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Algebra 1 Algebra 1 A Review and Summary A Review and Summary Gabriel Grahek Gabriel Grahek

Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

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Page 1: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Algebra 1 Algebra 1 A Review and SummaryA Review and Summary

Gabriel GrahekGabriel Grahek

Page 2: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

In the next slides you will review:In the next slides you will review:

Solving 1st power equations in one Solving 1st power equations in one variablevariable

A. Special cases where variables cancel to get A. Special cases where variables cancel to get {all reals} or {all reals} or

B.B. Equations containing fractional Equations containing fractional coefficientscoefficients

C.C. Equations with variables in the Equations with variables in the denominator denominator –(throw out answers that cause –(throw out answers that cause division by zero)division by zero)

Page 3: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

11stst Power Equations Power Equations

x+3=8x+3=8 4(x 4(x22-2)+8=36-2)+8=36

x+3-3=8x+3-3=8 (4x(4x22-8)+8=36-8)+8=36

x=5x=5 4x4x22-8=28-8=28

4x4x22=36=36

xx22=9=9

x=3x=3

• Any type of equation that has only one variable.

Note that the variable can be on both sides.

Page 4: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

11stst Power Equations Power Equations

3x-4=12+x3x-4=12+x 4y+16=244y+16=24

3x-4+4=12+4+x3x-4+4=12+4+x 4y+16-16=24-4y+16-16=24-1616

3x=16+x3x=16+x 4y=84y=8

3x-x=16+x-x3x-x=16+x-x y=2y=2

2x=162x=16

x=8x=8

Page 5: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

11stst Power Equations Power Equations• Special cases- Ø and {all reals}

x-(4-3)=x 4x+4=2(2x+2)

x-1=x 4x=4x

-1=0 x=x

Ø

Page 6: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

11stst Power Equations Power Equations•Equations containing fractional coefficients and with variables in the denominator.

2 416

24

162

4 32

8

x

x

x

x

2

2

2

2

2

2

5(7) 52

1030

210

30 2( 10)

30 2 20

50 2

25

5

x

x

x

x

x

x

x

Page 7: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

In the next slides you will In the next slides you will review:review:

Review all the Properties and Review all the Properties and then take a Quiz on then take a Quiz on

identifying the Property identifying the Property NamesNames

Page 8: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Addition Property (of Addition Property (of Equality)Equality)

Multiplication Property (of Equality)

Example: If a, b, and c, are any real numbers, and a=b, then a+c=b+c and c+a=c+bIf the same number is added to equal numbers, the sums are equal.

Example: If a, b, and c are real numbers, and a=b, then ca=cb and ac=bc. If equal numbers are multiplied by the same number, the products are equal.

Page 9: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Reflexive Property (of Reflexive Property (of Equality)Equality)

Symmetric Property (of Equality)

Transitive Property (of Equality)

Example: For all real numbers a, b, and c:a=a

Example: For all real numbers a, b, and c:If a=b, then b=a.

Example: For all real numbers a, b, and c:If a=b and b=c, then a=c.

Page 10: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Associative Property of Associative Property of AdditionAddition

Associative Property of Multiplication

Example: For all real numbers a, b, and c:(a+b)+c=a+(b+c) Example: (5+6)+7=5+(6+7)

Example: For all real numbers a, b, and c:

(ab)c=a(bc) Example: 2 3 4 2 3 4

Page 11: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Commutative Property of Commutative Property of AdditionAddition

Commutative Property of Multiplication

Example: For all real numbers a and b:a+b=b+a Example: 2+3=3+2

Example:For all real numbers a, b, and c:ab=ba Example: 4 5 5 4

Page 12: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Distributive Property (of Distributive Property (of Multiplication over Addition)Multiplication over Addition)

Example: For all real numbers a, b, and c:a(b+c)=ab+acand(b+c)a=ba+ca

Page 13: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Prop of Opposites or Inverse Prop of Opposites or Inverse Property of Addition Property of Addition

Prop of Reciprocals or Inverse Prop. of Multiplication

Example: For every real number a, there is a real number -a such thata+(-a)=0 and (-a)+a=0

Example: If we multiply a number times itsreciprocal, it will equal one. For example:

11x

x

Page 14: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Identity Property of Addition Identity Property of Addition

Identity Property of Multiplication

Example: There is a unique real number 0 such that for every real number a,

a + 0 = a and 0 + a = aZero is called the identity element of addition.

Example: There is a unique real number 1 such that for every real number a,

a · 1 = a and 1 · a = aOne is called the identity element of multiplication.

Page 15: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Multiplicative Property of Zero Multiplicative Property of Zero

Closure Property of Addition

Closure Property of Multiplication

Example: For every real number a,a · 0 = 0 and 0 · a = 0

Example: Closure property of real number addition states that the sum of any two real numbers equals another real number.

Example: Closure property of real number multiplication states that the product of any two real numbers equals another real number.

Page 16: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Product of Powers PropertyProduct of Powers Property

Power of a Product Property

Power of a Power Property

Example: This property states that to multiply powers having the same base, add the exponents.That is, for a real number non-zero a and two integers m and n, am × an = am+n.

Example: This property states that the power of a product can be obtained by finding the powers of each factor and multiplying them. That is, for any two non-zero real numbers a and b and any integer m, (ab)m = am × bm.

Example: This property states that the power of a power can be found by multiplying the exponents.That is, for a non-zero real number a and two integers m and n, (am)n = amn.

Page 17: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Quotient of Powers PropertyQuotient of Powers Property

Power of a Quotient Property

Example: This property states that to divide powers having the same base, subtract the exponents.That is, for a non-zero real number a and two integers m and n, .

Example: This property states that the power of a quotient can be obtained by finding the powers of numerator and denominator and dividing them. That is, for any two non-zero real numbers a and b and any integer m,

mm n

n

aa

a

m m

m

a a

b b

Page 18: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Zero Power Property Zero Power Property

Negative Power Property

Example: Any number raised to the zero power is equal to “1”.

Example: Change the number to its reciprocal.

0 1a

22

1x

x

Page 19: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Zero Product Property Zero Product Property

Example: Zero - Product Property states that if the product of two factors is zero, then at

least one of the factors must be zero.

If xy = 0, then x = 0 or y = 0.

Page 20: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Product of Roots Property Product of Roots Property

Quotient of Roots Property

For all positive real numbers a and b,

That is, the square root of the product is the same as the product of the square roots.

a b a b

For all positive real numbers a and b, b ≠ 0:

The square root of the quotient is the same as the quotient of the square roots.

a a

bb

Page 21: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Root of a Power Property Root of a Power Property

Power of a Root Property

Example:

Example:

4 2

2 2

2

( )

7 7

x x

x

2( 9) 9

Page 22: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Now you will take a quiz!Now you will take a quiz!Look at the sample problem and Look at the sample problem and give the name of the property give the name of the property

illustrated. illustrated.

