Holt McDougal Algebra 1

1-7 Solving Absolute-Value Equations1-7 Solving Absolute-Value Equations

Holt Algebra 1

Lesson QuizLesson Quiz

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Holt McDougal Algebra 1

Holt McDougal Algebra 1

1-7 Solving Absolute-Value Equations

Solve equations in one variable that contain absolute-value expressions.

Objectives

Holt McDougal Algebra 1

1-7 Solving Absolute-Value Equations

Holt McDougal Algebra 1

1-7 Solving Absolute-Value Equations

Additional Example 1A: Solving Absolute-Value Equations

Solve the equation.

|x| = 12|x| = 12

Case 1 x = 12

Case 2 x = –12

The solutions are {12, –12}.

Think: What numbers are 12 units from 0?

Rewrite the equation as two cases.

12 units 12 units

10 8 6 4 0 2 4 6 8 1012 2 12•••

Holt McDougal Algebra 1

1-7 Solving Absolute-Value Equations

Solve the equation.

Check It Out! Example 1a

|x| – 3 = 4|x| – 3 = 4

+ 3 +3|x| = 7

Case 1 x = 7

Case 2 x = –7

The solutions are {7, –7}.

Since 3 is subtracted from |x|, add 3 to both sides.

Think: What numbers are 7 units from 0?

Rewrite the equation as two cases.

Holt McDougal Algebra 1

1-7 Solving Absolute-Value Equations

Solve the equation.

Check It Out! Example 1b

8 =|x 2.5| Think: What numbers are

8 units from 0?

Case 18 = x 2.5

+2.5 +2.5

10.5 = x

+2.5 +2.55.5 = x

Case 2 8 = x 2.5

Rewrite the equations as two cases.

The solutions are {10.5, –5.5}.

8 =|x 2.5|

Since 2.5 is subtracted from x add 2.5 to both sides of each equation.

Holt McDougal Algebra 1

1-7 Solving Absolute-Value EquationsAdditional Example 1B: Solving Absolute-Value Equations

3|x + 7| = 24

|x + 7| = 8

The solutions are {1, –15}.

Case 1 x + 7 = 8

Case 2 x + 7 = –8

– 7 –7 – 7 – 7x = 1 x = –15

Since |x + 7| is multiplied by 3, divide both sides by 3 to undo the multiplication.

Think: What numbers are 8 units from 0?

Rewrite the equations as two cases. Since 7 is added to x subtract 7 from both sides of each equation.

Solve the equation.

Holt McDougal Algebra 1

1-7 Solving Absolute-Value EquationsAdditional Example 2A: Special Cases of Absolute-Value

Equations

Solve the equation. 8 = |x + 2| 8

8 = |x + 2| 8+8 + 8

0 = |x + 2|

0 = x + 22 22 = x

Since 8 is subtracted from |x + 2|, add 8 to both sides to undo the subtraction.

There is only one case. Since 2 is added to x, subtract 2 from both sides to undo the addition.

The solution is {2}.

Holt McDougal Algebra 1

1-7 Solving Absolute-Value EquationsAdditional Example 2B: Special Cases of Absolute-Value

Equations

Solve the equation.

3 + |x + 4| = 0

3 + |x + 4| = 03 3

|x + 4| = 3

Since 3 is added to |x + 4|, subtract 3 from both sides to undo the addition.

Absolute value cannot be negative.

This equation has no solution.

Holt McDougal Algebra 1

1-7 Solving Absolute-Value Equations

Remember!

Absolute value must be nonnegative because it represents a distance.

Holt McDougal Algebra 1

1-7 Solving Absolute-Value EquationsCheck It Out! Example 2a

Solve the equation.

2 |2x 5| = 7

2 |2x 5| = 7 2 2

|2x 5| = 5

Since 2 is added to –|2x – 5|, subtract 2 from both sides to undo the addition.

Absolute value cannot be negative.

|2x 5| = 5

This equation has no solution.

Since |2x – 5| is multiplied by negative 1, divide both sides by negative 1.

Holt McDougal Algebra 1

1-7 Solving Absolute-Value Equations

Check It Out! Example 2b

Solve the equation.

6 + |x 4| = 6

6 + |x 4| = 6+6 +6

|x 4| = 0

x 4 = 0+ 4 +4

x = 4

Since –6 is added to |x 4|, add 6 to both sides.

There is only one case. Since 4 is subtracted from x, add 4 to both sides to undo the addition.

Holt McDougal Algebra 1

1-7 Solving Absolute-Value EquationsLesson Quiz

Solve each equation.

1. 15 = |x| 2. 2|x – 7| = 14

3. |x + 1|– 9 = –9 4. |5 + x| – 3 = –2

5. 7 + |x – 8| = 6

–15, 15 0, 14

–1 –6, –4

no solution

6. Inline skates typically have wheels with a diameter of 74 mm. The wheels are manufactured so that the diameters vary from this value by at most 0.1 mm. Write and solve an absolute-value equation to find the minimum and maximum diameters of the wheels. |x – 74| = 0.1; 73.9 mm; 74.1 mm