# How to Solve a Linear Inequality

Linear inequalities are almost identical to linear equations. Instead of the answer being a number, the answer is an interval. When you solve an inequality, you must find what $x$ must be greater than, greater than or equal to, less than, or less than or equal to.

Rule

### InequalitySigns

$>$ means “greater than”

$\ge$ means “greater than or equal to”

$<$ means “less than”

$\le$ means “less than or equal to”

There is one important difference between solving linear inequalities and linear equations: When you multiply or divide by a negative number, you have to turn the inequality! Why? Take a look at the following example:

Example 1

You might agree that

 $5>-10$

because $-10$ is further left on the real line than 5. If you multiply both sides with $-1$, you get: $\begin{array}{llll}\hfill 5& >-10\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \left(-1\right)\cdot 5& >-10\cdot \left(-1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -5& >10\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Now it says that $-5$ is greater than 10. That doesn’t make sense! 10 is further to the right on the number line than $-5$, and is therefore a greater number.

Due to this issue with the sign, the inequality sign must be turned for the expression to still be true.

Rule

### MultiplyorDividebyNegativeNumbers

When you multiply or divide by a negative number, you have to TURN the inequality.

Example 2

Solve the inequality $x+3<8$

$\begin{array}{llll}\hfill x+3& <8\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& <8-3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& <5\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ The answer tells you that $x$ must be less than 5 for the inequality to be true.

Example 3

Solve the inequality $-x-12<14$

$\begin{array}{llll}\hfill -x-12& <14\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -x& <14+12\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \left(-1\right)\cdot -x& <26\cdot \left(-1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& >-26\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ The answer tells you that $x$ must be greater than $-26$ for the inequality to be true.

Example 4

Solve the inequality $2\left(x+2\right)-3x>-\left(-x-3\right)+4$

$\begin{array}{llll}\hfill 2\left(x+2\right)-3x& >-\left(-x-3\right)+4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 2x+4-3x& >x+3+4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 2x-3x-x& >3+4-4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -2x& >3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{\text{−2}x}{\text{−2}}& <\frac{3}{-2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& <-\frac{3}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ The answer tells you that $x$ must be less than $-\frac{3}{2}$ for the inequality to be true.

Example 5

Solve the inequality $2x-3\left(2x-5\right)\ge 2\left(3x-2\right)+5x$

$\begin{array}{cc}2x-3\left(2x-5\right)\ge 2\left(3x-2\right)+5x& \\ 2x-6x+15\ge 6x-4+5x& \\ 2x-6x-6x-5x\ge -4-15& \\ -15x\ge -19& \\ \frac{\text{−15}x}{\text{−15}}\ge \frac{-19}{-15}& \\ x\le \frac{19}{15}& \end{array}$

$\begin{array}{llll}\hfill 2x-3\left(2x-5\right)& \ge 2\left(3x-2\right)+5x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 2x-6x+15& \ge 6x-4+5x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 2x-6x-6x-5x& \ge -4-15\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -15x& \ge -19\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{\text{−15}x}{\text{−15}}& \ge \frac{-19}{-15}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& \le \frac{19}{15}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The answer tells you that $x$ must be less than or equal to $\frac{19}{15}$ for the inequality to be true.