
## Section1.3Basic Absolute Inequalities

##### Inequalities of form $\abs{ax+b} \lt c$ or $\abs{ax+b} \le c$

A graph of the function $y=\abs{x}$ is shown in Figure 1.3.1. The line $y=3$ is also. Note that the points where the absolute function has $y$-coordinates less than $3$ all have $x$-coordinates between $-3$ and $3\text{.}$ This indicates that the solution set to the inequality $\abs{x} \lt 3$ is the interval $(3,3)\text{.}$

In general, if $k \gt 0\text{,}$ then the solution set to an inequality of the form $\abs{ax+b} \lt k$ will be an open interval that is determined by solving a compound inequality. Specifically:

\begin{equation*} \abs{ax+b} \lt k;\,k \gt 0 \end{equation*}

is equivalent to the compound inequality

\begin{equation*} -k \lt ax+b \lt k\text{.} \end{equation*}
###### Example1.3.2

Determine the solution set to the inequality $\abs{3x+7} \le 14\text{.}$ State the solution set using interval notation (if possible).

Solution

We begin by writing an equivalent compound inequality that does not include an absolute value expression. We then solve that compound inequality.

\begin{alignat*}{2} -14 \amp\le 3x+7 \amp\amp\le 14\\ -14\subtractright{7} \amp\le 3x+7\subtractright{7} \amp\amp\le 14\subtractright{7}\\ -21 \amp\le 3x \amp\amp\le 7\\ \divideunder{-21}{3} \amp\le \divideunder{3x}{3} \amp\amp\le \divideunder{7}{3}\\ -7 \amp\le x \amp\amp\le \frac{7}{3} \end{alignat*}

The solution set to the give inequality is $\left[-7,\frac{7}{3}\right]\text{.}$

###### Example1.3.3

Determine the solution set to the inequality $\abs{5x+19} \lt -10\text{.}$ State the solution set using interval notation (if possible).

Solution

Let's pause for a moment and consider what's actually being asked. We are asked to determine what values of $x$ will cause an absolute value to be less than $-10\text{.}$ Since no absolute value is negative, no absolute value is ever less than $-10\text{.}$ So no value of $x$ makes the inequality $\abs{5x+19} \lt -10$ true, and the solution set to that inequality is $\emptyset\text{.}$

###### Example1.3.4

Determine the solution set to the inequality $\abs{-\frac{3}{7}x+6} \lt 9\text{.}$ State the solution set using interval notation (if possible).

Solution

We need to write and solve an equivalent compound inequality that does not include an absolute value expression. While solving, we need to remember that the direction of the inequality sign reverses anytime we multiply or divide both sides of the equation by a negative number,

\begin{alignat*}{2} -9 \amp\lt -\frac{3}{7}x+6 \amp\amp\lt 9\\ -9\subtractright{6} \amp\lt -\frac{3}{7}x+6\subtractright{6} \amp\amp\lt 9\subtractright{6}\\ -15 \amp\lt -\frac{3}{7}x \amp\amp\lt 3\\ \multiplyleft{-\frac{7}{3}}-15 \amp\gt \multiplyleft{-\frac{7}{3}}-\frac{3}{7}x \amp\amp\gt \multiplyleft{-\frac{7}{3}}3\\ 35 \amp\gt x \amp\amp\gt -7 \end{alignat*}

The solution set to the given inequality is $(-7,35)\text{.}$

##### Inequalities of form $\abs{ax+b} \gt c$ or $\abs{ax+b} \ge c\text{.}$

A graph of the function $y=\abs{x}$ is shown in Figure 1.3.5. The line $y=3$ is also. Note that the points where the absolute function has $y$-coordinates greater than $3$ have $x$-coordinates that are either to the left of $-3$ or to the right of $3\text{.}$ This indicates that the solution set to the inequality $\abs{x} \gt 3$ is the interval $(-\infty,-3) \cup (3,\infty)\text{.}$

In general, if $k \gt 0\text{,}$ then the solution set to an inequality of the form $\abs{ax+b} \gt k$ will be the union of two open intervals that are determined by solving a compound inequality. Specifically:

\begin{equation*} \abs{ax+b} \gt k;\,k \gt 0 \end{equation*}

is equivalent to the compound inequality

\begin{equation*} ax+b \lt -k \text{ or } ax+b \gt k\text{.} \end{equation*}
###### Example1.3.6

Determine the solution set to the inequality $\abs{3x+2} \gt 11\text{.}$ State the solution set using interval notation (if possible).

