith

th

eir

inve

nti

ons

by

mat

chin

g ea

ch s

yste

m w

ith

its

so

luti

on. (

Not

all

th

e so

luti

ons

wil

l b

e u

sed

.)

1. S

ara

Boo

ne

x +

y =

2 E

A

. (1

, 4)

auto

mat

ic t

raff

ic s

ign

al

x -

y =

10

2. S

arah

Goo

de

x =

2 -

y D

B

. (4

, -2)

eg

gbea

ter

2y

+ x

= 9

3. F

rede

rick

M.

y =

2x

+ 6

G

C.

(-2,

3)

fire

ext

ingu

ish

er

Jon

es

y =

-x

- 3

4. J

. L. L

ove

2x +

3y

= 8

F

D.

(-5,

7)

fold

ing

cabi

net

bed

2x -

y =

-8

5. T

. J. M

arsh

all

y -

3x

= 9

C

E.

(6, -

4)

iron

ing

boar

d

2y +

x =

4

6. J

an M

atze

lige

r y

+ 4

= 2

x J

F.

(-2,

4)

pen

cil

shar

pen

er

6x

- 3

y =

12

7. G

arre

tt A

. 3x

- 2

y =

-5

A

G.

(-3,

0)

port

able

x-r

ay m

ach

ine

Mor

gan

3y

- 4

x =

8

8. N

orbe

rt R

illi

eux

3x -

y =

12

I H

. (2

, -3)

pl

ayer

pia

no

y -

3x

= 1

5

I.

n

o so

luti

on

evap

orat

ing

pan

for

re

fin

ing

suga

r

J.

in

fin

itel

y la

stin

g (s

hap

ing)

m

any

mac

hin

e fo

r

so

luti

ons

man

ufa

ctu

rin

g sh

oes

6-4

028_

036_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

2912

/21/

10

8:43

AM



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 6

30

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

Ap

ply

ing

Syste

ms o

f Lin

ear

Eq

uati

on

s

Det

erm

ine

The

Bes

t M

eth

od

Yo

u h

ave

lear

ned

fiv

e m

eth

ods

for

solv

ing

syst

ems

of

lin

ear

equ

atio

ns:

gra

phin

g, s

ubs

titu

tion

, eli

min

atio

n u

sin

g ad

diti

on, e

lim

inat

ion

usi

ng

subt

ract

ion

, an

d el

imin

atio

n u

sin

g m

ult

ipli

cati

on. F

or a

n e

xact

sol

uti

on, a

n a

lgeb

raic

m

eth

od i

s be

st.

A

t a

bas

ebal

l ga

me,

Hen

ry b

ough

t 3

hot

dog

s an

d a

bag

of

chip

s fo

r $1

4. S

cott

bou

ght

2 h

otd

ogs

and

a b

ag o

f ch

ips

for

$10.

Eac

h o

f th

e b

oys

pai

d t

he

sam

e p

rice

for

th

eir

hot

dog

s, a

nd

th

e sa

me

pri

ce f

or t

hei

r ch

ips.

Th

e fo

llow

ing

syst

em o

f eq

uat

ion

s ca

n b

e u

sed

to

rep

rese

nt

the

situ

atio

n. D

eter

min

e th

e b

est

met

hod

to

solv

e th

e sy

stem

of

equ

atio

ns.

Th

en s

olve

th

e sy

stem

.

3x +

y =

14

2x +

y =

10

•

Sin

ce n

eith

er t

he

coef

fici

ents

of

x n

or t

he

coef

fici

ents

of

y ar

e ad

diti

ve i

nve

rses

, you

ca

nn

ot u

se e

lim

inat

ion

usi

ng

addi

tion

.

•

Sin

ce t

he

coef

fici

ent

of y

in

bot

h e

quat

ion

s is

1, y

ou c

an u

se e

lim

inat

ion

usi

ng

subt

ract

ion

. You

cou

ld a

lso

use

th

e su

bsti

tuti

on m

eth

od o

r el

imin

atio

n u

sin

g m

ult

ipli

cati

on

Th

e fo

llow

ing

solu

tion

use

s el

imin

atio

n b

y su

btra

ctio

n t

o so

lve

this

sys

tem

.

3x +

y =

14

Wri

te t

he

equ

atio

ns

in c

olu

mn

for

m a

nd

subt

ract

. (-

) 2x

+ (-

) y

= (-

)10

x

= 4

T

he

vari

able

y i

s el

imin

ated

.

3(4)

+ y

= 1

4 S

ubs

titu

te t

he

valu

e fo

r x

back

in

to t

he

firs

t eq

uat

ion

. y

= 2

S

olve

for

y.

Th

is m

ean

s th

at h

ot d

ogs

cost

$4

each

an

d a

bag

of c

hip

s co

sts

$2.

Exer

cise

sD

eter

min

e th

e b

est

met

hod

to

solv

e ea

ch s

yste

m o

f eq

uat

ion

s. T

hen

sol

ve

the

syst

em.

1. 5

x +

3y

= 1

6 2.

3x

- 5

y =

73x

- 5

y =

-4

2x

+ 5

y =

13

elim

inat

ion

(×

); (

2, 2

) e

limin

atio

n (

+);

(4,

1)

3. y

+ 3

x =

24

4. -

11x

- 1

0y =

17

5x -

y =

8

5x

– 7y

= 5

0el

imin

atio

n (

+);

(4,

12)

e

limin

atio

n (

×);

(3,

-5)

6-5

Exam

ple

028_

036_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

3012

/21/

10

8:43

AM


NA

ME

DAT

E

P

ER

IOD

Lesson 6-5

Cha

pte

r 6

31

Gle

ncoe

Alg

ebra

1

Ap

ply

Sys

tem

s O

f Li

nea

r Eq

uat

ion

s W

hen

app

lyin

g sy

stem

s of

lin

ear

equ

atio

ns

to

prob

lem

sit

uat

ion

s, i

t is

im

port

ant

to a

nal

yze

each

sol

uti

on i

n t

he

con

text

of

the

situ

atio

n.

B

USI

NES

S A

T-s

hir

t p

rin

tin

g co

mp

any

sell

s T-

shir

ts f

or $

15 e

ach

. Th

e co

mp

any

has

a f

ixed

cos

t fo

r th

e m

ach

ine

use

d t

o p

rin

t th

e T-

shir

ts a

nd

an

ad

dit

ion

al c

ost

per

T-

shir

t. U

se t

he

tab

le t

o es

tim

ate

the

nu

mb

er o

f T-

shir

ts t

he

com

pan

y m

ust

sel

l in

ord

er f

or t

he

inco

me

to e

qu

al e

xpen

ses.

Un

der

stan

d Y

ou k

now

th

e in

itia

l in

com

e an

d th

e in

itia

l ex

pen

se a

nd

the

rate

s of

ch

ange

of

each

qu

anti

ty w

ith

eac

h T

-sh

irt

sold

.

Pla

n

W

rite

an

equ

atio

n t

o re

pres

ent

the

inco

me

and

the

expe

nse

s. T

hen

sol

ve t

o fi

nd

how

man

y T-

shir

ts n

eed

to b

e so

ld f

or b

oth

val

ues

to

be e

qual

.

Sol

ve

L

et x

= t

he

nu

mbe

r of

T-s

hir

ts s

old

and

let

y =

th

e to

tal

amou

nt.

in

com

e y

=

0 +

15

x

ex

pen

ses

y

= 3

000

+

5x

Yo

u c

an u

se s

ubs

titu

tion

to

solv

e th

is s

yste

m.

y

= 1

5x

The

fi rs

t eq

uatio

n.

