# Solving equations with inequalities

In this section, we will be solving inequalities such as $2x + 1 \le 7$. Let's begin by taking a look at two equalities:

$6 + 1 = 4 + 3$
$2x + 1 = 7$

We know that both of these are equalities because each contains an equal sign. Therefore, we know that we can add, subtract, multiply, or divide both sides of the equal sign by any Real number, and the expression will remain true.

In place of the equal sign, the symbols used in an inequality are:

< less than
≤ less than or equal to
> greater than
≥ greater than or equal to

In an inequality, we can add, subtract, multiply (by a positive Real number), or divide (by a positive Real number), and the inequality will remain true. However, if we were to multiply or divide both sides of the inequality by a negative Real number, we would have to switch the sign around in order for the inequality to remain true.

#### Example:   Add a number to both sides of $4 \lt 8$

\begin{align} 4 &\lt 8 \\ 4 \, {\color{Red} + \, 5} &\lt 8 \, {\color{Red} + \, 5} \qquad \text{add 5 to both sides of the inequality} \\ 9 &\lt 13 \\ \end{align}

The inequality remains true.

#### Example:   Subtract a number from both sides of $4 \lt 8$

\begin{align} 4 &\lt 8 \\ 4 \, {\color{Red}- \, 3} &\lt 8 \, {\color{Red}- \, 3} \qquad \text{subtract 3 from both sides of the inequality} \\ 1 &\lt 5 \\ \end{align}

The inequality remains true.

#### Example:    Multiply both sides of $4 \lt 8$ by a positive number

\begin{align} 4 &\lt 8 \\ 4 \cdot {\color{Red} 5} &\lt 8 \cdot {\color{Red} 5} \qquad \text{multiply both sides of the inequality by 5} \\ 20 &\lt 40 \\ \end{align}

The inequality remains true.

#### Example:    Multiply both sides of $4 \lt 8$ by a negative number

\begin{align} 4 &\lt 8 \\ 4 \, {\color{Red}\cdot} \, {\color{Red}-5} &\lt 8 \, {\color{Red}\cdot} \, {\color{Red}-5} \qquad \text{multiply both sides of the inequality by --5} \\ -20 &\lt -40 \qquad \text{this equation is NOT true} \\ \end{align}

The inequality will be false unless we switch the < sign to a > sign, because we multiplied by a negative.

#### Example:    Divide both sides of $4 \lt 8$ by a negative number

\begin{align} 4 &\lt 8 \\ \frac{4}{{\color{Red} -2}} &\lt \frac{8}{{\color{Red} -2}} \qquad \text{divide both sides of the inequality by -2} \\ -2 &\lt -4 \qquad \text{this equation is NOT true} \\ \end{align}

The inequality will be false unless we switch the < sign to a > sign because we divided by a negative.

Whenever we multiply or divide both sides of an equality by a negative, the inequality sign must be switched around. Therefore,

< switches to >
≤ switches to ≥
and
> switches to <
≥ switches to ≤

## Solving an inequality

Solving an inequality follows the same principles as solving an equality with the exception that, we must switch the inequality sign around when multiplying or dividing both sides of the inequality by a negative. Let's try out some examples.

### Example 1:   $2x + 1 \lt 7$

\begin{align} 2x + 1 &\lt 7 \\ 2x + 1 \, {\color{Red}- \, 1} &\lt 7 \, {\color{Red}- \, 1} \qquad \text{to get the term with the x alone on one side of the inequality, subtract 1 from both sides of the} \lt \text{sign} \\ 2x &\lt 6 \qquad \text{combine like terms on each side of the} \lt \text{sign} \\ \frac{2x}{{\color{Red} 2}} &\lt \frac{6}{{\color{Red} 2}} \qquad \text{divide both sides of the} \lt \text{sign by 2; as 2 is positive, there is no need to switch the} \lt \text{sign} \\ x &\lt 3 \\ \end{align}

The solution set is all Real numbers less than three.

### Example 2:   $-5y - 3 \ge 17$

\begin{align} -5y - 3 &\ge 17 \\ -5y - 3 \, {\color{Red} + \, 3} &\ge 17 \, {\color{Red} + \, 3} \qquad \text{to get the term with the y alone, add 3 to both sides of the} \ge \text{sign} \\ -5y &\ge 20 \qquad \text{combine like terms on both sides of the} \ge \text{sign} \\ \frac{-5y}{{\color{Red} -5}} &{\color{Red}\le} \frac{20}{{\color{Red} -5}} \qquad \text{Divide both sides by -5 to get y; switch} \ge \text{to} \le \text{because we divided both sides of the} \ge \text{sign by a negative} \\ y &\le -4 \\ \end{align}

The solution set contains all Real numbers less than or equal to -4.

### Example 3:   $-3(2a + 5) \le 4a + 6$

The first step in solving this inequality will be to distribute the -3 on the left side to the two terms inside the parentheses. Note that because we are not multiplying both sides of the inequality by a negative, we do not need to switch the ≤ sign.

\begin{align} -3(2a + 5) &\le 4a + 6 \\ {\color{Red}-6a - 15} &\le 4a + 6 \qquad \text{simplify the left side of the inequality by multiplying through the parentheses by -3} \\ -6a \, {\color{Red}- \, 4a} - 15 &\le 4a \, {\color{Red}- \, 4a} + 6 \qquad \text{to get the terms with the a on one side of the inequality, subtract 4a from both sides} \\ -10a - 15 &\le 6 \qquad \text{combine like terms on both sides of the} \le \text{sign} \\ -10a - 15 \, {\color{Red}+ \, 15} &\le 6 \, {\color{Red}+ \, 15} \qquad \text{to get the term with the a alone, add 15 to both sides} \\ -10a &\le 21 \qquad \text{combine like terms} \\ \frac {-10a}{{\color{Red}-10}} &{\color{Red}\ge} \frac{21}{{\color{Red}-10}} \qquad \text{divide both sides by -10 to get a; switch to} \ge \\ a &\ge -\frac{21}{10} \\ \end{align}

The solution set is all Real numbers greater than or equal to $-\frac{21}{10}$

### Example 4:   $6 \lt 5x - 4$

\begin{align} 6 &\lt 5x - 4 \\ 6 \, {\color{Red}+ \, 4} &\lt 5x -4 \, {\color{Red} + \, 4} \qquad \text{to get the term with the variable alone on one side of the inequality, we add 4 to both sides of the} \lt \text{sign} \\ 10 &\lt 5x \qquad \text{combine like terms} \\ \frac{10}{{\color{Red}5}} &\lt \frac{5x}{{\color{Red}5}} \qquad \text{five is multiplied by x, so divide both sides of the inequality by 5 to get x by itelf} \\ 2 &\lt x \quad \text{OR} \quad x \gt 2 \\ \end{align}

We are solving for x, so we want to say that x is < or > some Real number. In this case, 2 is less than x. Therefore, x has to be greater than 2.