Solution

We begin by writing an equivalent compound inequality that does not include an absolute value expression. We then solve that compound inequality

\begin{align*} 3x+2 \amp\lt -11 \amp\amp\text{or}\amp 3x+ 2 \amp\gt 11\\ 3x+2\subtractright{2} \amp\lt -11\subtractright{2} \amp\amp\text{or}\amp 3x+2\subtractright{2} \amp\gt 11\subtractright{2}\\ 3x \amp\lt -13 \amp\amp\text{or}\amp 3x \amp\gt 9\\ \divideunder{3x}{3} \amp\lt \divideunder{-13}{3} \amp\amp\text{or}\amp \divideunder{3x}{3}\amp\gt \divideunder{9}{3}\\ x \amp\lt -\frac{13}{3} \amp\amp\text{or}\amp x \amp\gt 3 \end{align*}

The solution set to the given inequality is $\left(-\infty,-\frac{13}{3}\right) \cup (3,\infty)\text{.}$

###### Example1.3.7

Determine the solution set to the inequality $\abs{6-x} \ge 12\text{.}$ State the set using interval notation (if possible).

Solution

We need to write an equivalent compound inequality that does not include an absolute value expression and solve that compound inequality. We need to remember that when we multiply or divide both sides of an inequality by a negative value, the direction of the inequality is reversed.

\begin{align*} 6-x \amp\le -12 \amp\amp\text{or}\amp 6-x \amp\ge 12\\ 6-x\subtractright{6} \amp\le -12\subtractright{6} \amp\amp\text{or}\amp 6-x\subtractright{6} \amp\ge 12\subtractright{6}\\ -x \amp\le -18 \amp\amp\text{or}\amp -x \amp\ge 6\\ \multiplyleft{-1}-x \amp\ge -\multiplyleft{-1}-18 \amp\amp\text{or}\amp \multiplyleft{-1}-x \amp\le \multiplyleft{-1}6\\ x \amp\ge 18 \amp\amp\text{or}\amp x \amp\le -6 \end{align*}

The solution set is $(-\infty,-6] \cup [18,\infty)\text{.}$

###### Example1.3.8

Determine the solution set to the inequality $\abs{\frac{4x-8}{3}} \ge -9\text{.}$ Write the solution set using interval notation (if possible).

Solution

We need to be careful not to prematurely jump into rote mode. The smallest value the absolute value ever has is zero. Any value greater than or equal to zero is also greater than $-9\text{.}$ So the value of the absolute value expression will be greater than $-9$ regardless of the value of $x\text{,}$ and the solution set to $\abs{\frac{4x-8}{3}} \ge -9$ is $(-\infty,\infty)\text{.}$

### Subsection1.3.1Exercises

Determine the solution set to each of the following inequalities. Express each solution set using interval notation (where possible).

###### 1

$\abs{x+9} \lt 20$

Solution

We begin by writing an equivalent compound inequality that does not include an absolute value expression. We then solve that compound inequality.

\begin{alignat*}{2} -20 \amp\lt x+9 \amp\amp\lt 20\\ -20\subtractright{9} \amp\lt x+9\subtractright{9} \amp\amp\lt 20\subtractright{9}\\ -29 \amp\lt x \amp\amp\lt 11 \end{alignat*}

The solution set to the given inequality is $(-29,11)\text{.}$

###### 2

$\abs{2x+8} \lt -4$

Solution

There are no numbers whose absolute value is negative, hence there are no numbers whose absolute value is less than $-4\text{.}$ So the given inequality has no solutions and the solution set to the inequality is $\emptyset\text{.}$

###### 3

$\abs{-10-\frac{3x}{2}} \le 19$

Solution

We begin by writing an equivalent compound inequality that does not include an absolute value expression. We then solve that compound inequality.

\begin{alignat*}{2} -19 \amp\le -10-\frac{3x}{2} \amp\amp\le 19\\ -19\addright{10} \amp\le -10-\frac{3x}{2}\addright{10} \amp\amp\le 19\addright{10}\\ -9 -\amp\le \frac{3x}{2} \amp\amp\le 29\\ \multiplyleft{-\frac{2}{3}}-9 \amp\ge \multiplyleft{-\frac{2}{3}}-\frac{3x}{2} \amp\amp\ge \multiplyleft{-\frac{2}{3}}29\\ 6 \amp\ge x \amp\amp\ge -\frac{58}{3} \end{alignat*}

The solution set is $\left[-\frac{58}{3},6\right]\text{.}$

###### 4

$\abs{4x+5} \le 25$

Solution

We begin by writing an equivalent compound inequality that does not include an absolute value expression. We then solve that compound inequality.

\begin{alignat*}{2} -25 \amp\le 4x+5 \amp\amp\le 25\\ -25\subtractright{5} \amp\le 4x+5\subtractright{5} \amp\amp\le 25\subtractright{5}\\ -30 \amp\le 4x \amp\amp\le 20\\ \divideunder{-30}{4} \amp\le \divideunder{4x}{4} \amp\amp\le \divideunder{20}{4}\\ -\frac{15}{2} \amp\le x \amp\amp\le 5 \end{alignat*}