15

x =

300

0 +

5x

Sub

stitu

te t

he v

alue

for

y in

to t

he s

econ

d eq

uatio

n.

10

x =

300

0 S

ubtr

act

10x

from

eac

h si

de a

nd s

impl

ify.

x

= 3

00

Div

ide

each

sid

e by

10

and

sim

plify

.

T

his

mea

ns

that

if

300

T-sh

irts

are

sol

d, t

he

inco

me

and

expe

nse

s of

th

e T-

shir

t co

mpa

ny

are

equ

al.

Ch

eck

Doe

s th

is s

olu

tion

mak

e se

nse

in

th

e co

nte

xt o

f th

e pr

oble

m?

Aft

er s

elli

ng

300

T-sh

irts

, th

e in

com

e w

ould

be

abou

t 30

0 ×

$15

or

$450

0. T

he

cost

s w

ould

be

abou

t $3

000

+ 3

00 ×

$5

or $

4500

.

Exer

cise

sR

efer

to

the

exam

ple

ab

ove.

If

the

cost

s of

th

e T-

shir

t co

mp

any

chan

ge t

o th

e gi

ven

val

ues

an

d t

he

sell

ing

pri

ce r

emai

ns

the

sam

e, d

eter

min

e th

e n

um

ber

of

T-sh

irts

th

e co

mp

any

mu

st s

ell

in o

rder

for

in

com

e to

eq

ual

exp

ense

s.

1. p

rin

tin

g m

ach

ine:

$50

00.0

0;

2. p

rin

tin

g m

ach

ine:

$21

00.0

0;

T-sh

irt:

$10

.00

each

10

00

T-s

hir

t: $

8.00

eac

h

300

3. p

rin

tin

g m

ach

ine:

$88

00.0

0;

4. p

rin

tin

g m

ach

ine:

$12

00.0

0;T-

shir

t: $

4.00

eac

h

800

T-s

hir

t: $

12.0

0 ea

ch

400

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Ap

ply

ing

Syste

ms o

f Lin

ear

Eq

uati

on

s

6-5

T-sh

irt

Pri

nti

ng

Co

st

prin

ting

mac

hine

$300

0.00

blan

k T-

shir

t$5

.00

Exam

ple

tota

l am

ount

initi

al

amou

ntra

te o

f ch

ange

tim

es

num

ber

of T

-shi

rts

sold

028_

036_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

3112

/21/

10

8:43

AM



An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 6

32

Gle

ncoe

Alg

ebra

1

Det

erm

ine

the

bes

t m

eth

od t

o so

lve

each

sys

tem

of

equ

atio

ns.

T

hen

sol

ve t

he

syst

em.

1. 5

x +

3y

= 1

6 2.

3x

– 5y

= 7

3x –

5y

= -

4 2

x +

5y

= 1

3el

imin

atio

n (

×);

(2,

2)

elim

inat

ion

(+

); (

4, 1

)

3. y

= 3

x -

24

4. -

11x

– 10

y =

17

5x -

y =

8

5x

– 7y

= 5

0su

bst

ituti

on

; (-

8, -

48)

elim

inat

ion

(×

); (

3, -

5)

5. 4

x +

y =

24

6. 6

x –

y =

-14

55x

- y

= 1

2

x =

4 –

2y

elim

inat

ion

(+

); (

4, 8

) s

ub

stitu

tio

n;

(-22

, 13)

7. V

EGET

AB

LE S

TAN

D

A r

oads

ide

vege

tabl

e st

and

sell

s pu

mpk

ins

for

$5 e

ach

an

d sq

uas

hes

for

$3

each

. On

e da

y th

ey s

old

6 m

ore

squ

ash

th

an p

um

pkin

s, a

nd

thei

r sa

les

tota

led

$98.

Wri

te a

nd

solv

e a

syst

em o

f eq

uat

ion

s to

fin

d h

ow m

any

pum

pkin

s an

d sq

uas

h t

hey

sol

d?

y =

6 +

x a

nd

5x +

3y =

98;

10

pu

mp

kin

s, 1

6 sq

uas

hes

8. I

NC

OM

E R

amir

o ea

rns

$20

per

hou

r du

rin

g th

e w

eek

and

$30

per

hou

r fo

r ov

erti

me

on t

he

wee

ken

ds. O

ne

wee

k R

amir

o ea

rned

a t

otal

of

$650

. He

wor

ked

5 ti

mes

as

man

y h

ours

du

rin

g th

e w

eek

as h

e di

d on

th

e w

eeke

nd.

Wri

te a

nd

solv

e a

syst

em o

f eq

uat

ion

s to

det

erm

ine

how

man

y h

ours

of

over

tim

e R

amir

o w

orke

d on

th

e w

eeke

nd.

20

x +

30y

= 6

50 a

nd

x =

5y;

5 h

ou

rs

9. B

ASK

ETB

ALL

A

nya

mak

es 1

4 ba

sket

s du

rin

g h

er g

ame.

Som

e of

th

ese

bask

ets

wer

e w

orth

2-p

oin

ts a

nd

oth

ers

wer

e w

orth

3-p

oin

ts. I

n t

otal

, sh

e sc

ored

30

poin

ts. W

rite

an

d so

lve

a sy

stem

of

equ

atio

ns

to f

ind

how

2-p

oin

ts b

aske

ts s

he

mad

e.

x +

y =

14

and

2x +

3y =

30;

12

6-5

Skill

s Pr

acti

ceA

pp

lyin

g S

yste

ms o

f Lin

ear

Eq

uati

on

s

028_

036_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

3212

/21/

10

8:43

AM


NA

ME

DAT

E

P

ER

IOD

Lesson 6-5

Cha

pte

r 6

33

Gle

ncoe

Alg

ebra

1

Det

erm

ine

the

bes

t m

eth

od t

o so

lve

each

sys

tem

of

equ

atio

ns.

Th

en s

olve

th

e sy

stem

.

1. 1

.5x

– 1.

9y =

-29

2.

1.2

x –

0.8y

= -

6 3.

18x

–16

y =

-31

2x

– 0.

9y =

4.5

4

.8x

+ 2

.4y

= 6

0 7

8x –

16y

= 4

08su

bst

ituti

on

;

elim

inat

ion

(×

);

elim

inat

ion

(-

);

(63,

65)

(

5, 1

5)

(12

, 33)

4. 1

4x +

7y

= 2

17

5. x

= 3

.6y

+ 0

.7

6. 5

.3x

– 4y

= 4

3.5

14x

+ 3

y =

189

2

x +

0.2

y =

38.

4 x

+ 7

y =

78

elim

inat

ion

(-

);

su

bst

ituti

on

;

su

bst

ituti

on

; (1

2, 7

) (

18.7

, 5)

(15

, 9)

7. B

OO

KS

A l

ibra

ry c

onta

ins

2000

boo

ks. T

her

e ar

e 3

tim

es a

s m

any

non

-fic

tion

boo

ks a

s fi

ctio

n b

ooks

. Wri

te a

nd

solv

e a

syst

em o

f eq

uat

ion

s to

det

erm

ine

the

nu

mbe

r of

non

-fi

ctio

n a

nd

fict

ion

boo

ks.

x +

y =

20

00

and

x =

3y;

150

0 n

on

-fi c

tio

n,

500

fi ct

ion

8. S

CH

OO

L C

LUB

S T

he

ches

s cl

ub

has

16

mem

bers

an

d ga

ins

a n

ew m

embe

r ev

ery

mon

th. T

he

film

clu

b h

as 4

mem

bers

an

d ga

ins

4 n

ew m

embe

rs e

very

mon

th. W

rite

an

d so

lve

a sy

stem

of

equ

atio

ns

to f

ind

wh

en t

he

nu

mbe

r of

mem

bers

in

bot

h c

lubs

wil

l be

eq

ual

. y =

16

+ x

an

d y

= 4

+ 4

x;

4 m

on

ths

9. T

ia a

nd

Ken

eac

h s

old

snac

k ba

rs a

nd

mag

azin

e su

bscr

ipti

ons

for

a sc

hoo

l fu

nd-

rais

er, a

s sh

own

in

th

e ta

ble.