1. a + b = b + a

Click when you’re ready to see the answer.

Answer: Commutative Property (of Addition)

Page 23: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Now you will take a quiz!Now you will take a quiz!Look at the sample problem and Look at the sample problem and give the name of the property give the name of the property

illustrated. illustrated.

2. am × an = am+n

Click when you’re ready to see the answer.

Answer: Product of Powers

Page 24: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Now you will take a quiz!Now you will take a quiz!Look at the sample problem and Look at the sample problem and give the name of the property give the name of the property

illustrated. illustrated.

3. For every real number a,a · 0 = 0 and 0 · a = 0

Click when you’re ready to see the answer.

Answer: Multiplicative Property of Zero

Page 25: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Now you will take a quiz!Now you will take a quiz!Look at the sample problem and Look at the sample problem and give the name of the property give the name of the property

illustrated. illustrated.

4. the sum of any two real numbers equals another real number.

Click when you’re ready to see the answer.

Answer: Closure Property of Addition

Page 26: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Now you will take a quiz!Now you will take a quiz!Look at the sample problem and Look at the sample problem and give the name of the property give the name of the property

illustrated. illustrated.

5. There is a unique real number 1 such that for every real number a,a · 1 = a and 1 · a = a

Click when you’re ready to see the answer.

Answer: Identity property of Multiplication

Page 27: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Now you will take a quiz!Now you will take a quiz!Look at the sample problem and Look at the sample problem and give the name of the property give the name of the property

illustrated. illustrated.

6.

Click when you’re ready to see the answer.

Answer: Zero Power Property

0 1a

Page 28: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Now you will take a quiz!Now you will take a quiz!Look at the sample problem and Look at the sample problem and give the name of the property give the name of the property

illustrated. illustrated.

7.

Click when you’re ready to see the answer.

Answer: Quotient of Powers Property

mm n

n

aa

a

Page 29: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Now you will take a quiz!Now you will take a quiz!Look at the sample problem and Look at the sample problem and give the name of the property give the name of the property

illustrated. illustrated.

8. (ab)m = am

Click when you’re ready to see the answer.

Answer: Power of a Product

Page 30: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Now you will take a quiz!Now you will take a quiz!Look at the sample problem and Look at the sample problem and give the name of the property give the name of the property

illustrated. illustrated.

9.

Click when you’re ready to see the answer.

Answer: Negative Power Property

22

1x

x

Page 31: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Now you will take a quiz!Now you will take a quiz!Look at the sample problem and Look at the sample problem and give the name of the property give the name of the property

illustrated. illustrated.

10.

Click when you’re ready to see the answer.

Answer: Prop of Reciprocals or Inverse Prop. of Multiplication

11x

x

Page 32: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Now you will take a quiz!Now you will take a quiz!Look at the sample problem and Look at the sample problem and give the name of the property give the name of the property

illustrated. illustrated.

11. (ab)c=a(bc)

Click when you’re ready to see the answer.

Answer: Associative Property of Multiplication

Page 33: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Solving InequalitiesSolving Inequalities

Page 34: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Solving InequalitiesSolving Inequalities Remember the Multiplication Remember the Multiplication

Property of Inequality! If you Property of Inequality! If you multiply or divide by a negative, you multiply or divide by a negative, you must reverse the inequality sign.must reverse the inequality sign.

-4

-2x < 8-2x < 8 x > -4x > -4

Solution Set: {x: x > -4}Solution Set: {x: x > -4} Graph of the Solution:Graph of the Solution:

Page 35: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Solving InequalitiesSolving Inequalities Open endpoint for these symbols: > <Open endpoint for these symbols: > < Closed endpoint for these symbols: Closed endpoint for these symbols: ≥ or≥ or ≤≤ Conjunction must satisfy both conditionsConjunction must satisfy both conditions Conjunction = “AND”Conjunction = “AND”

-5 8

{x: -5 < x ≤ 8}Click to see solution graph

Page 36: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Solving InequalitiesSolving Inequalities Open endpoint for these symbols: > <Open endpoint for these symbols: > < Closed endpoint for these symbols: Closed endpoint for these symbols: ≥ or≥ or

≤≤ Disjunction must satisfy either one or Disjunction must satisfy either one or

both of the conditionsboth of the conditions Disjunction = “OR”Disjunction = “OR”

{x: x < -6 or x ≥ 8}Click to see solution graph

-6 8

Page 37: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Solving Inequalities – Special Solving Inequalities – Special CasesCases Watch for special casesWatch for special cases No solutions that work: Answer is No solutions that work: Answer is ØØ Every number works: Answer is Every number works: Answer is

{reals}{reals} When the disjunction goes the same When the disjunction goes the same

way you use one arrow.way you use one arrow.

{x: x > -6 or x ≥ 8}Click to see solution graph

-6 8

Page 38: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Solving Inequalities – Special Solving Inequalities – Special CasesCases

Watch for special cases:Watch for special cases: No solutions that work: Answer is No solutions that work: Answer is ØØ Every number works: Answer is {reals}Every number works: Answer is {reals}

{x: -2x < -4 and -9x ≥ 18}

Click to see solution

Ø

Page 39: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Solving InequalitiesSolving Inequalities Now you try this problemNow you try this problem

2x > 6 or -16x ≤ 32Click to see solution and graph

-2 3

-2 < x or x ≤ 3

Page 40: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Solving InequalitiesSolving Inequalities Now you try this problem.Now you try this problem.

4x-8 < 12 and -x < 10-4Click to see solution and graph

-6 < x < 5

-6 5

Page 41: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Type the answer here. Set to fade-in Type the answer here. Set to fade-in on clickon click

Type a sample problem here. Blah blah blah. You can duplicate this slide.

Type any needed explanation or tips here. Set to fade-in 3 seconds after the answer appears above.

Click when ready to see the answer.

Page 42: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

In the next slides you will In the next slides you will review:review:

Linear equations in two Linear equations in two variablesvariables

Lots to cover here: slopes of Lots to cover here: slopes of all types of lines; equations of all all types of lines; equations of all types of lines, standard/general types of lines, standard/general form, point-slope form, how to form, point-slope form, how to graph, how to find intercepts, graph, how to find intercepts, how and when to use the point-how and when to use the point-slope formula, etc. Remember slope formula, etc. Remember you can make lovely graphs in you can make lovely graphs in Geometer's Sketchpad and copy Geometer's Sketchpad and copy and paste them into PPT.and paste them into PPT.

Page 43: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Linear EquationsLinear Equations

Slope= Slope= Point-Slope Point-Slope

Formula= Formula= Slope-Intercept Slope-Intercept

Formula=Formula= Midpoint Formula= Midpoint Formula= Standard/ General Standard/ General

Form= Form= Ax+Bx=CAx+Bx=C Distance Between Distance Between

Two Points Formula= Two Points Formula=

2 1

2 1

y y

x x

1 1( )y y m x x

y mx b 1 2 1 2

,2 2

x x y y

2 1 2 2 1 2( ) ( )d x x y y

Page 44: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

SlopeSlope

(9,12) and (13, 20)(9,12) and (13, 20)

Would be negative if it Would be negative if it had a negative sign in had a negative sign in front of it. It would front of it. It would then be a falling line then be a falling line and not a rising line.and not a rising line.

