Copyright © 2014, 2010, 2006 Pearson Education, Inc. 1 Chapter 2 Linear Functions and Equations

Copyright © 2014, 2010, 2006 Pearson Education, Inc. 1

Chapter 2

Linear Functions and

Equations

2Copyright © 2014, 2010, 2006 Pearson Education, Inc.

Linear Equations

♦ Learn about equations and recognize a linear equation♦ Solve linear equations symbolically♦ Sole linear equations graphically and numerically♦ Solve problems involving percentages♦ Apply problem-solving strategies

2.2


Equations

An equation is a statement that two mathematical expressions are equal.

Some examples of equations are:

x 15 9x 1

x2 2x 12x

z 5 0

xy x2 y 3 x

1 3 4


Solutions to Equations

To solve an equation means to find all the values of the variable that make the equation a true statement.

Such values are called solutions.The set of all solutions is the solution set.Solutions to an equation satisfy the equation.Two equations are equivalent if they have the same solution set.


Types of Equations in One Variable

Contradiction – An equation for which there is no solution.• Example: 2x + 3 = 5 + 4x – 2x

•Simplifies to 2x + 3 = 2x + 5•Simplifies to 3 = 5 •FALSE statement – there are no values

of x for which 3 = 5. The equation has NO SOLUTION.



Identity – An equation for which every meaningful value of the variable is a solution.

• Example: 2x + 3 = 3 + 4x – 2x•Simplifies to 2x + 3 = 2x + 3•Simplifies to 3 = 3 •TRUE statement – no matter the value of

x, the statement 3 = 3 is true. The solution is ALL REAL NUMBERS



Conditional Equation – An equation that is satisfied by some, but not all, values of the variable.

• Example 1: 2x + 3 = 5 + 4x•Simplifies to 2x – 4x = 5 – 3•Simplifies to 2x = 2 •Solution of the equation is: x = 1

• Example 2: x2 = 1•Solutions of the equation are: x = 1, x = 1


Linear Equations in One Variable

A linear equation in one variable is an equation that can be written in the form

ax + b = 0,where a and b are real numbers with a ≠ 0.

If an equation is not linear, then we say it is a nonlinear equation.


Linear Equations in One Variable

Rules of algebra can be used to write any linear equation in the form ax + b = 0.A linear equation has exactly one solution

Examples of linear equations:

x

b

a.

x 12 0 2x 4 x

2 1 4x 16x x 5 3 x 1 0


Symbolic Solutions

Linear equations can be solved symbolically, and the solution is always exact.

To solve a linear equation symbolically, we usually apply the properties of equality to the given equation and transform it into an equivalent equation that is simpler.


Properties of Equality

Addition Property of EqualityIf a, b, and c are real numbers, then

a = b is equivalent to a + c = b + c.

Multiplication Property of EqualityIf a, b, and c are real numbers with c ≠ 0, then

a = b is equivalent to ac = bc.


Solve the equation 3(x – 4) = 2x – 1. Check your answer.

Solution

3 x 4 2x 1

Apply the distributive property

3x 12 2x 1

3x 2x 12 12 2x 2x 112

x 11

Example: Solving a linear equation symbolically


3 x 4 2x 1

The solution is 11. Check the answer.

?

3 4 2 111 11

2121

Example: Solving a linear equation symbolically


Solve the linear equation

Solution

2 1 1 5 3

4 3 12

tt t

To eliminate fractions, multiply each side (or term in the equation) by the LCD, 12.

Example: Eliminating fractions


12 t 2 4

12

3t 12 5 12

123 t

3 t 2 4t 60 3 t 3t 6 4t 60 3 t

t 6 57 t

2t 63

t

63

2

The solution

is

63

2.

Example: Eliminating fractions


Solve the linear equation.

Solution

To eliminate fractions, multiply each side (or term in the equation) by the LCD, 12.

0.03 3 0.5 2 1 0.23z z

Example: Eliminating decimals


To eliminate decimals, multiply each side (or term in the equation) by 100.

0.03 z 3 0.5 2z 1 0.23

3 z 3 50 2z 1 0.23

3z 9 100z 50 23

97z 59 23

97z 82

z

82

97

The solution

is

82

97.