T

ia e

arn

ed $

132

and

Ken

ear

ned

$19

0.

a. D

efin

e va

riab

le a

nd

form

ula

te a

sys

tem

of

lin

ear

equ

atio

n f

rom

th

is s

itu

atio

n.

L

et x

= t

he

cost

per

sn

ack

bar

an

d le

t y =

th

e co

st p

er m

agaz

ine

sub

scri

pti

on

s; 1

6x +

4y =

132

an

d 2

0x +

6y =

190

.

b.

Wh

at w

as t

he

pric

e pe

r sn

ack

bar?

Det

erm

ine

the

reas

onab

len

ess

of y

our

solu

tion

.

$2;

Sam

ple

an

swer

: $2

is a

rea

son

able

pri

ce f

or

a sn

ack

bar

, an

d t

he

solu

tio

n c

hec

ks if

th

e m

agaz

ine

sub

scri

pti

on

s ar

e $2

5.

Prac

tice

Ap

ply

ing

Syste

ms o

f Lin

ear

Eq

uati

on

s

6-5

Item

Nu

mb

er S

old

Tia

Ken

snac

k ba

rs16

20

mag

azin

e su

bscr

iptio

ns4

6

028_

036_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

3312

/21/

10

8:43

AM



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 6

34

Gle

ncoe

Alg

ebra

1

Wor

d Pr

oble

m P

ract

ice

Ap

ply

ing

Syste

ms o

f Lin

ear

Eq

uati

on

s

1. M

ON

EY V

eron

ica

has

bee

n s

avin

g di

mes

an

d qu

arte

rs. S

he

has

94

coin

s in

all

, an

d th

e to

tal

valu

e is

$19

.30.

How

man

y di

mes

an

d h

ow m

any

quar

ters

doe

s sh

e h

ave?

28

dim

es;

66 q

uar

ters

2. C

HEM

ISTR

Y H

ow m

any

lite

rs o

f 15

%

acid

an

d 33

% a

cid

shou

ld b

e m

ixed

to

mak

e 40

lit

ers

of 2

1% a

cid

solu

tion

?

Co

nce

ntr

atio

no

f S

olu

tio

nA

mo

un

t o

f S

olu

tio

n (

L)

Am

ou

nt

of A

cid

15%

x

33%

y

21%

40

26

2 −

3 L

of

15%

; 13

1

−

3 L o

f 33

%

3. B

UIL

DIN

GS

Th

e S

ears

Tow

er i

n C

hic

ago

is t

he

tall

est

buil

din

g in

Nor

th A

mer

ica.

T

he

tota

l h

eigh

t of

th

e to

wer

t a

nd

the

ante

nn

a th

at s

tan

ds o

n t

op o

f it

a i

s 17

29 f

eet.

Th

e di

ffer

ence

in

hei

ghts

be

twee

n t

he

buil

din

g an

d th

e an

ten

na

is

279

feet

. How

tal

l is

th

e S

ears

Tow

er?

1450

ft

4. P

RO

DU

CE

Rog

er a

nd

Tre

vor

wen

t sh

oppi

ng

for

prod

uce

on

th

e sa

me

day.

T

hey

eac

h b

ough

t so

me

appl

es a

nd

som

e po

tato

es. T

he

amou

nt

they

bou

ght

and

the

tota

l pr

ice

they

pai

d ar

e li

sted

in

th

e ta

ble

belo

w.

Wh

at w

as t

he

pric

e of

app

les

and

pota

toes

per

pou

nd?

ap

ple

s: $

1.49

p

er lb

; p

ota

toes

: $0

.99

per

lb

5. S

HO

PPIN

G T

wo

stor

es a

re h

avin

g a

sale

on

T-s

hir

ts t

hat

nor

mal

ly s

ell

for

$20.

S

tore

S i

s ad

vert

isin

g an

s p

erce

nt

disc

oun

t, a

nd

Sto

re T

is

adve

rtis

ing

at

doll

ar d

isco

un

t. R

ose

spen

ds $

63 f

or

thre

e T-

shir

ts f

rom

Sto

re S

an

d on

e fr

om

Sto

re T

. Man

ny

spen

ds $

140

on f

ive

T-sh

irts

fro

m S

tore

S a

nd

fou

r fr

om

Sto

re T

. Fin

d th

e di

scou

nt

at e

ach

sto

re.

S

tore

S:

20%

; S

tore

T:

$5

6. T

RA

NSP

OR

TATI

ON

A S

peed

y R

iver

ba

rge

bou

nd

for

New

Orl

ean

s le

aves

B

aton

Rou

ge, L

ouis

ian

a, a

t 9:

00 A

.M. a

nd

trav

els

at a

spe

ed o

f 10

mil

es p

er h

our.

A

Rai

l Tra

nsp

ort

frei

ght

trai

n a

lso

bou

nd

for

New

Orl

ean

s le

aves

Bat

on R

ouge

at

1:30

P.M

. th

e sa

me

day.

Th

e tr

ain

tra

vels

at

25

mil

es p

er h

our,

and

the

rive

r ba

rge

trav

els

at 1

0 m

iles

per

hou

r. B

oth

th

e ba

rge

and

the

trai

n w

ill

trav

el 1

00 m

iles

to

rea

ch N

ew O

rlea

ns.

a. H

ow f

ar w

ill

the

trai

n t

rave

l be

fore

ca

tch

ing

up

to t

he

barg

e?

75 m

i

b.

Wh

ich

sh

ipm

ent

wil

l re

ach

New

O

rlea

ns

firs

t? A

t w

hat

tim

e?

Th

e tr

ain

will

arr

ive

fi rs

t. It

will

ar

rive

in N

ew O

rlea

ns

at

5:30

P.M

. of

the

sam

e d

ay.

c. I

f bo

th s

hip

men

ts t

ake

an h

our

to

un

load

bef

ore

hea

din

g ba

ck t

o B

aton

R

ouge

, wh

at i

s th

e ea

rlie

st t

ime

that

ei

ther

on

e of

th

e co

mpa

nie

s ca

n b

egin

to

loa

d gr

ain

to

ship

in

Bat

on

Rou

ge?

at

10:3

0 P.M

. th

e sa

me

day

6-5

Ap

ple

s(I

b)

Po

tato

es(I

b)

Tota

l Co

st

($)

Rog

er8

718

.85

Trev

or2

1012

.88

028_

036_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

3412

/21/

10

8:43

AM


NA

ME

DAT

E

P

ER

IOD

Lesson 6-5

Cha

pte

r 6

35

Gle

ncoe

Alg

ebra

1

Cra

mer’

s R

ule

Cra

mer

’s R

ule

is

a m

eth

od f

or s

olvi

ng

a sy

stem

of

equ

atio

ns.

To

use

Cra

mer

’s R

ule

, set

up

a m

atri

x to

rep

rese

nt

the

equ

atio

ns.

A m

atri

x is

a w

ay o

f or

gan

izin

g da

ta.

S

olve

th

e fo

llow

ing

syst

em o

f eq

uat

ion

s u

sin

g C

ram

er’s

Ru

le.

2x +

3y

= 1

3

x +

y =

5

Ste

p 1

: S

et u

p a

mat

rix

repr

esen

tin

g th

e co

effi

cien

ts o

f x

and

y.

