PreCalculus
Worksheet: Inequalities and Absolute Value
Name: __________________________
The rules for solving linear inequalities are simple, and correspond to the same rules for solving
equalities except when you multiply or divide both sides by a negative, in which case you must
“flip” the inequality. However, when solving more complicated inequalities the situation is not
always as obvious. In particular, absolute-value inequalities cannot be done in the “obvious
way.”
A safer approach is to treat an inequality as an equality, solve that, use the solutions as
“boundary-values” to partition the number-line into “zones of possible solutions,” and finally test
numbers in each “zone.” The following example illustrates this procedure applied to an
absolute-value inequality:
Example: | 2 x  3 |  x  1
Solution:
Treat this as an equality, and solve for the boundary-values: | 2 x  3 |  x  1
Since | y |  y if y is positive or 0, and | y |   y if y is negative, we must solve
two equations in order to get the boundary-values:
2x  3  x  1
2x  3  ( x  1)
and
2
. Next we plot
3
these on a number line, and then pick three numbers to test, one in each zone:
Solving each of these gives the boundary-values x  4 and x 
test-numbers
0
2
5
2
3
4
boundary-values
Test x  0 : | 2  0  3 |  0  1 :
| 3 |  1 , TRUE
Test x  2 : | 2  2  3 |  2  1 , | 1 |  3 , FALSE
Test x  5 : | 2  5  3 |  5  1 , | 7 |  6 , TRUE
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Therefore, the solution is: x 
2
or x  4
3
Inequalities which contain algebraic fractions must be treated even more carefully. The
boundary-values not only contain values of the unknown that make the equality true, but they
may also contain values that make a fraction undefined (that is, its denominator is 0).
Furthermore, if the inequality includes an equality (that is, “  ” or “  ”) then a boundary-value
that causes a fraction to be undefined can not be included in the solution.
Example:
3
x3

x2
2
Solution:
One of the boundary-values is x  2 because this makes the fraction 0. Treating it
as an equality,
3
x3
, we cross-multiply to get: x2  x  6  6 . To

x2
2
solve this equation, we subtract 6 from both sides, then factor: x2  x  12  0 ,
( x  3)( x  4)  0 . The solutions of this equation are x  3 and x  4 . These
two boundary-values are also solutions of the original inequality since it includes an
equality, but the boundary-value x  2 is not a solution since it makes a fraction
undefined. We must now test numbers in 4 zones determined by these three
boundary-values:
test-numbers
–5
0
–4
2.5
2
4
3
boundary-values
3
5  3
3
,   1 , TRUE

5  2
2
8
3
03
3 3
Test 0:
,   , FALSE

02
2
2 2
2 .5  3
3
Test 2.5:
, 6  2.75 , TRUE

2 .5  2
2
3
43 3 7
Test 4:
,  , FALSE

42
2
2 2
Test –5:
Therefore, the solution is: x  4 or 2  x  3 .
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Exercises: Solve. Show all work, including test-values.
1.
|x  2|  3
2.
| x  3|  2
3.
| 3  2x |  5
4.
| 2 x  1 |  3
5.
| 3x  2 |  x  2
6.
|x|  x
7.
|3  x |  x  3
8.
|x  2|  x  2
9.
2x  | x  2 |
10.
| x  5|  x  5
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11.
1 1

x 5
12.
1
1
 
x
5
13.
2
1

x3 x2
14.
4
x5

x2
2
15.
x
2

2
x3
16.
3
x5

x
2
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