20 12

13 9

Pt-Slope Formula

8

4

(9,12) and (13, 20)

Use when you only have solution points.

812 ( 9)

4y x

12 2 18

2 6

y x

y x

Page 45: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Midpoint Midpoint

(9,12) and (13, 20)(9,12) and (13, 20)

Use to find the middle Use to find the middle point on a line.point on a line.

9 13 12 20,

2 2

Distance

(9,12) and (13, 20)

Use to find the Distance between to points.

22 32,

2 2

11,16

2 2(13 9) (20 12)d 2 2(4) (8)

16 64

80

4 5

d

d

d

d

Page 46: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Equations in Two VariablesEquations in Two Variables

The pairs of numbers that come out The pairs of numbers that come out for each variable can be written as for each variable can be written as an (x,y) value. (ordered pair)an (x,y) value. (ordered pair)

You give the solutions in You give the solutions in alphabetical order of the variables. alphabetical order of the variables. So, it would be (a,b) and not (b,a).So, it would be (a,b) and not (b,a).

Page 47: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Standard FormStandard Form

ax+by=cax+by=c All linear equations can be written in All linear equations can be written in

this form.this form. A, b, and c are real numbers and a and A, b, and c are real numbers and a and

b are non-zero. A, b, and c are b are non-zero. A, b, and c are integers.integers.

To change to slope intercept:To change to slope intercept:

Ax+bx=c bx=ax+cAx+bx=c bx=ax+c

Page 48: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

How to graphHow to graph

To graph the slope-intercept form: To graph the slope-intercept form: you can take the y intercept and use you can take the y intercept and use the slope to determine the points on the slope to determine the points on the line.the line.

To graph the standard form you have To graph the standard form you have to change it to slope-intercept, to change it to slope-intercept, explained in the last slide, and then explained in the last slide, and then graph it.graph it.

Page 49: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

To find the y-To find the y-interceptintercept

F(x)=mx+bF(x)=mx+b

Set x to 0Set x to 0

F=bF=b

Example:Example:

F(x)=4x+6F(x)=4x+6

F(0)=4(0)+6F(0)=4(0)+6

F=6F=6

To find the x-interceptF(x)=mx+b

Set f(x)=0

0=mx+b

Divide out

Example:

F(x)=4x-8

0=4x-8

8=4x

2=x

Page 50: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

In the next slides you will In the next slides you will review:review:

Linear SystemsLinear SystemsA. Substitution MethodA. Substitution MethodB. Addition/SubtractionB. Addition/Subtraction

Method (Elimination ) Method (Elimination ) C. Check for C. Check for

understanding of the terms understanding of the terms dependent, inconsistent and dependent, inconsistent and consistentconsistent

Page 51: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

When two line share solution points…When two line share solution points…

Null set (if they are parallel)Null set (if they are parallel)This will be called an This will be called an INCONSISTENT SYSTEMINCONSISTENT SYSTEM

One point (if they cross)One point (if they cross)This will be called a This will be called a CONSISTENT SYSTEMCONSISTENT SYSTEM

Infinite Set or All Pts on the Line (if same Infinite Set or All Pts on the Line (if same

line is used twice)line is used twice)This will be called a This will be called a DEPENDENT SYSTEMDEPENDENT SYSTEM (It is (It is

also consistent. Dependent is the better name also consistent. Dependent is the better name

for it than consistent.)for it than consistent.)

Page 52: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Solution: Null set (if they are Solution: Null set (if they are

parallel)parallel)INCONSISTENT SYSTEMINCONSISTENT SYSTEM

One point (if they cross)One point (if they cross)CONSISTENT SYSTEMCONSISTENT SYSTEM

Infinite Set or All Pts on the Line (if Infinite Set or All Pts on the Line (if

same line is used twice)same line is used twice)DEPENDENT SYSTEMDEPENDENT SYSTEM

4

2

-2

-4

4

2

-2

-4

4

2

-2

-4

Page 53: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

The SOLUTION of a The SOLUTION of a SYSTEMSYSTEM is is the the INTERSECTION SETINTERSECTION SET

Where do the two lines Meet intersect.

Page 54: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Two Different Equation on the Two Different Equation on the same graph are called a same graph are called a

SYSTEM OF EQUATIONS.SYSTEM OF EQUATIONS.

Think about:Think about:

2x y 8

x y 1

y 2x 5

3y x 2

2

Page 55: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Method 1: To estimate the solution Method 1: To estimate the solution of a system, you have to find out of a system, you have to find out

where they intersect.where they intersect.

y 2x 5

3y x 2

2

5

4

3

2

1

2 4

g x = -2x+5 f x = 3

2 x-2

Solution to this system is:

{(2, 1)}

Page 56: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Example: You can use Trace on a graphing Example: You can use Trace on a graphing calculator to help you estimate the solution of calculator to help you estimate the solution of

a system. It can find where they intersect.a system. It can find where they intersect.

3

2

1

-1

-2 2 4

g x = -1

3 x+1

f x = 9-x2

2

1y x 1

3

y 9 x

Page 57: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Example: You can use Trace on a graphing Example: You can use Trace on a graphing calculator to help you estimate the solution of calculator to help you estimate the solution of

a system. It can find where they intersecta system. It can find where they intersect

Solution to this system appears to include TWO pts:

{ (-2.38, 1.79), (3, 0) }

3

2

1

-1

-2 2 4

B: (3.00, 0.00)

A: (-2.38, 1.79)g x = -

1

3 x+1

f x = 9-x2

2

1y x 1

3

y 9 x

Page 58: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Summary of Method 1: Estimate Summary of Method 1: Estimate the SOLUTION of a the SOLUTION of a SYSTEMSYSTEM on a on a graph. graph. (Goal: Find intersection pts.)(Goal: Find intersection pts.)

Disadvantages: Might only give an estimate.It might not be possible to graph some equations yet.

Advantages:If the graph is easy, this is a good way to check.It is good for a quick answer.

Page 59: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Method 2: Substitution Method 2: Substitution MethodMethod

(Goal: Replace one variable in one equation (Goal: Replace one variable in one equation with the set from another.)with the set from another.)

Step 1: Look for a variable with a coefficient of one.

Step 2: Move everything else to the other side.

Equation A now becomes: y = 15-x

Step 3: SUBSTITUTE this expression into that variable in Equation B

Equation B now becomes 4x – 3( 15-x ) = 38

Step 4:Solve the equation.

