Table of Contents

(g) y−1 y+1
+ 4 > 3 2
(h) 4(x + 1
2
) − 2(x + 3
2
) ≤ 5
(i) 2x+1
5
− x+3
2
< 4x + 10
(j) 4(m + 3) + 5(2m − 1) > 7m + 6
Section 3 Absolute Values and Inequalities
The absolute value of a number was discussed in Worksheet 1.7. Recall that the absolute value
of a, written |a|, is the distance in units that a is away from the origin. For example | − 7| = 7.
Alternatively you could define it as |a| = + √ a 2 . So when we combine absolute values and
inequalities we are looking for all numbers that are either less than or greater than a certain
distance away from the origin.
Example 1 : The equation |x| < 3 represents all the numbers whose distance away
from the origin is less than 3 units. We could rewrite this inequality as −3 < x < 3.
If we draw this on the number line we get
✛
❝ ❝
-5 -4 -3 -2 -1 0 1 2 3 4 5
To solve inequalities involving absolute values we often need to rewrite the inequality without
the absolute value signs as we have just done in the above example.
Example 2 : |x| > 1 can be rewritten as
x > 1 or x < −1
In other words all the numbers whose distance away from the origin is greater than
1 unit.
✛
✛ ❝ ❝ ✲
✲
-5 -4 -3 -2 -1 0 1 2 3 4 5
Notice that the solution is just the converse of −1 ≤ x ≤ 1. Since |x| ≤ 1 is easier
to solve than |x| > 1 we can begin by solving −1 ≤ x ≤ 1 and the solution will be
its converse.
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