A

=

x y

Ste

p 2

: F

ind

the

dete

rmin

ant

of m

atri

x A

.

If a

mat

rix

A =

⎪ a c b

d

⎥ , th

en t

he

dete

rmin

ant,

det

(A) =

ad

– b

c.

det(

A) =

2(1

) –

1(3)

= -

1

Ste

p 3

: R

epla

ce t

he

firs

t co

lum

n i

n A

wit

h 1

3 an

d 5

and

fin

d th

e de

term

inan

t of

th

e n

ew

mat

rix.

A1 =

⎪ 13

5 3 1 ⎥ ;

det

(A1)

= 1

3(1)

– 5

(3)

= -

2

Ste

p 4

: T

o fi

nd

the

valu

e of

x i

n t

he

solu

tion

to

the

syst

em o

f eq

uat

ion

s,

dete

rmin

e th

e va

lue

of

det(

A1)

−

det(

A) .

det(

A1)

−

det(

A) =

-2

−

-1 o

r 2

Ste

p 5

: R

epea

t th

e pr

oces

s to

fin

d th

e va

lue

of y

. Th

is t

ime,

rep

lace

th

e se

con

d co

lum

n w

ith

13

and

5 an

d fi

nd

the

dete

rmin

ant.

A2 =

⎪ 2 1 13

5 ⎥ ;

det

(A2)

= 2

(5)

– 1(

13) =

-3

and

det(

A2)

−

det(

A)

= -

3 −

-

1 or

3.

So,

th

e so

luti

on t

o th

e sy

stem

of

equ

atio

ns

is (

2, 3

).

Exer

cise

s U

se C

ram

er’s

Ru

le t

o so

lve

each

sys

tem

of

equ

atio

ns.

1. 2

x +

y =

1

2. x

+ y

= 4

3.

x –

y =

43x

+ 5

y =

5

2x

– 3y

= -

2 3

x –

5y =

8(0

, 1)

(2,

2)

(6,

2)

4. 4

x –

y =

3

5. 3

x –

2y =

7

6. 6

x –

5y =

1x

+ y

= 7

2

x +

y =

14

3x

+ 2

y =

5(2

, 5)

(5,

4)

(1,

1)

Enri

chm

ent

6-5

Exam

ple

⎪ 2 1 3 1 ⎥

028_

036_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

3512

/21/

10

8:43

AM



An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 6

36

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

Syste

ms o

f In

eq

ualiti

es

Syst

ems

of

Ineq

ual

itie

s T

he

solu

tion

of

a sy

stem

of

ineq

ual

itie

s is

th

e se

t of

all

or

dere

d pa

irs

that

sat

isfy

bot

h i

neq

ual

itie

s. I

f yo

u g

raph

th

e in

equ

alit

ies

in t

he

sam

e co

ordi

nat

e pl

ane,

th

e so

luti

on i

s th

e re

gion

wh

ere

the

grap

hs

over

lap.

S

olve

th

e sy

stem

of

ineq

ual

itie

s

by

grap

hin

g.y

> x

+ 2

y ≤

-2x

- 1

Th

e so

luti

on i

ncl

ude

s th

e or

dere

d pa

irs

in t

he

inte

rsec

tion

of

the

grap

hs.

Th

is r

egio

n i

s sh

aded

at

the

righ

t. T

he

grap

hs

of y

= x

+ 2

an

d y

= -

2x -

1 a

re b

oun

dari

es o

f th

is r

egio

n. T

he

grap

h o

f y

= x

+ 2

is

dash

ed a

nd

is n

ot i

ncl

ude

d in

th

e gr

aph

of

y >

x +

2.

S

olve

th

e sy

stem

of

ineq

ual

itie

s

by

grap

hin

g.x

+ y

> 4

x +

y <

-1

Th

e gr

aph

s of

x +

y =

4 a

nd

x +

y =

-1

are

para

llel

. Bec

ause

th

e tw

o re

gion

s h

ave

no

poin

ts i

n c

omm

on, t

he

syst

em o

f in

equ

alit

ies

has

no

solu

tion

.

Exer

cise

sS

olve

eac

h s

yste

m o

f in

equ

alit

ies

by

grap

hin

g.

1. y

> -

1 2.

y >

-2x

+ 2

3.

y <

x +

1x

< 0

y

≤ x

+ 1

3

x +

4y

≥ 1

2

x

y

O

y =

-1

x =

0

x

y

O

y =

-2x

+ 2

y =

x +

1

x

y

O

3x +

4y

= 1

2

y =

x +

1

4. 2

x +

y ≥

1

5. y

≤ 2

x +

3

6. 5

x -

2y

< 6

x -

y ≥

-2

y ≥

-1

+ 2

x y

> -

x +

1

x

y

O

2x +

y =

1

x -

y =

-2

x

y

O

y =

2x

+ 3

y =

-1

+ 2

x

x

y

O

y =

-x

+ 1

5x -

2y

= 6

x

y O

x +

y =

1

x +

y =

4

x

y O

y =

x +

2

y =

-2x

- 1

6-6

Exam

ple

1

Exam

ple

2

028_

036_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

3612

/21/

10

8:43

AM


NA

ME

DAT

E

P

ER

IOD

Lesson 6-6

Cha

pte

r 6

37

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Syste

ms o

f In

eq

ualiti

es

Ap

ply

Sys

tem

s o

f In

equ

alit

ies

In r

eal-

wor

ld p

robl

ems,

som

etim

es o

nly

wh

ole

nu

mbe

rs m

ake

sen

se f

or t

he

solu

tion

, an

d of

ten

on

ly p

osit

ive

valu

es o

f x

and

y m

ake

sen

se.

BU

SIN

ESS

AA

A G

em C

omp

any

pro

du

ces

n

eck

lace

s an

d b

race

lets

. In

a 4

0-h

our

wee

k, t

he

com

pan

y h

as 4

00 g

ems

to u

se. A

nec

kla

ce r

equ

ires

40

gem

s an

d a

b

race

let

req

uir

es 1

0 ge

ms.

It

tak

es 2

hou

rs t

o p

rod

uce

a

nec

kla

ce a

nd

a b

race

let

req

uir

es o

ne

hou

r. H

ow m

any

of

each

typ

e ca

n b

e p

rod

uce

d i

n a

wee

k?

Let

n =

th

e n

um

ber

of n

eckl

aces

th

at w

ill

be p

rodu

ced

and

b =

th

e n

um

ber

of b

race

lets

th

at w

ill

be p

rodu

ced.

Nei

ther

n o

r b

can

be

a n

egat

ive

nu

mbe

r, so

th

e fo

llow

ing

syst

em o

f in

equ

alit

ies

repr

esen

ts

the

con

diti

ons

of t

he

prob

lem

s.

n ≥

0b

≥ 0

b +

2n

≤ 4

010

b +

40n

≤ 4

00T

he

solu

tion

is

the

set

orde

red

pair

s in

th

e in

ters

ecti

on o

f th

e gr

aph

s. T

his

reg

ion

is

shad

ed

at t

he

righ

t. O

nly

wh

ole-

nu

mbe

r so

luti

ons,

su

ch a

s (5

, 20)

mak

e se

nse

in

th

is p

robl

em.

Exer

cise

sF

or e

ach

exe

rcis

e, g

rap

h t

he

solu

tion

set

. Lis

t th

ree

pos

sib

le s

olu

tion

s to

th

e p

rob

lem

.

1. H

EALT

H M

r. F

low

ers

is o

n a

res

tric

ted

2.