Step 5:Back-substitute this coordinate into Step 2 to find the other coordinate. (Or plug into any equation but step 2 is easiest!)

x y 15

4x 3y 38

Page 60: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Method 2: Substitution Method 2: Substitution (Goal: replace one variable(Goal: replace one variablewith an equal expression.)with an equal expression.)

Step 1: Look for a variable with a coefficient of one.

Step 2: Isolate that variable

Equation A now becomes:

y = 3x + 1

Step 3: SUBSTITUTE this expression into that variable in Equation B

Equation B becomes

7x – 2( 3x + 1 ) = - 4

Step 4: Solve for the remaining variable

Step 5: Back-substitute this coordinate into Step 2 to find the other coordinate. (Or plug into any equation but step 2 is easiest!)

3x y 1

7x 2y 4

y

7x 2( ) 4

7x 6x 2 4

x 2 4

x 2 y

3x 1

3x 1

Solution : {(

3( 2) 1

y 5

2, 5)}

Page 61: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Example: Substitution Example: Substitution (Goal: replace one variable with (Goal: replace one variable with the set of another equation.)the set of another equation.)

Step 1: Look for a variable with a coefficient of one.

Step 2: Move everything else to the other side.

Step 3: SUBSTITUTE this expression into that variable in Equation B

Step 4: Solve for the remaining variable

Step 5: Back-substitute this coordinate into Step 2 to find the other coordinate. (Or plug into any equation but step 2 is easiest!)

x 5y 8

2x 3y 3

A : x 5y 8 x

B : 2( ) 3y 3

10y 16 3y 3

13y 16 3

13y 13

y 1 x 5

5y 8

5y 8

y 8

Page 62: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Example: Substitution Example: Substitution (Goal: replace one variable(Goal: replace one variablewith an equal expression.)with an equal expression.)

Step 1: Look for a variable with a coefficient of one.

Step 2: Isolate that variable

Step 3: SUBSTITUTE this expression into that variable in Equation B

Step 4: Solve for the remaining variable

Step 5: Back-substitute this coordinate into Step 2 to find the other coordinate. (Or plug into any equation but step 2 is easiest!)

x 5y 8

2x 3y 3

A : x 5y 8 x

B : 2( ) 3y 3

10y 16 3y 3

13y 16 3

13y

5y 8

5y 8

Solut

1

ion

3

y

:{(3,1)

1 x 5y 8

x 5(1) 3

}

8

Page 63: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Method 2 Summary: Method 2 Summary: Substitution MethodSubstitution Method

(Goal: replace one variable with an equal (Goal: replace one variable with an equal expression.)expression.)

Disadvantages: Avoid this method when it requires messy fractions Avoid IF no coefficient equals one.

Advantages:This is the algebra method to use when degrees of the equations are not equal.

Page 64: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Step 1: Look for the LCM of the coefficients on either x or y. (Opposite signs are recommended to avoid errors.)

Here: -3y and +2y could be turned into -6y and +6y

Step 2: Multiply each equation by the necessary factor.

Equation A now becomes: 10x – 6y = 10

Equation B now becomes: 9x + 6y = -48

Step 3: ADD the two equations if using opposite signs (if not, subtract)

Step 4:Solve the equation.

Step 5:Back-substitute this coordinate into any equation to find the other coordinate. (Look for easiest coefficients to work with.)

Method 3: Elimination Methodor Addition/Subtraction Method(Goal: Combine equations to cancel out one variable.)

5x 3y 5

3x 2y 16

Page 65: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Step 1: Look for the LCM of the coefficients on either x or y. (Opposite signs are recommended to avoid errors.)

Here: -3y and + 2y could be turned into -6y and + 6y

Step 2: Multiply each equation by the necessary factor.

A becomes: 10x – 6y = 10

B becomes: 9x + 6y = -32

Step 3: ADD the two equations if using opposite signs (if not, subtract)

Step 4: Solve for the remaining variable

Step 5: Back-substitute this coordinate into any equation to find the other coordinate. (Look for easiest coefficients to work with.)

Method 3: Elimination or Addition/Subtraction Method(Goal: Combine equations to cancel out one variable.)

6y

6y

5x 3y 5

3x 2y 16

10x 10

9x 48

19x 38

x 2 Are wedone?

+

Page 66: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Step 1: Look for the LCM of the coefficients on either x or y. (Opposite signs are recommended to avoid errors.)

Step 2: Multiply each equation by the necessary factor.

Step 3: ADD the two equations if using opposite signs (if not, subtract)

Step 4: Solve for the remaining variable

Step 5: Back-substitute this coordinate into any equation to find the other coordinate. (Look for easiest coefficients to work with.)

Method 3: Elimination or Addition/Subtraction Method(Goal: Combine equations to cancel out one variable.)

+

5x 3y 5

3x 2y 16

10x 10

9x 48

19x 38

x 2 5 3y 5

5( ) 3y 5

10

6y

6y

x

2

Solution :{( 2, 5)

3y 5

3y 15

y 5

}

Page 67: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Step 1: Look for the LCM of the coefficients on either x or y. (Opposite signs are recommended to avoid errors.)

Step 2: Multiply each equation by the necessary factor.

Step 3: ADD the two equations if using opposite signs (if not, subtract)

Step 4: Solve for the remaining variable

Step 5: Back-substitute this coordinate into any equation to find the other coordinate. (Look for easiest coefficients to work with.)

Example: Elimination or Addition/Subtraction Method(Goal: Combine equations to cancel out one variable.)

+

2x 3y 2

5x 7y 34

Page 68: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Step 1: Look for the LCM of the coefficients on either x or y. (Opposite signs are recommended to avoid errors.)

Step 2: Multiply each equation by the necessary factor.

Step 3: ADD the two equations if using opposite signs (if not, subtract)

Step 4: Solve for the remaining variable

Step 5: Back-substitute this coordinate into any equation to find the other coordinate. (Look for easiest coefficients to work with.)

Example: Elimination or Addition/Subtraction Method(Goal: Combine equations to cancel out one variable.)

+

2x 3y 2

5x 7y 34

10x 15y 10

10x 14y 68

29y 58

y 2 2x 3(2) 2

2x

Solution :{(

6 2

2x

4,2)

8

}

x 4

Page 69: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Method 3 Summary: Method 3 Summary: Elimination MethodElimination Method

or Addition/Subtraction Methodor Addition/Subtraction Method(Goal: Combine equations to (Goal: Combine equations to

cancel out one variable.)cancel out one variable.)

Disadvantages: Avoid this method if degrees and/or formats of the equations do not match.

Advantages:Similar to getting an LCD, so this is intuitive, and uses only integers until the end of the problem.

Page 70: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Three MethodsThree Methods

Method 1: Graphing MethodMethod 1: Graphing Method

Method 2: Substitution MethodMethod 2: Substitution Method

Method 3: Elimination MethodMethod 3: Elimination Method

or Addition/Subtraction or Addition/Subtraction

MethodMethod

2

x 3y 11 8x 3y 11 x y 1A. B. C.

