Solving Linear Inequalities

• An inequality is an equation with ≤,<,≥ or > instead of an equal sign
• A linear inequality is an inequality that is linear in form (there are no roots, higher powers,
rationals, etc)
• Solving inequalities is the same as solving regular equations, but you need to be careful of the
direction of the inequality sign. Multiplying or dividing the equation by a negative swaps the
direction of the inequality sign (i.e. > would become <)
• A compound inequality is formed by putting together two inequalities with the word
− and (meaning intersection, both have to be satisfied to be a solution)
− or (meaning union, either has to be satisfied to be a solution)
• The compound inequality with and could be joined in one statement called a conjunction in the
manner below
x > 5 and x ≤12 5 < x ≤12
• The compound inequality with or is called a disjunction and cannot be rewritten in shorter form

Recap of the Properties:
• Adding any quantity to both sides won’t change the inequality
i.e. a < b and k is any real, =>a + k < b + k
• Multiplying both sides by a POSITIVE constant won’t change the inequality
i.e. a < b, k > 0=>ak < bk
• Multiplying both sides by a NEGATIVE constant WILL change the direction of the inequality
i.e. a < b, k < 0=>ak > bk
• Taking the reciprocal WILL change the direction of the inequality
i.e.

Describing the Intervals:
• There are several ways to describe intervals
– interval notation uses parenthesis, “(” and brackets, “]”
– set notation uses inequality symbols, “>”
– graphing uses the number line with open and closed circles connected with lines or shading
• An interval can be open (meaning you don’t actually “contain” the endpoint). The open end is
described with “)” “>” and/or open circles
• Or an interval can be closed (meaning you do contain the endpoint). The closed end is described
with “]” “ ≥ ” and/or closed circles
• Example. To describe the values for x greater than 5
(5,∞)
x > 5

Some Common Mistakes (with NON-linear equations):
• When solving for a product of two quantities, you must take into account that a negative times a
negative is a positive.
i.e. to solve for (x − 3)(x + 2) < 0 , you can’t just take when both are negative. You have to do
what is called a “sign diagram” This can be done in one of two ways…
1) Look at the sign of x – 3 and x + 2 separately on a chart

You can see that (x − 3)(x + 2) < 0 when –2 < x < 3

2) Still do a chart, but just “test” values to the left and right of your “zeros”
To the left of –2 (say –4), (−4 − 3)(−4 + 2) > 0
Between –2 and 3 (say 0), (0 − 3)(0 + 2) < 0
To the right of 3 (say +4), (4 − 3)(4 + 2) > 0

Which gives the same result as above

• Don’t ever just divide by x if you’re trying to solve an inequality for x. Not only are you neglecting
values, you may be inadvertently dividing by a negative.
i.e. x² < x does not reduce to x <1.
Solving correctly you get x² − x < 0, x(x −1) < 0 so (with sign diagram) 0 < x < 1

• When solving inequalities that contain a squared term, you can’t just take the square root as
normal. You have to include the +/- with the variable, NOT the constant term.
i.e. to solve
 ,

Some Exercise Problems:
• Example. Solve 5y −5+ y ≤ 2 − 6y −8

• Example. Solve (x +1)(x + 2) > x(x +1)

• Example. Solve −3 <1− 2x ≤ 3

• Example. Solve 5x +11≤ −4 or 5x +11≥ 4

Absolute Value with Inequalities:
• in other words, − k ≤ x ≤ k

• Example. Solve