1. 1-7 Solving Absolute-Value Equations Warm Up Lesson Presentation
Holt Algebra 1
Warm Up
Lesson Presentation
Lesson Quiz
Holt McDougal Algebra 1

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

4. Additional Example 1A: Solving Absolute-Value Equations
Solve the equation.
|x| = 12
Think: What numbers are 12 units from 0?
|x| = 12
12 units
10 8 6 4 2 4 6 8 10
12 2 12 •
Case 1
x = 12
Case 2
x = –12
Rewrite the equation as two cases.
The solutions are {12, –12}.

5. Check It Out! Example 1a
Solve the equation.
|x| – 3 = 4
Since 3 is subtracted from |x|, add 3 to both sides.
|x| – 3 = 4
|x| = 7
Think: What numbers are 7 units from 0?
Case 1
x = 7
Case 2
x = –7
Rewrite the equation as two cases.
The solutions are {7, –7}.

6. Check It Out! Example 1b
Solve the equation.
8 =|x  2.5|
Think: What numbers are 8 units from 0?
8 =|x  2.5|
Rewrite the equations as two cases.
Case 1
8 = x  2.5
Case 2
 8 = x  2.5
Since 2.5 is subtracted from x add 2.5 to both sides of each equation.
10.5 = x
5.5 = x
The solutions are {10.5, –5.5}.

7. Additional Example 1B: Solving Absolute-Value Equations
Solve the equation.
3|x + 7| = 24
Since |x + 7| is multiplied by 3, divide both sides by 3 to undo the multiplication.
Think: What numbers are 8 units from 0?
|x + 7| = 8
Case 1
x + 7 = 8
Case 2
x + 7 = –8
– 7 –7
– 7 – 7
x = 1
x = –15
Rewrite the equations as two cases. Since 7 is added to x subtract 7 from both sides of each equation.
The solutions are {1, –15}.

8. Additional Example 2A: Special Cases of Absolute-Value Equations
Solve the equation.
8 = |x + 2|  8
Since 8 is subtracted from |x + 2|, add 8 to both sides to undo the subtraction.
8 = |x + 2|  8
0 = |x + 2|
There is only one case. Since 2 is added to x, subtract 2 from both sides to undo the addition.
0 = x + 2
 2
2 = x
The solution is {2}.

9. Additional Example 2B: Special Cases of Absolute-Value Equations
Solve the equation.
3 + |x + 4| = 0
Since 3 is added to |x + 4|, subtract 3 from both sides to undo the addition.
3 + |x + 4| = 0
 3
|x + 4| = 3
Absolute value cannot be negative.
This equation has no solution.

10. Remember!
Absolute value must be nonnegative because it represents a distance.

11. Check It Out! Example 2a
Solve the equation.
2  |2x  5| = 7
Since 2 is added to –|2x – 5|, subtract 2 from both sides to undo the addition.
2  |2x  5| = 7
 2
 |2x  5| = 5
Since |2x – 5| is multiplied by negative 1, divide both sides by negative 1.
|2x  5| = 5
Absolute value cannot be negative.
This equation has no solution.

12. Check It Out! Example 2b
Solve the equation.
6 + |x  4| = 6
Since –6 is added to |x  4|, add 6 to both sides.
6 + |x  4| = 6
|x  4| = 0
x  4 = 0
x = 4
There is only one case. Since 4 is subtracted from x, add 4 to both sides to undo the addition.

13. Lesson Quiz
Solve each equation.
1. 15 = |x| |x – 7| = 14
3. |x + 1|– 9 = – |5 + x| – 3 = –2
|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