7x 4y 6 7x 4y 6 7x 4y 6

Page 71: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

FactoringFactoring

1.1. Factor GCF Factor GCF for any # terms for any # terms

2.2. Difference of Squares Difference of Squares binomials binomials

3.3. Sum or Difference of Cubes Sum or Difference of Cubes binomialsbinomials

4.4. PST (Perfect Square Trinomial) PST (Perfect Square Trinomial) trinomialstrinomials

5.5. Reverse of FOIL Reverse of FOIL trinomials trinomials

6.6. Factor by Grouping Factor by Grouping usually for usually for 4 or more terms 4 or more terms

Page 72: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

GCFGCF3x2-9x23x2-9x2

3x2(1-3)3x2(1-3)

Take out what the two side Take out what the two side share in common to share in common to

simplify.simplify.

Page 73: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

GCFGCF

Hint: Remember that a “glob” can be part of your GCF.

3(1-6)2(1+3)

3(1-6)2+6(1-6)2

You can take out a glob and then combine with the other globs.

Page 74: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Difference of Difference of SquaresSquares

25x2-4x225x2-4x2

(5x-2)(5x+2)(5x-2)(5x+2)

Recall these binomials are called conjugates.

Page 75: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

IMPORTANT!IMPORTANT!

Remember that the Remember that the difference of squares difference of squares

factors into conjugates . . . factors into conjugates . . .

The SUM of squares is PRIME The SUM of squares is PRIME – cannot be factored.– cannot be factored.aa22 + b + b2 2 PRIME PRIME

aa22 –b –b22 (a + b)(a – b) (a + b)(a – b)

Page 76: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Sum/Difference of Sum/Difference of CubesCubes

x3-y3x3-y3

(x-y)(x2+xy+y2)(x-y)(x2+xy+y2)

Page 77: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

xx33 - y - y33

((x - yx - y) ( )) ( )Cube roots w/ original Cube roots w/ original sign in the middlesign in the middle

Sum/Difference of CubesSum/Difference of Cubes

Page 78: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Sum/Difference of Sum/Difference of CubesCubes

xx33 - y - y33

(x - y) ((x - y) (xx22 + + yy22))

Squares of those cube roots.Squares of those cube roots.

Note that squares will always be positive.Note that squares will always be positive.

Page 79: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Sum/Difference of Sum/Difference of CubesCubes

xx33 - y - y33

(x - y) (x(x - y) (x22 + xy+ xy + y + y22))The opposite of the productThe opposite of the product

of the cube rootsof the cube roots

Page 80: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

xx33 - 8 - 8(x - 2)(x - 2)

Cube roots of each Squares of those cube Cube roots of each Squares of those cube roots & roots &

with same sign opp of product of roots with same sign opp of product of roots in middlein middle

Sum/Difference of Sum/Difference of CubesCubes

+ 2x(x2 + 4)

Page 81: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Special Case- Special Case- * 1* 1stststep: Diff of step: Diff of Squares Squares * 2 * 2ndnd step: Sum/Diff of step: Sum/Diff of CubesCubesxx66 – 64y – 64y66

( ) ( ) ( ) ( ) ( )( ) ( )( ( )( ) ( )(

) )

Page 82: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Special CaseSpecial Case * 1* 1stststep: Diff of step: Diff of

Squares Squares * 2 * 2ndnd step: Sum/Diff of step: Sum/Diff of

CubesCubesxx66 – 64y – 64y66

(x(x33 – 8y – 8y33) (x) (x33 + 8y + 8y33) )

( )( ) ( )( ( )( ) ( )( ) )

Page 83: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Special CaseSpecial Case * 1* 1stststep: Diff of step: Diff of

Squares Squares * 2 * 2ndnd step: Sum/Diff of step: Sum/Diff of

CubesCubesxx66 – 64y – 64y66

(x(x33 – 8y – 8y33)) (x(x33 + 8y + 8y33))

(x–2y)(x(x–2y)(x22+2xy+4y+2xy+4y22)) (x+2y)(x(x+2y)(x22--2xy+4y2xy+4y22))

Page 84: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

PSTPST

x2-10x+25x2-10x+25

(x-5)(x-5)(x-5)(x-5)

(x-5)(x-5)22

Recall PST test: Recall PST test:

If 1st & 3rd terms are squares and If 1st & 3rd terms are squares and the middle term is twice the the middle term is twice the

product of their square roots.product of their square roots.

Page 85: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

PSTPST

xx22-24+16-24+16

(3x-4)(3x-4)(3x-4)(3x-4)

ConjugatesConjugates

Page 86: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Reverse FOIL Reverse FOIL (Trial & Error)(Trial & Error)

12x2-45x+4212x2-45x+42

(3x-6)(4x-7)(3x-6)(4x-7)

ConjugatesConjugates

Page 87: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Reverse FOILReverse FOIL(Trial & Error)(Trial & Error)

Hint: don’t forget to read the “signs”Hint: don’t forget to read the “signs”axax22 + bx + c + bx + c ( + )( + ) ( + )( + )axax22 – bx + c – bx + c ( – )( – ) ( – )( – )axax22 + bx – c + bx – c ( + )( – ) ( + )( – )

positive product has positive product has larger valuelarger value

axax22 – bx – c – bx – c ( + )( – ) ( + )( – ) negative product has negative product has

larger valuelarger value

Page 88: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Example 11: Factor Example 11: Factor by Groupingby Grouping

(4 or more terms)(4 or more terms)

a(x-y)+x-y(x-y)a(x-y)+x-y(x-y)

(x-y)(x-y)33(a)(a)

Page 89: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Example 11: Factor Example 11: Factor by Groupingby Grouping

(4 or more terms)(4 or more terms)

8x-2y+16x-4y8x-2y+16x-4y

2(4x-y)+4(4x-y)2(4x-y)+4(4x-y)

(4x-y)2(2+4)(4x-y)2(2+4)2X2

Page 90: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Factor by Grouping Factor by Grouping 3 X 13 X 1

xx22+xy+y+xy+y22-4x2-4x2

(x-y)(x-y)-4x(x-y)(x-y)-4x22

[(x-y)-2x][(x-y)-2x][(x-y)-2x][(x-y)-2x]

•3X1 (PST)

Page 91: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Factor by Grouping Factor by Grouping 3 X 13 X 1

x2-10+25-4x2

(x-5)(x-5)-4x2

[(x-5)-2x][(x-5)-2x]

Page 92: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Rational ExpressionsRational Expressions

Page 93: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Rational NumbersRational Numbers

Thinking back to when you were dealing with whole-number fractions, one Thinking back to when you were dealing with whole-number fractions, one of the first things you did was simplify them: You "cancelled off" factors of the first things you did was simplify them: You "cancelled off" factors which were in common between the numerator and denominator. You which were in common between the numerator and denominator. You could do this because dividing any number by itself gives you just "1", and could do this because dividing any number by itself gives you just "1", and you can ignore factors of "1".you can ignore factors of "1".

