Solve linear equations using a variety of methods.Solve linear inequalities.

2-1Objectives

equationsolution set of an equationlinear equation in one variableidentifycontradictioninequality

Vocabulary

An equation is a mathematical statement that two expressions are equivalent. The solution set of an equation is the value or values of the variable that make the equation true. A linear equation in one variable can be written in the form ax = b, where a and b are constants and a ≠ 0.

Linear Equations in One variable

Nonlinear Equations

4x = 8

3x – = –9

2x – 5 = 0.1x +2

+ 1 = 32 + 1 = 41

3 – 2x = –5

The local phone company charges $12.95 a month for the first 200 of air time, plus $0.07 for each additional minute. If Nina’s bill for the month was $14.56, how many additional minutes did she use?

Example 1: Consumer Application

Solve.

12.95 + 0.07m = 14.56

0.07m = 1.610.07 0.07

m = 23

Nina used 23 additional minutes.

Example 1

–12.95 –12.95

Check It Out! Example 1

Stacked cups are to be placed in a pantry. One cup is 3.25 in. high and each additional cup raises the stack 0.25 in. How many cups fit between two shelves 14 in. apart?

Check It Out! Example 1 Continued

3.25 + 0.25c = 14.00

0.25c = 10.750.25 0.25

c = 43

44 cups fit between the 14 in. shelves.

Solve.

–3.25 –3.25

Example 2: Solving Equations with the Distributive Property

Solve 4(m + 12) = –36

Method 1The quantity (m + 12) is multiplied by 4, so divide by 4 first.

4(m + 12) = –364 4

m + 12 = –9

m = –21 –12 –12

Check 4(m + 12) = –364(–21 + 12) –36

4(–9) –36–36 –36

Example 2 Continued

Example 2 Continued

Distribute before solving. 4m + 48 = –36

4m = –84 –48 –48

=4m –84 4 4

m = –21

Solve 4(m + 12) = –36

Method 2

If there are variables on both sides of the equation, (1) simplify each side. (2) collect all variable terms on one side and all constants terms on the other side. (3) isolate the variables as you did in the previous problems.

Example 3: Solving Equations with Variables on Both Sides

–11k + 25 = –6k – 10+11k +11k

25 = 5k – 10

35 = 5k5 5

7 = k

+10 + 10

Solve 3k– 14k + 25 = 2 – 6k – 12.

You have solved equations that have a single solution. Equations may also have infinitely many solutions or no solution.An equation that is true for all values of the variable, such as x = x, is an identity. An equation that has no solutions, such as 3 = 5, is a contradiction because there are no values that make it true.

Solve 3v – 9 – 4v = –(5 + v).

Example 4A: Identifying Identities and Contractions

3v – 9 – 4v = –(5 + v) –9 – v = –5 – v + v + v

–9 ≠ –5 x Contradiction

The equation has no solution. The solution set is the empty set, which is represented by the symbol .

Solve 2(x – 6) = –5x – 12 + 7x.

Example 4B: Identifying Identities and Contractions

2(x – 6) = –5x – 12 + 7x 2x – 12 = 2x – 12

–2x –2x–12 = –12 Identity

The solutions set is all real number, or .

An inequality is a statement that compares two expressions by using the symbols <, >, ≤, ≥, or ≠. The graph of an inequality is the solution set, the set of all points on the number line that satisfy the inequality.The properties of equality are true for inequalities, with one important difference. If you multiply or divide both sides by a negative number, you must reverse the inequality symbol.

These properties also apply to inequalities expressed with >, ≥, and ≤.

To check an inequality, test• the value being compared with x • a value less than that, and• a value greater than that.

Helpful Hint

Solve and graph 8a –2 ≥ 13a + 8.

Example 5: Solving Inequalities

8a – 2 ≥ 13a + 8–13a –13a –5a – 2 ≥ 8 +2 +2

–5 –5 –5a ≤ 10

a ≤ –2

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1

•

Example 5 Continued

Check Test values in the original inequality. –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

•

Test x = –4 Test x = –2 Test x = –18(–4) – 2 ≥ 13(–4) + 8 8(–2) – 2 ≥ 13(–2) + 8 8(–1) – 2 ≥ 13(–1) + 8

–34 ≥ –44

So –4 is a solution.

So –1 is not a solution.

So –2 is a solution.

–18 ≥ –18 –10 ≥ –5 x

Solve and graph 8a – 2 ≥ 13a + 8.

Lesson Quiz: Part I

1. Alex pays $19.99 for cable service each month. He also pays $2.50 for each movie he orders through the cable company’s pay-per-view service. If his bill last month was $32.49, how many movies did Alex order?5 movies

Apply proportional relationships to rates, similarity, and scale.

2-2Objective

ratioproportionratesimilarindirect measurement

Vocabulary

Recall that a ratio is a comparison of two numbers by division and a proportion is an equation stating that two ratios are equal. In a proportion, the cross products are equal.

