# inequalities

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```					    Inequalities and Absolute Value problems
Laurel Benn
November 25, 2009

This set of notes will help clear up a lot of the problems we    had with
inequalities. We’ll divide up the inequalities into two types. The   ﬁrst type
is Linear Inequalities and the other type we’ll call Non-Linear      Inequali-
ties(quadratic and/or rational). We will then take a look at some     absolute
value problems.

Linear Inequalities
We’ll solve this type of inequality by using the same technique as that used
in solving regular linear equations. However, we need to remember that when
multiplying(or dividing) both sides of an inequality by a negative number ,
the direction of the inequality symbol changes.

Example 1 Solve the inequality. x + 5 ≤ 4.
Here we will add −5 to both sides to obtain, x ≤ 4 − 5 or x ≤ −1.

Example 2 Solve the inequality. 2 − 5x ≥ 7.
Here we will add -2 to both sides to obtain, −5x ≥ 5.
1
Multiplying both sides by − 5 yields x ≤ −1.
Remember that multiplying or dividing both sides of an inequality by a
negative number changes the direction of the inequality symbol.

Example 3 Solve the inequality. 2x + 7 < 3x − 2.
Solution: 2x + 7 < 3x − 2
⇒ 2x − 3x < −2 − 7

1
⇒ −x < −9
⇒x>9
Therefore the solution set is (9, +∞).

Example 4 Solve the inequality. 2x−1 > x+1 .
3       2
Solution: Here the LCM of the denominators is 6.
So we have 6( 2x−1 ) > 6( x+1 ).
3          2
⇒ 2(2x − 1) > 3(x + 1)
⇒ 4x − 2 > 3x + 3
⇒ 4x − 3x > 3 + 2
⇒x>5
Using interval notation the solution set is therefore (5, +∞).
Note: This is a classic example in which cross-multiplication can be used
with no problems.

Non-Linear Inequalities
These will be inequalities that contain quadratics or rationals with variable
factors. For these inequalities we will do the following in order:

• Rewrite the inequality with zero on the right hand side.

• Find the numbers that makes the inequality zero or undeﬁned.

• Place the numbers found on a number line to create intervals.

• Investigate the intervals to see when the given inequality is valid.
4
Example 1 Solve the inequality 2−x ≤ 1
Solution: Subtract 1 from each side to get:
4
2−x
2+x
− 1 ≤ 0 ⇒ 4−2+x ≤ 0 ⇒ 2−x ≤ 0
2−x
2+x
Now,-2 makes 2−x zero and 2 makes it undeﬁned. we therefore place -2 and
2 on the number line creating the intervals (−∞, −2),(−2, 2) and (2, +∞).
Investigating these intervals we see that the inequality holds if x > 2 or
x ≤ −2. Therefore, the solution set is (−∞, −2] ∪ (2, +∞).

In the following examples I will not be as detailed.

2
2      3
Example 2 Solve for x: x <   x−4

Solution: Subtract the value on the right from both sides to get:
2
x
− x−4 < 0 ⇒ 2(x−4)−3x < 0 ⇒ x(x−4) < 0
3
x(x−4)
−8−x

By inspection , we see that if x > 4 or −8 < x < 0 the inequality holds ⇒
the solution is (−8, 0) ∪ (4, +∞)
Example 3 Solve the inequality 8x3 − 4x2 − 2x + 1 < 0
Solution: Factoring we get 4x2 (2x − 1) − (2x − 1) < 0
⇒ (2x − 1)(4x2 − 1) < 0 ⇒ (2x − 1)2 (2x + 1) < 0.
1
This inequality is valid only if x < − 1 . That means that (−∞, − 2 ) is the
2
solution set.
Example 4 Solve for x: (3x + 2)(x − 3) ≥ 0.
Solution: (3x + 2)(x − 3) ≥ 0 ⇒ x ≥ 3 or x ≤ − 2 3
The solution set is therefore (−∞, − 2 ] ∪ [3, +∞)
3

Absolute Values
Here are some absolute value problems and their solutions.    I’ll add some
related inequality problems later.
Example 1 Find the real values of x which satisfy the equation |3x| = 2x+5.
Solution: Here either 3x = 2x + 5 or −3x = 2x + 5. Solving 3x = 2x + 5 we
get x = 5 and solving −3x = 2x + 5 we get −5x = 5 or x = −1.This implies
that the solution set is {-1,5}.
Example 2 Solve for x: |6x − 7| = |3 + 2x|
Solution: Here either 6x − 7 = 3 + 2x or 6x − 7 = −(3 + 2x)
5
Solving 6x − 7 = 3 + 2x we get 4x = 10 ⇒ x = 2
Solving 6x − 7 = −3 − 2x we get 8x = 4 ⇒ x = 1 .  2
This implies that { 1 , 2 } is the solution set.
2
5

Example 3 Solve for x: |x + 1| = |2x − 2| ⇒ either x + 1 = 2x − 2 or
x + 1 = 2 − 2x. Solving both cases we get x = 3 or x = 1 .
3

Example 4 Solve for x: | x+5 | = 6
2−x
Solution: We have that either (i) 2−x = 6 or (ii) 2−x = −6. Solving (i) we
x+5             x+5

get x+5 = 6(2−x) ⇒ 7x = 7 ⇒ x = 1. Solving (ii) we get x+5 = −6(2−x)
which gives 5x = 17 ⇒ x = 17 . The solution set is therefore { 17 , 1}
5                                   5

3
Example 5 Solve for x: |x + 5| < 3
Solution: For this we must have that −3 < x + 5 < 3
⇒ −8 < x < −2
The solution set is therefore (−8, −2).

Example 6 Solve for x: |x + 5| < |2x − 1|
Solution: Taking |2x − 1| to the left side we get
|x + 5| − |2x − 1| < 0
Now for the LHS to be zero we must have x + 5 = 2x − 1 or x + 5 = 1 − 2x
That is x = 6 or x = −4/3. Using −4/3 and 6 to divide up the real
number line into intervals , we ﬁnd by inspection that
(−∞, − 4 ) ∪ (6, +∞) is the solution set.
3

Alternate Solution: Let’s solve this a diﬀerent way.
Squaring both sides of the inequality we get ,
(x + 5)2 < (2x − 1)2 (Note this is valid since both sides were +ve)
⇒ x2 + 10x + 25 < 4x2 − 4x + 1
⇒ −3x2 + 14x + 24 < 0
⇒ 3x2 − 14x − 24 > 0
Factoring we get (3x + 4)(x − 6) > 0. Which gives (−∞, − 4 ) ∪ (6, +∞)
3
as the solution set. Now if I had said before that squaring both sides would
be quicker then I would like to take that back.

Example 7 Solve for x: |x + 5| < |2x − 1| + 3
I know we did not do this type but I would like to see how you would approach
or solve it.

4

```
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