Using the same reasoning and methods, let's simplify some rational Using the same reasoning and methods, let's simplify some rational expressions.expressions.

Simplify the following expression:Simplify the following expression:    To simplify a numerical fraction, I would cancel off any common numerical To simplify a numerical fraction, I would cancel off any common numerical

factors. For this rational expression (this polynomial fraction), I can factors. For this rational expression (this polynomial fraction), I can similarly cancel off any common numerical similarly cancel off any common numerical or variableor variable factors. factors.

The numerator factors as (2)(The numerator factors as (2)(xx); the denominator factors as (); the denominator factors as (xx)()(xx). ). Anything divided by itself is just "1", so I can cross out any factors common Anything divided by itself is just "1", so I can cross out any factors common to both the numerator and the denominator. Considering the factors in this to both the numerator and the denominator. Considering the factors in this particular fraction, I get:particular fraction, I get:

Then the simplified form of the expression is: Then the simplified form of the expression is:     

Page 94: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Additoin and Subtraction of Additoin and Subtraction of Rational NumbersRational Numbers

    3           5               8               (2)(4)            23           5               8               (2)(4)            2       +              =              =                   =                +              =              =                   =          20         20             20              (4)(5)            5 20         20             20              (4)(5)            5

Notice the steps we have done to solve this Notice the steps we have done to solve this problem.  We first combined the numerators since problem.  We first combined the numerators since the denominators are the same.  Then we the denominators are the same.  Then we factored both the numerator and denominator factored both the numerator and denominator and finally we cross cancelled. and finally we cross cancelled.

Page 95: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Multiplication And Division of Multiplication And Division of Rational NumbersRational Numbers

  x2 - y2      x2 - y2                          is a rational expression.                    is a rational expression.(x - y)2  (x - y)2  

To simplify, we just factor and cancel:To simplify, we just factor and cancel:

(x - y)(x + y)            x  + y(x - y)(x + y)            x  + y                         =                                         =                    (x - y)2                  x - y    (x - y)2                  x - y

Page 96: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Quadratic EquationsQuadratic Equations

Page 97: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Quadratic EquationsQuadratic Equations

A quadratic equation is an equation that can A quadratic equation is an equation that can be written in this form. be written in this form.

ax2+bx+c=0 ax2+bx+c=0 The a,b, and c here represent real number The a,b, and c here represent real number

coefficients. So this means we are talking coefficients. So this means we are talking about an equation that is a constant times about an equation that is a constant times the variable squared plus a constant times the variable squared plus a constant times the variable plus a constant equals zero, the variable plus a constant equals zero, where the coefficient a on the variable where the coefficient a on the variable squared can't be zero, because if it were then squared can't be zero, because if it were then it would be a linear equation. it would be a linear equation.

Page 98: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Completing the SquareCompleting the Square

aa² + 2² + 2abab + + bb²=(²=(aa + + bb)².The technique is valid only when 1 is the )².The technique is valid only when 1 is the coefficient of coefficient of xx².².

  1)  Transpose the constant term to the right:1)  Transpose the constant term to the right:xx² + 6² + 6xx  =  −2  =  −2   2)  Add a square number to both sides.  Add the square of 2)  Add a square number to both sides.  Add the square of halfhalf the the

coefficient of coefficient of xx.  In this case, add the square of 3:.  In this case, add the square of 3:xx² + 6² + 6xx + 9  =  −2 + 9  =  −2 + 9.+ 9.

The left-hand side is now the perfect square of  (The left-hand side is now the perfect square of  (xx + 3). + 3). ((xx + 3)²  =  7. + 3)²  =  7. 3 is 3 is halfhalf of the coefficient 6. of the coefficient 6. This equation has the formThis equation has the form aa² = ² = bb   which implies a = ± .          Therefore,   which implies a = ± .          Therefore,xx + 3 = ±   + 3 = ±   xx = −3 ± . = −3 ± . That is, the solutions toThat is, the solutions to xx² + 6² + 6xx + 2  =  0 + 2  =  0 are the conjugate pair,are the conjugate pair, −−3 + ,  −3 − .3 + ,  −3 − .

Page 99: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Quadratic FormulaQuadratic Formula

Page 100: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Quadratic FormulaQuadratic Formula

on multiplying both on multiplying both cc and and aa by 4 by 4aa, thus making the , thus making the denominators the same (denominators the same (Lesson 23Lesson 23),),

This is the quadratic formula.This is the quadratic formula.    

Page 101: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

DiscriminantDiscriminant

The The radicandradicand    bb² − 4² − 4acac  is called the  is called the discriminant.  If the discriminant isdiscriminant.  If the discriminant is

a)   a)   PositivePositive:The roots are real and :The roots are real and conjugate. conjugate. 

b) b) NegativeNegative:  The roots are complex :  The roots are complex and conjugate. and conjugate. 

c)   c)   ZeroZero:The roots are rational and :The roots are rational and equal -- i.e. a double root.equal -- i.e. a double root.

Page 102: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

FactoringFactoring

Problem 2.Problem 2.   Find the roots of each    Find the roots of each quadratic by factoring.quadratic by factoring.

a)  a)  xx² − 3² − 3xx + 2 + 2 b)  b)  xx² + 7² + 7xx + + 1212

((xx − 1)( − 1)(xx − 2) − 2)   ((xx + 3)( + 3)(xx + 4) + 4) xx = 1  or  2. = 1  or  2.   xx = −3  or  −4. = −3  or  −4.

Page 103: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

In the next slides you will review:In the next slides you will review:

FunctionsFunctions D. Quadratic functions – D. Quadratic functions – explain everything we know about how explain everything we know about how to graph a parabolato graph a parabola

Page 104: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

FunctionsFunctions

A function is an operation on numbers of A function is an operation on numbers of some set (some set (domaindomain) that gives (calculates) ) that gives (calculates) one number for every number from the one number for every number from the domain. For example, function 3x is defined domain. For example, function 3x is defined for all numbers and its result is a number for all numbers and its result is a number multiplied by 3. The notation is y=f(x). x is multiplied by 3. The notation is y=f(x). x is called the called the argumentargument of the function, and y of the function, and y is the is the valuevalue of the function. The inverse of the function. The inverse function of f(x), is a different function of y function of f(x), is a different function of y that finds x that gives f(x) = y. that finds x that gives f(x) = y.

Page 105: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

FunctionsFunctions We have seen that a function is a special relation. In the same sense, real function is a special function. We have seen that a function is a special relation. In the same sense, real function is a special function.