REC

REA

TIO

N M

aria

had

$15

0 in

gif

t di

et t

hat

all

ows

him

to

hav

e be

twee

n

cer

tifi

cate

s to

use

at

a re

cord

sto

re. S

he

1600

an

d 20

00 C

alor

ies

per

day.

His

b

ough

t fe

wer

th

an 2

0 re

cord

ings

. Eac

h

dail

y fa

t in

take

is

rest

rict

ed t

o be

twee

n

tap

e co

st $

5.95

an

d ea

ch C

D c

ost

$8.9

5.

45 a

nd

55 g

ram

s. W

hat

dai

ly C

alor

ie

How

man

y of

eac

h t

ype

of r

ecor

din

g m

igh

t an

d fa

t in

take

s ar

e ac

cept

able

? s

he

hav

e bo

ugh

t?

60 50 40 30 20 10

Cal

ori

es

Fat Grams

1000

2000

3000

0

S

amp

le a

nsw

ers:

160

0 C

alo

ries

,

Sam

ple

an

swer

s: 1

0 ta

pes

, 9 C

Ds;

45

fat

gra

ms;

180

0 C

alo

ries

, 50

fat

0

tap

es, 1

6 C

Ds;

14

tap

es, 5

CD

sg

ram

s; 2

00

0 C

alo

ries

, 55

fat

gra

ms

30 25 20 15 10 5

Tap

es

Compact Discs

510

1520

2530

0

5.95

t + 8

.95c

= 1

50

t + c

= 2

0

Nec

klac

es

Bracelets

1020

3040

500

50 40 30 20 10

10b

+ 4

0n =

400

b +

2n

= 4

0

6-6

Exam

ple

037_

044_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

3712

/21/

10

8:43

AM



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF 2nd



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 6

38

Gle

ncoe

Alg

ebra

1

Skill

s Pr

acti

ceS

yste

ms o

f In

eq

ualiti

es

Sol

ve e

ach

sys

tem

of

ineq

ual

itie

s b

y gr

aph

ing.

1. x

> -

1 2.

y >

2

3. y

> x

+ 3

y ≤

-3

x <

-2

y ≤

-1

x

y

O

x =

-1

y =

-3

x

y

O

y =

2

x =

-2

x

y

O

y =

x +

3

y =

-1

4. x

< 2

5.

x +

y ≤

-1

6. y

- x

> 4

y -

x ≤

2

x +

y ≥

3 Ø

x

+ y

> 2

x

y

O

y -

x =

2x

= 2

y

xO

y =

-x

+ 3

y =

-x

- 1

x

y

O

x +

y =

2

y -

x =

4

7. y

> x

+ 1

8.

y ≥

-x

+ 2

9.

y <

2x

+ 4

y ≥

-x

+ 1

y

< 2

x -

2

y ≥

x +

1

x

y

O

y =

-x

+ 1

y =

x +

1

x

y

O

y =

2x

- 2

y =

-x

+ 2

x

y

O

y =

x +

1

y =

2x

+ 4

Wri

te a

sys

tem

of

ineq

ual

itie

s fo

r ea

ch g

rap

h.

10.

x

y

O

11

.

x

y

O

12

.

x

y

O

y

≤ x

+ 2

, y ≥

x -

3

y >

-x, y

> x

y ≥

x +

1, y

< 1

6-6

037_

044_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

3812

/24/

10

6:11

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 6-6

Cha

pte

r 6

39

Gle

ncoe

Alg

ebra

1

Prac

tice

Syste

ms o

f In

eq

ualiti

es

$4 S

ton

es

Turq

uo

ise S

ton

es

$6 Stones

13

24

56

706 5 4 3 2 1

Gym

(h

ou

rs)

Die

go

’s R

ou

tin

e

Walking (miles)

13

24

56

78

0

16 14 12 10 8 6 4 2

6-6

Sol

ve e

ach

sys

tem

of

ineq

ual

itie

s b

y gr

aph

ing.

1. y

> x

- 2

2.

y ≥

x +

2

3. x

+ y

≥ 1

y ≤

x

y >

2x

+ 3

x

+ 2

y >

1

x

y

Oy

= x

y =

x -

2

x

y

O

y =

2x

+ 3

y =

x +

2

x

y

Ox

+ 2

y =

1

x +

y =

1

4. y

< 2

x -

1

5. y

> x

- 4

6.

2x

- y

≥ 2

y >

2 -

x

2x

+ y

≤ 2

x

- 2

y ≥

2

x

y

O

y =

2 -

x

y =

2x

- 1

x

y

O

2x +

y =

2

y =

x -

4

x

y

O

2x -

y =

2

x -

2y

= 2

7. F

ITN

ESS

Die

go s

tart

ed a

n e

xerc

ise

prog

ram

in

wh

ich

eac

h

wee

k h

e w

orks

ou

t at

th

e gy

m b

etw

een

4.5

an

d 6

hou

rs a

nd

wal

ks b

etw

een

9 a

nd

12 m

iles

.

a.

Mak

e a

grap

h t

o sh

ow t

he

nu

mbe

r of

hou

rs D

iego

wor

ks

out

at t

he

gym

an

d th

e n

um

ber

of m

iles

he

wal

ks p

er

wee

k.

b.

Lis

t th

ree

poss

ible

com

bin

atio

ns

of w

orki

ng

out

and

wal

kin

g th

at m

eet

Die

go’s

goa

ls.

8. S

OU

VEN

IRS

Em

ily

wan

ts t

o bu

y tu

rqu

oise

sto

nes

on

her

tr

ip t

o N

ew M

exic

o to

giv

e to

at

leas

t 4

of h

er f

rien

ds. T

he

gift

sh

op s

ells

sto

nes

for

eit

her

$4

or $

6 pe

r st

one.

Em

ily

has

n

o m

ore

than

$30

to

spen

d.

a. M

ake

a gr

aph

sh

owin

g th

e n

um

bers

of

each

pri

ce o

f st

one

Em

ily

can

pu

rch

ase.

b.

Lis

t th

ree

poss

ible

sol

uti

ons.

Sam

ple

an

swer

s:

gym

5 h

, wal

k 9

mi;

gym

6 h

, wal

k 10

mi,

gym

5.

5 h

, wal

k 11

mi

Sam

ple

an

swer

: o

ne

$4 s

ton

e an

d f

ou

r $6

sto

nes

; th

ree

$4

sto

nes

an

d t

hre

e $6

sto

nes

; fi

ve $

4 st

on

es a

nd

on

e $6

sto

ne

037_

044_

ALG

1_A

_CR

M_C

06_C

R_6

6028

0.in

dd

3912

/21/

10

8:43

AM


A01_A21_ALG1_A_CRM_C06_AN_660280.indd A18A01_A21_ALG1_A_CRM_C06_AN_660280.indd A18 12/24/10 6:23 PM12/24/10 6:23 PM

An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 6

40

Gle

ncoe

Alg

ebra

1

Wor

d Pr

oble

m P

ract

ice

Syste

ms o

f In

eq

ualiti

es

1. P

ETS

Ren

ée’s

Pet

Sto

re n

ever

has

mor

e th

an a

com

bin

ed t

otal

of

20 c

ats

and

dogs

an

d n

ever

mor

e th

an 8

cat

s. T

his

is

repr

esen

ted

by t

he

ineq

ual

itie

s x

≤ 8

an

d x

+ y

≤ 2

0. S

olve

th

e sy

stem

of

ineq

ual

itie

s by

gra

phin

g.

2. W

AG

ES T

he

min

imu

m w

age

for

one

grou

p of

wor

kers

in

Tex

as i

s $7

.25

per

hou

r ef

fect

ive

Sep

t. 1

, 200

8. T

he

grap

h

belo

w s

how

s th

e po

ssib

le w

eekl

y w

ages

fo

r a

pers

on w

ho

mak

es a

t le

ast

min

imu

m w

ages

an

d w

orks

at

mos

t 40

hou

rs. W

rite

th

e sy

stem

of

ineq

ual

itie

s fo

r th

e gr

aph

.

