## Using the Properties of Inequalities

When we work with inequalities, we can usually treat them similarly to but not exactly as we treat equalities. We can use the addition property and the multiplication property to help us solve them. The one exception is when we multiply or divide by a negative number; doing so reverses the inequality symbol.

### A General Note: Properties of Inequalities

$\begin{array}{ll}\text{Addition Property}\hfill& \text{If }a< b,\text{ then }a+c< b+c.\hfill \\ \hfill & \hfill \\ \text{Multiplication Property}\hfill & \text{If }a< b\text{ and }c> 0,\text{ then }ac< bc.\hfill \\ \hfill & \text{If }a< b\text{ and }c< 0,\text{ then }ac> bc.\hfill \end{array}$

These properties also apply to $a\le b$, $a>b$, and $a\ge b$.

### Example 3: Demonstrating the Addition Property

Illustrate the addition property for inequalities by solving each of the following:

a. $x - 15<4$
b. $6\ge x - 1$
c. $x+7>9$

### Solution

The addition property for inequalities states that if an inequality exists, adding or subtracting the same number on both sides does not change the inequality.

a. $\begin{array}{ll}x - 15<4\hfill & \hfill \\ x - 15+15<4+15 \hfill & \text{Add 15 to both sides.}\hfill \\ x<19\hfill & \hfill \end{array}$

b. $\begin{array}{ll}6\ge x - 1\hfill & \hfill \\ 6+1\ge x - 1+1\hfill & \text{Add 1 to both sides}.\hfill \\ 7\ge x\hfill & \hfill \end{array}$

c. $\begin{array}{ll}x+7>9\hfill & \hfill \\ x+7 - 7>9 - 7\hfill & \text{Subtract 7 from both sides}.\hfill \\ x>2\hfill & \hfill \end{array}$

### Try It 3

Solve $3x - 2<1$.

Solution

### Example 4: Demonstrating the Multiplication Property

Illustrate the multiplication property for inequalities by solving each of the following:

1. $3x<6$
2. $-2x - 1\ge 5$
3. $5-x>10$

### Solution

a. $\begin{array}{l}3x<6\hfill \\ \frac{1}{3}\left(3x\right)<\left(6\right)\frac{1}{3}\hfill \\ x<2\hfill \end{array}$

b. $\begin{array}{ll}-2x - 1\ge 5\hfill & \hfill \\ -2x\ge 6\hfill & \hfill \\ \left(-\frac{1}{2}\right)\left(-2x\right)\ge \left(6\right)\left(-\frac{1}{2}\right)\hfill & \text{Multiply by }-\frac{1}{2}.\hfill \\ x\le -3\hfill & \text{Reverse the inequality}.\hfill \end{array}$

c. $\begin{array}{ll}5-x>10\hfill & \hfill \\ -x>5\hfill & \hfill \\ \left(-1\right)\left(-x\right)>\left(5\right)\left(-1\right)\hfill & \text{Multiply by }-1.\hfill \\ x<-5\hfill & \text{Reverse the inequality}.\hfill \end{array}$

### Try It 4

Solve $4x+7\ge 2x - 3$.

Solution

## Solving Inequalities in One Variable Algebraically

As the examples have shown, we can perform the same operations on both sides of an inequality, just as we do with equations; we combine like terms and perform operations. To solve, we isolate the variable.

### Example 5: Solving an Inequality Algebraically

Solve the inequality: $13 - 7x\ge 10x - 4$.

### Solution

Solving this inequality is similar to solving an equation up until the last step.

$\begin{array}{ll}13 - 7x\ge 10x - 4\hfill & \hfill \\ 13 - 17x\ge -4\hfill & \text{Move variable terms to one side of the inequality}.\hfill \\ -17x\ge -17\hfill & \text{Isolate the variable term}.\hfill \\ x\le 1\hfill & \text{Dividing both sides by }-17\text{ reverses the inequality}.\hfill \end{array}$

The solution set is given by the interval $\left(-\infty ,1\right]$, or all real numbers less than and including 1.

### Try It 5

Solve the inequality and write the answer using interval notation: $-x+4<\frac{1}{2}x+1$.

Solution

### Example 6: Solving an Inequality with Fractions

Solve the following inequality and write the answer in interval notation: $-\frac{3}{4}x\ge -\frac{5}{8}+\frac{2}{3}x$.

### Solution

We begin solving in the same way we do when solving an equation.

$\begin{array}{ll}-\frac{3}{4}x\ge -\frac{5}{8}+\frac{2}{3}x\hfill & \hfill \\ -\frac{3}{4}x-\frac{2}{3}x\ge -\frac{5}{8}\hfill & \text{Put variable terms on one side}.\hfill \\ -\frac{9}{12}x-\frac{8}{12}x\ge -\frac{5}{8}\hfill & \text{Write fractions with common denominator}.\hfill \\ -\frac{17}{12}x\ge -\frac{5}{8}\hfill & \hfill \\ x\le -\frac{5}{8}\left(-\frac{12}{17}\right)\hfill & \text{Multiplying by a negative number reverses the inequality}.\hfill \\ x\le \frac{15}{34}\hfill & \hfill \end{array}$

The solution set is the interval $\left(-\infty ,\frac{15}{34}\right]$.

### Try It 6

Solve the inequality and write the answer in interval notation: $-\frac{5}{6}x\le \frac{3}{4}+\frac{8}{3}x$.

Solution