The special about real function is that its domain and range are subsets of real numbers “R”. In The special about real function is that its domain and range are subsets of real numbers “R”. In mathematics, we deal with functions all the time – but with a difference. We drop the formal notation, mathematics, we deal with functions all the time – but with a difference. We drop the formal notation, which involves its name, specifications of domain and co-domain, direction of relation etc. Rather, we work which involves its name, specifications of domain and co-domain, direction of relation etc. Rather, we work with the rule alone. For example, with the rule alone. For example,

f(x) =x2+2x+3f(x) =x2+2x+3 This simplification is based on the fact that domain, co-domain and range are subsets of real numbers. In This simplification is based on the fact that domain, co-domain and range are subsets of real numbers. In

case, these sets have some specific intervals other than “R” itself, then we mention the same with a case, these sets have some specific intervals other than “R” itself, then we mention the same with a semicolon (;) or a comma(,) or with a combination of them : semicolon (;) or a comma(,) or with a combination of them :

f(x) =\f(x) =\(x+1) 2−1;x<−2,x≥0(x+1) 2−1;x<−2,x≥0 Note that the interval “ Note that the interval “ x<−2,x≥0x<−2,x≥0 ” specifies a subset of real number and defines the domain of function. ” specifies a subset of real number and defines the domain of function.

In general, co-domain of real function is “R”. In some cases, we specify domain, which involves exclusion of In general, co-domain of real function is “R”. In some cases, we specify domain, which involves exclusion of certain value(s), like : certain value(s), like :

f(x) =f(x) = 11−x,x≠111−x,x≠1 This means that domain of the function is This means that domain of the function is R−{1{ R−{1{ . Further, we use a variety of ways to denote a subset of . Further, we use a variety of ways to denote a subset of

real numbers for domain and range. Some of the examples are : real numbers for domain and range. Some of the examples are : x>1:x>1: denotes subset of real number greater than “1”. denotes subset of real number greater than “1”. R−{0,1{ R−{0,1{ denotes subset of real number that excludes integers “0” and “1”. denotes subset of real number that excludes integers “0” and “1”. 1<x<2:1<x<2: denotes subset of real number between “1” and “2” excluding end points. denotes subset of real number between “1” and “2” excluding end points. (1,2]: (1,2]: denotes subset of real number between “1” and “2” excluding end point “1”, but denotes subset of real number between “1” and “2” excluding end point “1”, but

including end point “2”.including end point “2”. Further, we may emphasize the meaning of following inequalities of real numbers as the same will be used Further, we may emphasize the meaning of following inequalities of real numbers as the same will be used

frequently for denoting important segment of real number line : frequently for denoting important segment of real number line : Positive number means x > 0 (excludes “0”). Positive number means x > 0 (excludes “0”). Negative number means x < 0 (excludes “0”). Negative number means x < 0 (excludes “0”). Non - negative number means x ≥ 0 (includes “0”). Non - negative number means x ≥ 0 (includes “0”). Non – positive number means x ≤ 0 (includes “0”).Non – positive number means x ≤ 0 (includes “0”).

Page 106: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Quadratic FunctionsQuadratic Functions

A A quadratic functionquadratic function is one of the form is one of the form f(x) = ax2 + bx + cf(x) = ax2 + bx + c, , where where aa,, b b, and , and cc are numbers with are numbers with aa not equal to zero. not equal to zero.

The graph of a quadratic function is a curve called a The graph of a quadratic function is a curve called a parabolaparabola. . Parabolas may open upward or downward and vary in "width" or Parabolas may open upward or downward and vary in "width" or "steepness", but they all have the same basic "U" shape. All "steepness", but they all have the same basic "U" shape. All parabolas are symmetric with respect to a line called the parabolas are symmetric with respect to a line called the axis of axis of symmetrysymmetry. A parabola intersects its axis of symmetry at a point . A parabola intersects its axis of symmetry at a point called the called the vertexvertex of the parabola. of the parabola.

You know that two points determine a line. This means that if you You know that two points determine a line. This means that if you are given any two points in the plane, then there is one and only are given any two points in the plane, then there is one and only one line that contains both points. A similar statement can be one line that contains both points. A similar statement can be made about points and quadratic functions.made about points and quadratic functions.

Given three points in the plane that have different first Given three points in the plane that have different first coordinates and do not lie on a line, there is exactly one quadratic coordinates and do not lie on a line, there is exactly one quadratic function f whose graph contains all three points. To graph simplify function f whose graph contains all three points. To graph simplify (see quadratic equations) and plot the points.(see quadratic equations) and plot the points.

Page 107: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Simplifying expressions with Simplifying expressions with exponentsexponents

Page 108: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Simplifying ExponentsSimplifying Exponents

Use the Power of a Power Property, Use the Power of a Power Property, the Product of a Power Property, the the Product of a Power Property, the Quotient of a Power Property, the Quotient of a Power Property, the Power of a Quotient Property, the Power of a Quotient Property, the

Page 109: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Simplifying exponentsSimplifying exponents

Use These(4

Page 110: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Simplifying ExponentsSimplifying Exponents

3322 3 31515=3=31717

6

4

8

828

2 4 8( )x x

2 2 2(4 5) 4 5

2 2

2

4 4

5 5

0

0

1

7 1

a

22

1 16

6 36

Page 111: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Simplifying expressions with Simplifying expressions with radicalsradicals

Page 112: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Simplifying radicalsSimplifying radicals

When presented with a problem like , we don’t have too much difficulty saying that the When presented with a problem like , we don’t have too much difficulty saying that the answer 2 (since 2x2=4 ). Our trouble usually occurs when we either can’t easily see the answer answer 2 (since 2x2=4 ). Our trouble usually occurs when we either can’t easily see the answer or if the number under our radical sign is not a perfect square or a perfect cube. or if the number under our radical sign is not a perfect square or a perfect cube.

A problem like may look difficult because there are no two numbers that multiply together A problem like may look difficult because there are no two numbers that multiply together to give 24. However, the problem can be simplified. So even though 24 is not a perfect square, it to give 24. However, the problem can be simplified. So even though 24 is not a perfect square, it can be broken down into smaller pieces where one of those pieces might be perfect square. So can be broken down into smaller pieces where one of those pieces might be perfect square. So now we have now we have

Simplifying a radical expression can also involve variables as well as numbers. Just as you were Simplifying a radical expression can also involve variables as well as numbers. Just as you were able to break down a number into its smaller pieces, you can do the same with variables. When able to break down a number into its smaller pieces, you can do the same with variables. When the radical is a square root, you should try to have terms raised to an even power (2, 4, 6, 8, the radical is a square root, you should try to have terms raised to an even power (2, 4, 6, 8, etc). When the radical is a cube root, you should try to have terms raised to a power of three (3, etc). When the radical is a cube root, you should try to have terms raised to a power of three (3, 6, 9, 12, etc.). For example, 6, 9, 12, etc.). For example,

4

24

24 4 6 2 6

3 2x x x x x http://www.algebralab.org/lessons/lesson.aspx?file=Algebra_radical_simplify.xml

Page 113: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Simplifying RadicalsSimplifying Radicals

Use the root of a power, power of a Use the root of a power, power of a root, product of a root, and quotient root, product of a root, and quotient of a root properties to solve.of a root properties to solve.