3. F

UN

D R

AIS

ING

Th

e C

amp

Cou

rage

C

lub

plan

s to

sel

l ti

ns

of p

opco

rn a

nd

pean

uts

as

a fu

ndr

aise

r. T

he

Clu

b m

embe

rs h

ave

$900

to

spen

d on

pro

duct

s to

sel

l an

d w

ant

to o

rder

up

to 2

00 t

ins

in a

ll. T

hey

als

o w

ant

to o

rder

at

leas

t as

m

any

tin

s of

pop

corn

as

tin

s of

pea

nu

ts.

Eac

h t

in o

f po

pcor

n c

osts

$3

and

each

tin

of

pea

nu

ts c

osts

$4.

Wri

te a

sys

tem

of

equ

atio

ns

to r

epre

sen

t th

e co

ndi

tion

s of

th

is p

robl

em.

x +

y ≤

20

0; x

≥ y

;

3x +

4y ≤

90

0S

tud

ents

may

als

o in

clu

de

x ≥

0

and

y ≥

0.

4. B

USI

NES

S F

or m

axim

um

eff

icie

ncy

, a

fact

ory

mu

st h

ave

at l

east

100

wor

kers

, bu

t n

o m

ore

than

200

wor

kers

on

a s

hif

t.

Th

e fa

ctor

y al

so m

ust

man

ufa

ctu

re a

t le

ast

30 u

nit

s pe

r w

orke

r.

a. L

et x

be

the

nu

mbe

r of

wor

kers

an

d le

t y

be

the

nu

mbe

r of

un

its.

Wri

te

fou

r in

equ

alit

ies

expr

essi

ng

the

con

diti

ons

in t

he

prob

lem

giv

en

abov

e.

x ≤

20

0;

x ≥

10

0; y

≥ 3

00

0; y

≥ 3

0x

b.

Gra

ph t

he

syst

ems

of i

neq

ual

itie

s.

c. L

ist

at l

east

th

ree

poss

ible

sol

uti

ons.

S

amp

le a

nsw

er:

(110

, 341

0),

(150

, 510

0), (

180,

630

0)

O

Dogs

Cat

s

46 28101214161820y

64

210

1416

1820

812

x

x

y

(40,

290

)

255075100

Wages ($)

125

150

175

200

225

250

275

300

Ho

urs

510

1520

2530

3540

4550

OUnits (1000)

23 14567

Wo

rker

s

y

150

100

5025

035

020

030

040

0x

x ≤

40,

y ≥

7.2

5x

6-6

037_

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ME

DAT

E

P

ER

IOD

Lesson 6-6

Cha

pte

r 6

41

Gle

ncoe

Alg

ebra

1

Enri

chm

ent

Descri

bin

g R

eg

ion

s

Th

e sh

aded

reg

ion

in

side

th

e tr

ian

gle

can

be

desc

ribe

d

wit

h a

sys

tem

of

thre

e in

equ

alit

ies.

y <

-x

+ 1

y >

1 −

3 x

- 3

y >

-9x

- 3

1

Wri

te s

yste

ms

of i

neq

ual

itie

s to

des

crib

e ea

ch r

egio

n. Y

ou m

ay

firs

t n

eed

to

div

ide

a re

gion

in

to t

rian

gles

or

qu

adri

late

rals

.

1.

x

y

O

2.

x

y

O

y

≤ 5 −

2 x

+ 1

2

y ≥

-x -

2

y ≤

x +

2

y

≤ -

5 −

2 x

+ 19

−

2

y ≥

x -

2

y ≤

-x +

2

y

≤ 2

y

≥ -

3

y ≤

5

y ≥

-5

3.

x

y

O

6-6

x

y

O

top

: y

≤ 3 −

2 x

+ 7

, y ≤

- 3 −

2 x

+ 7

, y ≥

4

mid

dle

: y ≤

4, y

≥ 0

, y ≥

- 4 −

3 x

- 4

,

y ≥

4 −

3 x

- 4

bo

tto

m le

ft:

y ≤

4x +

12,

y ≥

1 −

2 x

- 2

,y ≤

0, x

≤ 0

bo

tto

m r

igh

t: y

≤ -

4x +

12,

y ≥

- 1 −

2 x

- 2

, y ≤

0, x

≥ 0

037_

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ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 6

42

G

lenc

oe A

lgeb

ra 1

Gra

phin

g Ca

lcul

ator

Act

ivit

yS

had

ing

Ab

so

lute

Valu

e I

neq

ualiti

es

Abs

olu

te v

alu

e in

equ

alit

ies

of t

he

form

y ≤

a|

x -

h

| +

k o

r y

≥ a|

x -

h|

+ k

ca

n b

e gr

aph

ed u

sin

g th

e S

HA

DE

com

man

d or

by

usi

ng

the

shad

ing

feat

ure

in

th

e Y

= sc

reen

.

G

rap

h y

≤

2|

x -

3|

+

2.

Met

hod

1

SH

AD

E c

omm

and

En

ter

the

bou

nda

ry e

quat

ion

y =

2 |

x -

3|

+

2

into

Y1.

Fro

m t

he

hom

e sc

reen

, use

SH

AD

E t

o sh

ade

the

regi

on u

nde

r th

e bo

un

dary

eq

uat

ion

bec

ause

th

e in

equ

alit

y u

ses

<. U

se Y

min

as

the

low

er

bou

nda

ry a

nd

the

bou

nda

ry e

quat

ion

, Y1,

as

the

upp

er b

oun

dary

.K

eyst

roke

s:

2nd

[D

RA

W]

7 V

AR

S E

NT

ER

4

,

VA

RS

E

NT

ER

E

NT

ER

) E

NT

ER

.

Met

hod

2 Y

=

On

th

e Y

= sc

reen

, mov

e th

e cu

rsor

to

the

far

left

an

d re

peat

edly

pr

ess

EN

TE

R t

o to

ggle

th

rou

gh t

he

grap

h c

hoi

ces.

Be

sure

to

sele

ct

the

opti

on t

o sh

ade

belo

w

. Th

en e

nte

r th

e bo

un

dary

equ

atio

n a

nd

grap

h.

Th

e sa

me

tech

niq

ues

can

be

use

d to

sol

ve a

sys

tem

of

abso

lute

va

lue

ineq

ual

itie

s.

Exer

cise

sS

olve

eac

h s

yste

m o

f in

equ

alit

ies

by

grap

hin

g.

1. y

≤

2 |

x -

4|

+ 1

2.

y <

1 −

2 |x

- 3|

+ 2

3.

y ≤

4|

x -

4|

+ 3

4.

y ≤

1 −

4 |x

+ 1

| +

1

y

> |

x +

3|

+ 6

y

≤ |

x +

1|

+ 4

y

≥ 3

|x -

4|

- 2

y

≥ 2

|x +

3|

+ 5

S

olve

th

e sy

stem

y ≥

|x

+ 5

| -

4

and

y ≤

3|

x -

6|

+ 1.

En

ter

y =

|x

+ 5|

- 4

in

Y1

and

set

the

grap

h t

o sh

ade

abov

e.

The

n en

ter

y =

3 |

x -

6|

+ 1

in Y

2 an

d se

t th

e gr

aph

to s

hade

bel

ow.

