114 CHAPTER 2 Linear Equations, Inequalities, and Applications
Solving an Absolute Value EquationSolve Graph the solution set.
For to equal 7, must be 7 units from 0 on the number line. This
can happen only when or This is Case 1 in the preceding
box. Solve this compound equation as follows.
Subtract 1.
Divide by 2.Check by substituting 3 and then into the original absolute value equation to ver-
ify that the solution set is 5 - 4, 3 6 .The graph is shown in FIGURE 32.
- 4
x= 3 or x=- 4
2 x= 6 or 2 x=- 8
2 x+ 1 = 7 or 2 x+ 1 = - 7
2 x+ 1 = 7 2 x+ 1 =-7.
| 2 x+ 1 | 2 x+ 1
| 2 x+ 1 | =7.
NOW TRY EXAMPLE 1
EXERCISE 1
Solve| 4 x- 1 |=11.
–5 –4 –3 –2 –1 01234
FIGURE 32NOW TRY
EXERCISE 2
Solve| 4 x- 1 | 7 11.
–5 –4 –3 –2 –1 01234
FIGURE 33NOW TRYOBJECTIVE 3 Solve inequalities of the form and of the
form for.
Solving an Absolute Value Inequality withSolve Graph the solution set.
By Case 2 described in the previous box, this absolute value inequality is rewrit-
ten as
because must represent a number that is morethan 7 units from 0 on either
side of the number line. Now, solve the compound inequality.
Subtract 1.
Divide by 2.Check these solutions. The solution set is See FIGURE 33. Notice
that the graph is a disjoint interval.
1 - q, - 42 ́ 1 3, q 2.
x 7 3 or x6- 4
2 x 76 or 2 x6- 8
2 x+ 177 or 2 x+ 1 6- 7
2 x+ 1
2 x+ 177 or 2 x+ 1 6- 7 ,
| 2 x+ 1 | 7 7.
EXAMPLE 2 7
|axb|>k, k> 0
|axb|<k
NOW TRYSolving an Absolute Value Inequality withSolve Graph the solution set.
The expression must represent a number that is less than 7 units from 0 on
either side of the number line. That is, must be between and 7. As Case 3
in the previous box shows, that relationship is written as a three-part inequality.
Subtract 1 from each part.- 46 x 63 Divide each part by 2.
- 86 2 x 66
- 76 2 x+ 167
2 x+ 1 - 7
2 x+ 1
| 2 x+ 1 | 6 7.
EXAMPLE 3 <
NOW TRY ANSWERS
- A-q, -^52 B ́ 1 3, q 2
E-^52 , 3F