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EXAMPLE 1 Solve a simple absolute value equation
Solve |x – 5| = 7. Graph the solution.
SOLUTION
| x – 5 | = 7
x – 5 = – 7 or x – 5 = 7
x = 5 – 7 or x = 5 + 7
x = –2 or x = 12
Write original equation.
Write equivalent equations.
Solve for x.
Simplify.
EXAMPLE 1
The solutions are –2 and 12. These are the values of x that are 7 units away from 5 on a number line. The graph is shown below.
ANSWER
Solve a simple absolute value equation
EXAMPLE 2 Solve an absolute value equation
| 5x – 10 | = 45
5x – 10 = 45 or 5x –10 = – 45
5x = 55 or 5x = – 35
x = 11 or x = – 7
Write original equation.
Expression can equal 45 or – 45 .
Add 10 to each side.
Divide each side by 5.
Solve |5x – 10 | = 45.
SOLUTION
EXAMPLE 2 Solve an absolute value equation
The solutions are 11 and –7. Check these in the original equation.
ANSWER
Check:| 5x – 10 | = 45
| 5(11) – 10 | = 54?
|45| = 45?
45 = 45
| 5x – 10 | = 45
| 5(– 7 ) – 10 | = 54?
45 = 45
| – 45| = 45?
EXAMPLE 3
| 2x + 12 | = 4x
2x + 12 = 4x or 2x + 12 = – 4x
12 = 2x or 12 = – 6x
6 = x or –2 = x
Write original equation.
Expression can equal 4x or – 4 x
Add – 2x to each side.
Solve |2x + 12 | = 4x. Check for extraneous solutions.
SOLUTION
Solve for x.
Check for extraneous solutions
EXAMPLE 3
| 2x + 12 | = 4x
| 2(– 2) +12 | = 4(–2)?
|8| = – 8?
8 = –8
Check the apparent solutions to see if either is extraneous.
Check for extraneous solutions
| 2x + 12 | = 4x
| 2(6) +12 | = 4(6)?
|24| = 24?
24 = 24
The solution is 6. Reject – 2 because it is an extraneous solution.
ANSWER
CHECK
GUIDED PRACTICE
Solve the equation. Check for extraneous solutions.
1. | x | = 5
SOLUTION
| x | = 5
| x | = – 5 or | x | = 5
x = –5 or x = 5
Write original equation.
Write equivalent equations.
Solve for x.
for Examples 1, 2 and 3
GUIDED PRACTICE for Examples 1, 2 and 3
The solutions are –5 and 5. These are the values of x that are 5 units away from 0 on a number line. The graph is shown below.
ANSWER
– 3
– 4
– 2
– 1
0
1 2
3
4
5
6
7
– 5
– 6
– 7
5 5
GUIDED PRACTICE
Solve the equation. Check for extraneous solutions.
2. |x – 3| = 10
SOLUTION
| x – 3 | = 10
x – 3 = – 10 or x – 3 = 10
x = 3 – 10 or x = 3 + 10
x = –7 or x = 13
Write original equation.
Write equivalent equations.
Solve for x.
Simplify.
for Examples 1, 2 and 3
GUIDED PRACTICE
The solutions are –7 and 13. These are the values of x that are 10 units away from 3 on a number line. The graph is shown below.
ANSWER
– 3
– 4
– 2
– 1
0
1
2
3
4
5
6
7
– 5
– 6
– 7
8
9
10
11
12
13
10 10
for Examples 1, 2 and 3
GUIDED PRACTICE
Solve the equation. Check for extraneous solutions.
SOLUTION
| x + 2 | = 7
x + 2 = – 7 or x + 2 = 7
x = – 7 – 2 or x = 7 – 2
x = –9 or x = 5
Write original equation.
Write equivalent equations.
Solve for x.
Simplify.
3. |x + 2| = 7
for Examples 1, 2 and 3
GUIDED PRACTICE
The solutions are –9 and 5. These are the values of x that are 7 units away from – 2 on a number line.
ANSWER
for Examples 1, 2 and 3
GUIDED PRACTICE
Solve the equation. Check for extraneous solutions.
4. |3x – 2| = 13
SOLUTION
3x – 2 = 13 or 3x – 2 = – 13
Write original equation.
Solve for x.
Simplify.
|3x – 2| = 13
Write equivalent equations.
x = or x = 5 3–3 2
for Examples 1, 2 and 3
x =
–13 + 23
or x =
13 + 23
GUIDED PRACTICE
The solutions are – and 5.
ANSWER
332
for Examples 1, 2 and 3
GUIDED PRACTICE
Solve the equation. Check for extraneous solutions.
5. |2x + 5| = 3x
| 2x + 5 | = 3x
2x + 5 = – 3x or 2x + 5 = 3x
x = 1 or x = 5
Write original equation.
Write Equivalent equations.
Simplify
SOLUTION
for Examples 1, 2 and 3
2x + 3x = 5 or 2x – 3x = –5
GUIDED PRACTICE
The solution of is 5. Reject 1 because it is an extraneous solution.
ANSWER
for Examples 1, 2 and 3
| 2x + 12 | = 4x
| 2(1) +12 | = 4(1)?
|14| = 4?
14 = –8
Check the apparent solutions to see if either is extraneous.
| 2x + 5 | = 3x
| 2(5) +5 | = 3(5)?
|15| = 15?
15 = 15
CHECK
GUIDED PRACTICE
Solve the equation. Check for extraneous solutions.
6. |4x – 1| = 2x + 9
SOLUTION
4x – 1 = – (2x + 9) or 4x – 1 = 2x + 9
Write original equation.
Solve For xx = or x = 5 311
|4x – 1| = 2x + 9
Write equivalent equations.
for Examples 1, 2 and 3
4x + 2x = – 9 + 1 or 4x – 2x = 9 + 1 Rewrite equation.
GUIDED PRACTICE
ANSWER
The solutions are – and 5. 311
for Examples 1, 2 and 3
| 4x – 1 | = 2x + 9
| 4(5) – 1 | = 2(5) + 9?
|19| = 19?
19 = 19
Check the apparent solutions to see if either is extraneous.
| 4x – 1 | = 2x + 9
| 4( ) – 1 | = 2( ) + 93–11
3–11 ?
CHECK
| | = ? 3–
19319
= 3– 19
319
