Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

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

Learning Targets

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

equationsolution set of an equationlinear equation in one variableidentifycontradictioninequality

Vocabulary

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

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.

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Linear Equations in One variable

Nonlinear Equations

4x = 8

3x – = –9

2x – 5 = 0.1x +2

Notice that the variable in a linear equation is not under a radical sign and is not raised to a power other than 1. The variable is also not an exponent and is not in a denominator.

Solving a linear equation requires isolating the variable on one side of the equation by using the properties of equality.

+ 1 = 32

+ 1 = 41

3 – 2x = –5

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

To isolate the variable, perform the inverse or opposite of every operation in the equation on both sides of the equation. Do inverse operations in the reverse order of operations.

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

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

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Example 1 Continued

monthly charge plus

additional minute charge

times

12.95 0.07

number of additional minutes

total charge

+

=

Let m represent the number of additional minutes that Nina used.

m 14.56* =

Model

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Solve.

12.95 + 0.07m = 14.56

0.07m = 1.610.07 0.07

m = 23

Subtract 12.95 from both sides.

Divide both sides by 0.07.

Nina used 23 additional minutes.

Example 1 Continued

–12.95 –12.95

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

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?

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Check It Out! Example 1 Continued

Let c represent the number of additional cups needed.

one cup plusadditional

cup height

times

3.25 0.25

number of additional

cups

total height

+

=

c 14.00* =

Model

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Check It Out! Example 1 Continued

3.25 + 0.25c = 14.00

0.25c = 10.75

0.25 0.25

c = 43

44 cups fit between the 14 in. shelves.

Solve.

Subtract 3.25 from both sides.

Divide both sides by 0.25.

–3.25 –3.25

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Example 2: Solving Equations with the Distributive Property

Solve 4(m + 12) = –36

Divide both sides by 4.

Method 1

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

4(m + 12) = –364 4

m + 12 = –9

m = –21

–12 –12 Subtract 12 from both sides.

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Check 4(m + 12) = –36

4(–21 + 12) –36

4(–9) –36–36 –36

Example 2 Continued

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Example 2 Continued

Distribute 4.

Distribute before solving.

4m + 48 = –36

4m = –84

–48 –48 Subtract 48 from both sides.

Divide both sides by 4.=4m –84 4 4

m = –21

Solve 4(m + 12) = –36

Method 2

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Divide both sides by –3.

Method 1

The quantity (5 – 4r) is multiplied by –3, so divide by –3 first.

–3(5 – 4r) –9–3 –3

=

Check It Out! Example 2b

Solve –3(5 – 4r) = –9.

Subtract 5 from both sides.

–4r = –2

5 – 4r = 3 –5 –5

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Check It Out! Example 2b Continued

Divide both sides by –4.=–4 –4–4r –2

r =

–9 –9

Check –3(5 –4r) = –9

–3(5 – 4• ) –9

–3(5 – 2) –9

–3(3) –9

Solve –3(5 – 4r) = –9.

Method 1

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Distribute 3.

Distribute before solving.

–15 + 12r = –9

12r = 6

+15 +15Add 15 to both sides.

Divide both sides by 12.=12r 612 12

Check It Out! Example 2b Continued

r =

Solve –3(5 – 4r) = –9.

Method 2

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

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.

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Example 3: Solving Equations with Variables on Both Sides

Simplify each side by combining like terms.

–11k + 25 = –6k – 10

Collect variables on the right side.

Add.

Collect constants on the left side.

Isolate the variable.

+11k +11k

25 = 5k – 10

35 = 5k

5 5

7 = k

+10 + 10

Solve 3k– 14k + 25 = 2 – 6k – 12.

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

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.

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Solve 3v – 9 – 4v = –(5 + v).

Example 4A: Identifying Identities and Contractions

3v – 9 – 4v = –(5 + v)

Simplify.–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 .

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Solve 2(x – 6) = –5x – 12 + 7x.

Example 4B: Identifying Identities and Contractions

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

–2x –2x

–12 = –12 Identity

The solutions set is all real number, or .

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Solve 5(x – 6) = 3x – 18 + 2x.

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

Check It Out! Example 4a

5(x – 6) = 3x – 18 + 2x

Simplify.5x – 30 = 5x – 18

–5x –5x

–30 ≠ –18 x Contradiction

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Solve 3(2 –3x) = –7x – 2(x –3).

3(2 –3x) = –7x – 2(x –3)

Simplify.6 – 9x = –9x + 6

+ 9x +9x

6 = 6 Identity

The solutions set is all real numbers, or .

Check It Out! Example 4b

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

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.

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

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

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

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

Helpful Hint

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Solve and graph 8a –2 ≥ 13a + 8.

Example 5: Solving Inequalities

Subtract 13a from both sides.8a – 2 ≥ 13a + 8

–13a –13a

–5a – 2 ≥ 8Add 2 to both sides. +2 +2

–5a ≥ 10Divide both sides by –5 and reverse the inequality.

–5 –5

–5a ≤ 10

a ≤ –2

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

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 = –1

8(–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.

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Solve and graph x + 8 ≥ 4x + 17.

Subtract x from both sides.x + 8 ≥ 4x + 17

–x –x

8 ≥ 3x +17Subtract 17 from both sides.–17 –17

–9 ≥ 3xDivide both sides by 3.

3 3

–9 ≥ 3x

–3 ≥ x or x ≤ –3

Check It Out! Example 5

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Check Test values in the original inequality.

Test x = –6 Test x = –3 Test x = 0

–6 + 8 ≥ 4(–6) + 17 –3 +8 ≥ 4(–3) + 17 0 +8 ≥ 4(0) + 17

2 ≥ –7

So –6 is a solution.

So 0 is not a solution.

So –3 is a solution.

5 ≥ 5 8 ≥ 17 x

Check It Out! Example 5 Continued

Solve and graph x + 8 ≥ 4x + 17.

–6 –5 –4 –3 –2 –1 0 1 2 3

•

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

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

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Lesson Quiz: Part II

y = –4

x = 6

all real numbers, or

Solve.

2. 2(3x – 1) = 34

3. 4y – 9 – 6y = 2(y + 5) – 3

4. r + 8 – 5r = 2(4 – 2r)

5. –4(2m + 7) = (6 – 16m)

no solution, or

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Lesson Quiz: Part III

5. Solve and graph.

12 + 3q > 9q – 18 q < 5

–2 –1 0 1 2 3 4 5 6 7

°