If a proportion contains a variable, you can cross multiply to solve for the variable. When you set the cross products equal, you create a linear equation that you can solve by using the skills that you learned in Lesson 2-1.

In a ÷ b = c ÷ d, b and c are the means, and a and d are the extremes. In a proportion, the product of the means is equal to the product of the extremes.

Reading Math

Solve each proportion.

Example 1: Solving Proportions

A.

206.4 = 24p

=

=16 24 p 12.9

16 24 p 12.9

206.4 24p 24 24

8.6 = p

14 c 88 132 =

=

=

B.

14 c 88 132

88c = 1848

=88c 184888 88

c = 21

Percent is a ratio that means per hundred.For example:30% = 0.30 =

Remember!

30100

Because percents can be expressed as ratios, you can use the proportion

to solve percent problems.

A poll taken one day before an election showed that 22.5% of voters planned to vote for a certain candidate. If 1800 voters participated in the poll, how many indicated that they planned to vote for that candidate?

Example 2: Solving Percent Problems

You know the percent and the total number of voters, so you are trying to find the part of the whole (the number of voters who are planning to vote for that candidate).

Example 2 Continued

Method 1 Use a proportion.

So 405 voters are planning to vote for that candidate.

Method 2 Use a percent equation.22.5% 0.225 Divide the percent

by 100.

Percent (as decimal) whole = part0.225 1800 = x

405 = x

x = 405

22.5(1800) = 100x

At Clay High School, 434 students, or 35% of the students, play a sport. How many students does Clay High School have?

You know the percent and the total number of students, so you are trying to find the part of the whole (the number of students that Clay High School has).

Check It Out! Example 2

Check It Out! Example 2 ContinuedMethod 1 Use a proportion.

Cross multiply.

Solve for x.

Clay High School has 1240 students.

Method 2 Use a percent equation.

Divide the percent by 100.

0.35x = 434

35% = 0.35

x = 1240

Percent (as decimal) whole = part

x = 1240

100(434) = 35x

A rate is a ratio that involves two different units. You are familiar with many rates, such as miles per hour (mi/h), words per minute (wpm), or dollars per gallon of gasoline. Rates can be helpful in solving many problems.

Ryan ran 600 meters and counted 482 strides. How long is Ryan’s stride in inches? (Hint: 1 m ≈ 39.37 in.)

Example 3: Fitness Application

Use a proportion to find the length of his stride in meters.

600 m 482 strides

x m 1 stride=

600 = 482x

x ≈ 1.24 m

Write both ratios in the form .

metersstrides

Example 3: Fitness Application continued

Convert the stride length to inches.

Ryan’s stride length is approximately 49 inches.

is the conversion factor. 39.37 in.1 m

≈ 1.24 m1 stride length

39.37 in.1 m

49 in.1 stride length

Use a proportion to find the length of his stride in meters.

Check It Out! Example 3

Luis ran 400 meters in 297 strides. Find his stride length in inches.

x ≈ 1.35 m400 = 297x

400 m 297 strides

x m 1 stride= Write both ratios in the form .

metersstrides

Convert the stride length to inches.

Luis’s stride length is approximately 53 inches.

Check It Out! Example 3 Continued

is the conversion factor. 39.37 in.1 m

≈ 1.35 m1 stride length

39.37 in.1 m

53 in.1 stride length

Similar figures have the same shape but not necessarily the same size. Two figures are similar if their corresponding angles are congruent and corresponding sides are proportional.

The ratio of the corresponding side lengths of similar figures is often called the scale factor.

Reading Math

Example 4: Scaling Geometric Figures in the Coordinate Plane

∆XYZ has vertices X(0, 0), Y(–6, 9) and Z(0, 9).

∆XAB is similar to ∆XYZ with a vertex at B(0, 3).

Graph ∆XYZ and ∆XAB on the same grid.

Step 1 Graph ∆XYZ. Then draw XB.

x

yz

B

Example 4 Continued

= height of ∆XAB width of ∆XAB

height of ∆XYZ width of ∆XYZ

=3 x9 69x = 18

Step 2 To find the width of ∆XAB, use a proportion.

x = 2

Example 4 Continued

The width is 2 units, and the height is 3 units, so the coordinates of A are (–2, 3).

BA

X

YZ

Step 3 To graph ∆XAB, first find the coordinate of A.

Example 5: Nature ApplicationThe tree in front of Luka’s house casts a 6-foot shadow at the same time as the house casts a 22-foot shadow. If the tree is 9 feet tall, how tall is the house?Sketch the situation. The triangles formed by using the shadows are similar, so Luka can use a proportion to find h the height of the house.

=6 9 h

22=Shadow of tree

Height of treeShadow of house Height of house

6h = 198h = 33

The house is 33 feet high.

9 ft

6 ft

h ft

22 ft