Page 114: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

In the next slides you will In the next slides you will review:review:

Minimum of four word problems Minimum of four word problems of various types. You can mix of various types. You can mix these in among the topics these in among the topics above or put them all together above or put them all together in one section. (Think what in one section. (Think what types you expect to see on types you expect to see on your final exam.)your final exam.)

Page 115: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Suppose that it takes Janet 6 hours to paint Suppose that it takes Janet 6 hours to paint her room if she works alone and it takes her room if she works alone and it takes

Carol 4 hours to paint the same room if she Carol 4 hours to paint the same room if she works alone. How long will it take them to works alone. How long will it take them to

paint the room if they work together?paint the room if they work together? First, we will let First, we will let xx be the amount of time it takes to paint the be the amount of time it takes to paint the

room (in hours) if the two work together. room (in hours) if the two work together.    Janet would need 6 hours if she did the entire job by herself, Janet would need 6 hours if she did the entire job by herself,

so her working rate is of the job in an hour. Likewise, so her working rate is of the job in an hour. Likewise, Carol’s rate is of the job in an hour. Carol’s rate is of the job in an hour.

   In In xx hours, Janet paints of the room and Carol paints hours, Janet paints of the room and Carol paints

of the room. Since the two females will be working of the room. Since the two females will be working together, we will add the two parts together. The sum equals together, we will add the two parts together. The sum equals one complete job and gives us the following equation: one complete job and gives us the following equation:

  

1

6

Click to see answer.

1

41

6 6

xx 1

4 4

xx

16 4

x x

Page 116: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Continued answerContinued answer

16 4

12 12 12 16 4

2 3 12

5 12

12 22

5 5

x x

x x

x x

x

x hours

Multiply each term Multiply each term of the equation by of the equation by the common the common denominator 12 denominator 12

Simplify

Collect like terms

Solve for x

Page 117: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Suppose Kirk has taken three tests and made 88, 90, and 84. Suppose Kirk has taken three tests and made 88, 90, and 84. Kirk’s teacher tells the class that each test counts the same Kirk’s teacher tells the class that each test counts the same amount. Kirk wants to know what he needs to make on the amount. Kirk wants to know what he needs to make on the fourth test to have an overall average of 90 so he can make fourth test to have an overall average of 90 so he can make

an A in the class.an A in the class.

88 90 8490

488 90 84 360

262 360

98

88 90 84 98 360: 90

4 4

x

x

x

x

Check

Steps

Page 118: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Suppose a bank is offering its customers 3% Suppose a bank is offering its customers 3% interest on savings accounts. If a customer interest on savings accounts. If a customer deposits $1500 in the account, how much deposits $1500 in the account, how much

interest does the customer earn in 5 years?interest does the customer earn in 5 years? I I is the amount of interest the is the amount of interest the

account earns. account earns. PP is the principle or the amount of is the principle or the amount of

money that is originally put into an money that is originally put into an account. account.

rr is the interest rate and must is the interest rate and must ALWAYS be in a decimal form rather ALWAYS be in a decimal form rather than a percent. than a percent.

tt is the amount of time the money is the amount of time the money is in the account earning interest. is in the account earning interest.

  

(1500)(.03)(5)

225

I P r t

I

I

If we want to find out the total amount in the account, we would need to add the interest to the original amount. In this case, there would be $1725 in the account

Page 119: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

The Smith’s have a rectangular pool that measure The Smith’s have a rectangular pool that measure 12 feet by 20 feet. They are building a walkway 12 feet by 20 feet. They are building a walkway

around it of uniform width.around it of uniform width. The length of the larger rectangle is , which simplifies toThe length of the larger rectangle is , which simplifies to    Length larger rectangle = Length larger rectangle =    The width of the larger rectangle is , which simplifies toThe width of the larger rectangle is , which simplifies to    Width larger rectangle =Width larger rectangle =    The area for the larger rectangle then becomesThe area for the larger rectangle then becomes    Area larger rectangle = Area larger rectangle =    The pool itself has an area of square feetThe pool itself has an area of square feet

Page 120: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Answer continuedAnswer continued

Rearrange the terms for Rearrange the terms for easier multiplication and easier multiplication and find the sum of 68 and 240. find the sum of 68 and 240.

Multiply the binomials. Multiply the binomials. Combine like terms and Combine like terms and

subtract 308 from each subtract 308 from each side. side.

Factor. Factor. Solve each factor. Solve each factor. Since dimensions of a pool Since dimensions of a pool

and a walkway around a and a walkway around a pool cannot be negativepool cannot be negativeour answer is that the width our answer is that the width of the walkway is 1 foot. of the walkway is 1 foot.

(20 2 )(12 2 ) 68 240

(2 20)(2 12) 308

4 2 24 40 240 308

4 2 64 68 0

4( 17)( 1) 0

17; 1

1

x x

x x

x x x

x x

x x

x x

x

Page 121: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Line of Best Fit or Regression Line of Best Fit or Regression LineLine

Page 122: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Line of best fitLine of best fit

A A line of best fitline of best fit is a straight line  is a straight line that best represents the data on a that best represents the data on a scatter plot.  scatter plot.  This line may pass through some of This line may pass through some of the points, none of the points, or all the points, none of the points, or all of the points.of the points.

Page 123: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Line of Best FitLine of Best Fit

You can find the line of best fit by You can find the line of best fit by estimation or by using graphing estimation or by using graphing calculators.calculators.

The line of best fit is good for The line of best fit is good for estimating the average of the points estimating the average of the points on the graph.on the graph.

Page 124: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Line of best fit (how to Line of best fit (how to solve)solve)

1. Separate the data into three groups of equal size according to the values of the

horizontal coordinate. 2. Find the summary point for each group based on the median x-value and the median y-value. 3. Find the equation of the line (Line L) through the summary points of the outer groups. 4. Slide L one-third of the way to the middle summary point. a. Find the y-coordinate of the point on L with the same x-coordinate as the middle summary point. b. Find the vertical distance between the middle summary point and the line by subtracting y-values. c. Find the coordinates of the point P one-third of the way from the line L to the middle summary point. 5. Find the equation of the line through the point P that is parallel to line L.

Page 125: Algebra 1 A Review and Summary Gabriel Grahek. In the next slides you will review: Solving 1st power equations in one variable A. Special cases where

Line of Best FitLine of Best Fit

Try to find the line of best fit for the Try to find the line of best fit for the points (3,9)(4,8)(6,6)(7,5)(9,7)(11,9)points (3,9)(4,8)(6,6)(7,5)(9,7)(11,9)(13,12)(14,17)(13,19).(13,12)(14,17)(13,19).