Gra

ph t

he in

equa

litie

s. N

otic

e th

at a

dif

fere

nt p

atte

rn is

use

d fo

r th

e se

cond

ineq

ualit

y. T

he r

egio

n w

here

the

tw

o pa

tter

ns o

verl

ap is

the

so

luti

on t

o th

e sy

stem

.

[-10, 10]

scl:

1 b

y [

-10, 10]

scl:

1

[-10, 10]

scl:

1 b

y [

-10, 10]

scl:

1

[-20, 20]

scl:

2 b

y [

-20, 20]

scl:

2

[-3

0, 3

0]

scl:

5 b

y [

-10

, 5

0]

scl:

5[-

10

, 10

] s

cl:

1 b

y [

-10

, 10

] s

cl:

1[-

5, 15

] s

cl:

1 b

y [

-5

, 15

] s

cl:

1[-

10

, 10

] s

cl:

1 b

y [

-10

, 10

] s

cl:

1

6-6

Exam

ple

1

Exam

ple

2

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IOD

Lesson 6-6

Cha

pte

r 6

43

G

lenc

oe A

lgeb

ra 1

6-6

Spre

adsh

eet

Act

ivit

yS

yste

ms o

f In

eq

ualiti

es

Exer

cise

sU

se t

he

spre

adsh

eet

to d

eter

min

e w

het

her

Top

Sp

ort

can

mak

e th

e fo

llow

ing

com

bin

atio

ns

of s

hoe

s.

1. 1

000

Ru

nn

ers,

110

0 F

lyer

s yes

2. 1

200

Ru

nn

ers,

100

0 F

lyer

s yes

3. 2

000

Ru

nn

ers,

500

Fly

ers

no

4.

300

Ru

nn

ers,

150

0 F

lyer

s yes

4. 9

50 R

un

ner

s, 1

400

Fly

ers

yes

5.

180

0 R

un

ner

s, 1

300

Fly

ers

no

6. I

f Top

Spo

rt c

an e

ith

er b

uy

anot

her

cu

ttin

g m

ach

ine

or h

ire

mor

e as

sem

bler

s, w

hic

h w

ould

mak

e m

ore

com

bin

atio

ns

of s

hoe

pro

duct

ion

po

ssib

le?

Exp

lain

. H

ire m

ore

assem

ble

rs. E

ach

ord

ere

d p

air

th

at

was n

ot

a s

olu

tio

n o

f th

e

syste

m w

as t

rue f

or

the fi

rst

ineq

uality

. C

han

gin

g t

he s

eco

nd

in

eq

uality

m

ay m

ake s

om

e o

f th

ose o

rdere

d p

air

s s

olu

tio

ns o

f th

e s

yste

m.

T

opS

por

t S

hoe

Com

pan

y h

as a

tot

al o

f 96

00 m

inu

tes

of m

ach

ine

tim

e ea

ch w

eek

to

cut

the

mat

eria

ls f

or t

he

two

typ

es o

f at

hle

tic

shoe

s th

ey m

ake.

Th

ere

are

a to

tal

of 2

8,00

0 m

inu

tes

of w

ork

er

tim

e p

er w

eek

for

ass

emb

ly. I

t ta

kes

3 m

inu

tes

to c

ut

and

12

min

ute

s to

ass

emb

le a

pai

r of

Ru

nn

ers

and

2 m

inu

tes

to c

ut

and

10

min

ute

s to

as

sem

ble

a p

air

of F

lyer

s. I

s it

pos

sib

le f

or t

he

com

pan

y to

mak

e 12

00 p

airs

of

Ru

nn

ers

and

140

0 p

airs

of

Fly

ers

in a

wee

k?

Ste

p 1

R

epre

sent

the

sit

uati

on u

sing

a s

yste

m o

f in

equa

litie

s. L

et r

rep

rese

nt

the

nu

mbe

r of

Ru

nn

ers

and

f re

pres

ent

the

nu

mbe

r of

Fly

ers.

3r +

2f

≤

96

00

12

r +

10

f ≤ 28

,000

Ste

p 2

C

olu

mn

s A

an

d B

con

tain

th

e va

lues

of

r an

d f.

Col

um

ns

C a

nd

D

con

tain

th

e fo

rmu

las

for

the

ineq

ual

itie

s. T

he

form

ula

s w

ill

retu

rn

TR

UE

or

FA

LS

E. I

f bo

th i

neq

ual

itie

s ar

e tr

ue

for

an o

rder

ed p

air,

then

th

e or

dere

d pa

ir i

s a

solu

tion

to

the

syst

em o

f in

equ

alit

ies.

Sin

ce o

ne

of t

he

ineq

ual

itie

s is

fal

se f

or (

1200

, 140

0), t

he

orde

red

pair

is n

ot p

art

of t

he

solu

tion

set

. The

com

pany

ca

nn

ot m

ake

1200

pai

rs o

f R

un

ner

s an

d 14

00 p

airs

of

Fly

ers

in a

wee

k.Th

e fo

rmul

a in

cel

l D2

is12

*A2

+ 1

0*B

2 <

= 2

8,00

0.

+≤

+≤

Exam

ple

037_

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Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

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cGraw

-Hill C

om

panies, Inc.

PDF Pass


Chapter 6 Assessment Answer KeyQuiz 1 (Lessons 6-1 and 6-2) Quiz 3 (Lessons 6-5) Mid-Chapter Test

Page 47 Page 48 Page 49

Quiz 2 (Lessons 6-3 and 6-4)

Page 47

Quiz 4 (Lessons 6-6)

Page 48

1.

2.

3.

4.

5.

y

x

x - 2y = 3

x - 2y = -2

y

x

y = -x + 5

y = x32

1.

2.

3.

4.

5.

(1, 5)

infi nitely many solutions

A

one solution; (2, 3)

no solution

(

5 1 −

2 , -1 1 −

2 )

(-1 1 − 2 , 2)

(2, -3)

(2, -1)

D

7.

8.

9.

10.

11.

12.

y

x

x + 3y = 9

x - 3y = -3

(-9, -5)

(1, -4)

(

3 3 −

5 , 3 −

5 )

(-1, -4)

34 games


y

x

x + 3y = 9

x - 3y = -3

1. B

2. A

3. F

4. C

5. G

6. D

1.

2.

3.

x

y

y

x

D

1.

2.

3.

4.

5.

substitution; (5, 2)

elimination (×); (2, 3)

elimination (-); (-2, 2)

elimination

(+); (

- 1 −

4 , 1

)

D


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ight

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Chapter 6 Assessment Answer KeyVocabulary Test Form 1Page 50 Page 51 Page 52

1.

2.

3.

4.

5.

6.

Sample answer:

A system of equations

that has exactly one

solution is called

independent.

system of equations

consistent

dependent

eliminationSample answer:

Substitution is a

method of solving a

system of equations

where one variable is

solved in terms of the

other variable.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11. D

H

B

G

C

G

A

H

C

F

G

J

C

A

J

AB 12.

13.

14.

15.

16.

17.

B: substitution; (0, 0)


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Chapter 6 Assessment Answer KeyForm 2A Form 2BPage 53 Page 54 Page 55 Page 56

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11. J

B

H

A

H

C

F

A

J

B

A

12.

13.

14.

15.

16.

17. J

J

C

F

A

(

-7, -3 1 −

2 )

B:

D

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11. H

C

F

D

H

A

J

H

B

B

C

12.

13.

14.

15.

16.

17.

C

G

A

J

C

F

Manuel is 15 years old; his sister is 7 years old.