Example: Eliminating decimals


Intersection-of-Graphs Method

The intersection-of-graphs method can be used to solve an equation graphically.

STEP 1: Set y1 equal to the left side of the equation, and set y2 equal to the right side of the equation.

STEP 2: Graph y1 and y2.STEP 3: Locate any points of intersection. The

x-coordinates of these points correspond to solutions to the equation.


Solve graphically and symbolically.

Solution

2x 1

1

2x 2

Graph

and y

2

1

2x 2

y12x 1

Intersect at (2, 3), solution is 2.

Example: Solving an equation graphically and symbolically


Solution is 2, agrees with graphical.

2x 1

1

2x 2

2x

1

2x 3

3

2x 3

2

33

2x

2

33

x 2

Example: Solving an equation graphically and symbolically


Solve numerically to the nearest tenth.

3 2x 1

3x 0

Enter Y

1 3 2x X / 3

Solution

Make a table for y1, incrementing by 1.

This will show the solution is located in the interval 1 < x < 2.

Example: Solving an equation numerically


Solution lies in 1.43 < x < 1.44

Make a table for y1, start at 1, increment by 0.1.

Make a table for y1, start at 1.4, increment by 0.01.

Solution lies in 1.4 < x < 1.5

Solution is 1.4 to the nearest tenth

Example: Solving an equation numerically


The area of a trapezoid with bases a and b and height h is given by

Solve this equation for b.

A

1

2h a b .

Multiply each side by 2

Solution

Divide each side by h

Subtract a from each side isolates b

2A bh a

2A

ha b

2A

h a b

Example: Solving for a variable


Solving Application Problems

STEP 1: Read the problem and make sure you understand it. Assign a variable to what you are being asked. If necessary, write other quantities in terms of the variable.

STEP 2: Write an equation that relates the quantities described in the problem. You may need to sketch a diagram and refer to known formulas.

STEP 3: Solve the equation and determine the solution.

STEP 4: Look back and check your solution. Does it seem reasonable?


In 1 hour an athlete traveled 10.1 miles by running first at 8 miles per hour and then at 11 miles per hour. How long did the athlete run at each speed?

STEP 1: Let x represent the time in hours running at 8 mph, then 1 – x represents the time spent running at 11 mph.

x: Time spent running at 11 miles per hour1 – x: Time spent running at 9 miles per hour

Solution

Example: Solving an application involving motion


STEP 2: d = rt; total distance is 10.1

d r1t1 r

2t2

10.1 8 11 1x x

3x 0.9

x 0.3

STEP 3: Solve symbolically

10.1 8 11 1x x



STEP 3: The athlete runs 0.3 hour (18 min) at 8 miles per hour and 0.7 hour (42 min) at 11 miles per hour.

This sounds reasonable. The runner’s average speed was 10.1 miles per hour so the runner must have run longer at 11 miles per hour than at 8 miles per hour.

STEP 4: check the solution

8 0.3 11 0.7 10.1



Pure water is being added to 153 milliliters of a 30% solution of hydrochloric acid. How much water should be added to dilute the solution to a 13% mixture?

STEP 1: Let x be the amount of pure water added to the 153 ml of 30% acid to make a 13% solution

x: Amount of pure water to be addedx + 153: Final volume of 13% solution

Solution

Example: Mixing acid in chemistry


STEP 2: Pure water contains no acid, so the amount of acid before the water is added equals the amount of acid after water is added. Pure acid before is 30% of 153, pure acid after is 13% of x + 153.The equation is:

0.13 153 0.30 153x



STEP 3: Solve: divide each side by 0.13

0.13 153 0.30 153x

x

0.30 153 0.13

153

x 200.08

x 153

0.30 153 0.13



STEP 3: We should add about 200 milliliters of water.

Concentration or ≈ 13%

STEP 4: Initially the solution contains 0.30(153) = 45.9 milliliters of pure acid. If we add 200 milliliters of water to the 153 milliliters, the final solution is 353 milliliters, which includes 45.9 milliliters of pure acid.

45.9

3530.13


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Copyright © 2014, 2010, 2006 Pearson Education, Inc. 1 Chapter 2 Linear Functions and Equations