B:


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An

swer

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Chapter 6 Assessment Answer KeyForm 2CPage 57 Page 58

y

xO

2x - y = -3

6x - 3y = -9

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

(7, -1)

(-3, 2)

(4, -1)

(-8, 0)

(3, 5)

(1, 3)

y

xO

y = -x + 4

y = x - 4

one solution

no solution



elimination (-); (2, -7)


14 dimes; 19 nickels

186 student tickets;

135 adult tickets

23 and -6

16.

x

y

17.

x

y

18.

x = test score, y = family

income; x ≥ 75, y ≤ 40,000 ;

sample answer: (80, 38,000)

Mavis is 15 years old;

her brother is 10 years

old B:


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Chapter 6 Assessment Answer KeyForm 2DPage 59 Page 60

1.

2. 3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

(-2, 5)

(-1, 4)

(2, 2)

(5, 3)


(2, 1)

(2, 4)

y

x

4x - 2y = 10

2x - y = 5

y

x

y = x - 3

y = -x + 3

one solution



18 and -2

substitution; (-2, -5)

elimination (-);

(-1, 1)

16 dimes;

7 quarters

34 hours

16.

x

y

17.

x

y

18.

x = credit score, y = cost

of car; x ≥ 600, y ≥

5000; sample answer:

(650, 8000)

(-1, -3) B:


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PDF 2nd


Chapter 6 Assessment Answer KeyForm 3Page 61 Page 62

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

(-32, 15)

(-2, -6)

(-14, 7)

no solution

(1, 2)

no solution

(4, 1)

y

x

y = x 13

y + x + 4 = 0

one solution; (-1, -3)

no solution

y

x

3y = -x + 9

x + 3y = 3


6, 10

3 1 −

2 , 8 1 −

2

10 lb of $2.45 mix; 20 lb of $2.30 mix

(3, - 3 −

7 )

16.

17.

x

y

18.

x

y

19.

B:

5 mph

x = women’s shoes

sold; y = men’s shoes

sold; x ≥ 10, y ≥ 5,

2.5x + 3y ≥ 60; sample

answer: (15, 10)

10 square unitselimination (×);

(-

1 −

2 , 1 −

2 )


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Chapter 6 Assessment Answer Key Page 63, Extended-Response Test Scoring Rubric

Score General Description Specifi c Criteria

4 Superior

A correct solution that

is supported by well-

developed, accurate

explanations

• Shows thorough understanding of the concepts of solving systems of equations.

• Uses appropriate strategies to solve problems.

• Computations are correct.

• Written explanations are exemplary.

• raphs are accurate and appropriate.

• Goes beyond requirements of some or all problems.

3 Satisfactory

A generally correct solution,

but may contain minor fl aws in

reasoning or computation

• Shows an understanding of the concepts of solving systems of equations.

• Uses appropriate strategies to solve problems.

• Computations are mostly correct.

• Written explanations are effective.

• Graphs are mostly accurate and appropriate.

• Satisfi es all requirements of problems.

2 Nearly Satisfactory

A partially correct

interpretation and/or solution

to the problem

• Shows an understanding of most of the concepts of solving systems of equations.

• May not use appropriate strategies to solve problems.

• Computations are mostly correct.

• Written explanations are satisfactory.

• Graphs are mostly accurate.

• Satisfi es the requirements of most of the problems.

1 Nearly Unsatisfactory

A correct solution with no

supporting evidence or

explanation

• Final computation is correct.

• No written explanations or work is shown to substantiate

the fi nal computation.

• Graphs may be accurate but lack detail or explanation.

• Satisfi es minimal requirements of some of the problems.

0 Unsatisfactory

An incorrect solution indicating

no mathematical

understanding of the concept

or task, or no solution is given

• Shows little or no understanding of most of the concepts of

solving systems of equations.• Does not use appropriate strategies to solve problems.

• Computations are incorrect.

• Written explanations are unsatisfactory.

• Graphs are inaccurate or inappropriate.

• Does not satisfy requirements of problems.

• No answer may be given.


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Chapter 6 Assessment Answer Key Page 63, Extended-Response Test Sample Answers

1a. Sample answer: (5000, 5000); $600

1b. (15,000, -5000); To represent a possible investment both x and y must be positive. Thus, there is no interpretation for a negative value for y.

2a. The student should recognize that the total cost of a one-day rental of a compact car from the car rental company is modeled by the equation y = A + Bx, where y is the total cost and x is the number of miles driven during the day. Likewise, the total cost of a one-day rental of a compact car from the leading competitor is modeled by the equation y = 15 + 0.25x. The solution to the system is the number of miles driven during the day that makes the cost of renting a compact car from the two companies for one day equal, and the corresponding total cost.

2b. The value of A determines if the one-day fee for the rental of the compact car from the car rental company will be less than, greater than, or equal to the one-day fee from their leading competitor. The value of B determines whether the number of miles driven will keep the total cost less than, greater than, or equal to the total cost of renting from the leading competitor, or change which company has the higher total cost for the rental.

3a. x + y = S2.5x + 0.75y = PBoth S and P must be positive. The student may recognize that the maximum profit is 2.5 times the weekly sales, and the minimum profit is 0.75 times the weekly sales. i.e., 0.75S ≤ P ≤ 2.5S.

3b. Sample answer:

The graph of this system of equations shows the ordered pair that corresponds to weekly sales of 250 and a weekly profit of $450. In order to have a profit of $450 from magazine and book sales, the store would have to sell 150 books and 100 magazines.

4a. a + b = 21a - 5 b = 3

4b. 17 adults; 4 babies

Mag

azin

es

100

0

200

400

600

Books100 200 300 400

y

x

2.5x + .75y = 450

x + y = 250

In addition to the scoring rubric found on page A28, the following sample answers may be used as guidance in evaluating extended-response assessment items.


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Chapter 6 Assessment Answer KeyStandardized Test PracticePage 64 Page 65

1. A B C D

2.

3. A B C D

4. F G H J

5. A B C D

6. F G H J

7. A B C D

8.

9. A B C D

10.

F G H J

F G H J

F G H J

11. A B C D

12. F G H J

13. A B C D

14. F G H J

15. A B C D

16. F G H J

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

2 5

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

8 17. 18.


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Chapter 6 Assessment Answer KeyStandardized Test PracticePage 66

19.

20.

21.

22.

23.

24.

25.

26.

27.

28a.

28b. -12, 4

Sample answer: Let

x = fi rst number

and y = second number;

3x - y = -40

x + 2y = -4

y

xOx - 2y = 4

2x - y = 4

-1-2-3-4 0 1 2 43

{r | r < 3}

d = 55t ; 275 miles

yes; common difference is 3

-2

3

2

10

-1-2

decrease; 25%

- 8 −

7

Let n = the number; 2n - 7 > 13 or 2n - 7 ≤ -5; {n | n > 10 or n ≤ 1}

y = 3x + 5


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Chapter 6 Assessment Answer KeyUnit 2 TestPage 67 Page 68

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

y = -3x - 6

3x + 2y = -10

y = 5x - 2

y = 2x + 9

d = 40t

-41

y

xO

no solution

0

40

0

80

120

160

1 2 3 4

Dis

tan

ce

(m

ile

s)

Time (hours)

positive; sample

answer: y = 130 −

3 x + 20

−

3

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23. y

xO

-3-2-1 0 1 2 4 53

no solution

{w | w < -6 or w > -4}

elimination (×); (3, -5)

elimination (-); (10, -4)

elimination (+); (3, -2)


{v | v < -3 or v > 11}

{t | -1 < t < 2}

{a | a ≤ 8}

{x | x > -5}

-7-6-5-4-8 -3-2-1 0

about $2.25

{

b | 2 −

3 ≤ b ≤ 4

}


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Chapter 6 Resource Masters - commackschools.org 6 Worksheets.pdf · Chapter 6 Resource Masters Chapter Resources Student-Built Glossary (pages 1–2) These masters are a student study