The solution set is $\left[-\frac{15}{2},5\right]\text{.}$

###### 5

$\abs{-9x} \le 63$

Solution

We begin by writing an equivalent compound inequality that does not include an absolute value expression. We then solve that compound inequality.

\begin{alignat*}{2} -63 \amp\le -9x \amp\amp\le 63\\ \divideunder{-63}{-9} \amp\ge \divideunder{-9x}{-9} \amp\amp\ge \divideunder{63}{-9}\\ 7 \amp\ge x \amp\amp\ge -7 \end{alignat*}

The solution set is $[-7,7]\text{.}$

###### 6

$\abs{2+10x} \lt 0$

Solution

No number has an absolute value that is negative, so no number has an absolute value that is less than $0\text{.}$ Therefore, the given inequality has no solutions and the solution set for the inequality is $\emptyset\text{.}$

###### 7

$\abs{2x+8} \gt 64$

Solution

We begin by writing an equivalent compound inequality and proceed to solve that compound inequality.

\begin{align*} 2x+8 \amp\lt -64 \amp\amp\text{or}\amp 2x+8 \amp\gt 64\\ 2x+8\subtractright{8} \amp\lt -64\subtractright{8} \amp\amp\text{or}\amp 2x+8\subtractright{8} \amp\gt 64\subtractright{8}\\ 2x \amp\lt -72 \amp\amp\text{or}\amp 2x \amp\gt 56\\ \divideunder{2x}{2} \amp\lt \divideunder{-72}{2} \amp\amp\text{or}\amp \divideunder{2x}{2} \amp\gt \divideunder{56}{2}\\ x \amp\lt -36 \amp\amp\text{or}\amp x\amp\gt 28 \end{align*}

The solution set is $(-\infty,-36) \cup (28,\infty)\text{.}$

###### 8

$\abs{7-x} \ge 16$

Solution

We begin by writing an equivalent compound inequality and proceed to solve that compound inequality. We need to remember that when we multiply or divide both sides of an inequality by a negative number, the direction of the inequality sign is reversed.

\begin{align*} 7-x \amp\le -16 \amp\amp\text{or}\amp 7-x \amp\ge 16\\ 7-x\subtractright{7} \amp\le -16\subtractright{7} \amp\amp\text{or}\amp 7-x\subtractright{7} \amp\ge 16\subtractright{7}\\ -x \amp\le -23 \amp\amp\text{or}\amp -x \amp\ge 9\\ \multiplyleft{-1}-x \amp\ge \multiplyleft{-1}-23 \amp\amp\text{or}\amp \multiplyleft{-1}-x \amp\le \multiplyleft{-1}9\\ x \amp\ge 23 \amp\amp\text{or}\amp x \amp\le -9 \end{align*}

The solution set is $(-\infty,-9] \cup [23,\infty)\text{.}$

###### 9

$\abs{-9x+14} \gt 83$

Solution

We begin by writing an equivalent compound inequality and proceed to solve that compound inequality. We need to remember that when we multiply or divide both sides of an inequality by a negative number, the direction of the inequality sign is reversed.

\begin{align*} -9x+14 \amp\lt -83 \amp\amp\text{or}\amp -9x+14 \amp\gt 83\\ -9x+14\subtractright{14} \amp\lt -83\subtractright{14} \amp\amp\text{or}\amp -92+14\subtractright{14} \amp\gt 83\subtractright{14}\\ -9x \amp\lt -97 \amp\amp\text{or}\amp -9x \amp\gt 69\\ \divideunder{-9x}{-9} \amp\gt \divideunder{-97}{-9} \amp\amp\text{or}\amp \divideunder{-9x}{-9} \amp\lt \divideunder{69}{-9}\\ x \amp\gt \frac{97}{9} \amp\amp\text{or}\amp x \amp\lt -\frac{23}{3} \end{align*}

The solution set is $\left(-\infty,-\frac{23}{3}\right) \cup \left(\frac{97}{9},\infty\right)\text{.}$

###### 10

$\abs{4x+15} \ge -10$

Solution

We begin by observing that every number has an absolute value that is greater than (or equal to) $-10\text{.}$ So we conclude by noting that every value of $x$ is a solution to the given inequality and that the solution set to the inequality is $(-\infty,\infty)\text{.}$

###### 11

$\abs{3x-18} \gt 14$

Solution

We begin by writing an equivalent compound inequality and proceed to solve that compound inequality.

\begin{align*} 3x-18 \amp\lt -14 \amp\amp\text{or}\amp 3x-18 \amp\gt 14\\ 3x-18\addright{18} \amp\lt -14\addright{18} \amp\amp\text{or}\amp 3x-18\addright{18} \amp\gt 14\addright{18}\\ 3x \amp\lt 4 \amp\amp\text{or}\amp 3x \amp\gt 32\\ \divideunder{3x}{3} \amp\lt \divideunder {4}{3} \amp\amp\text{or}\amp \divideunder{3x}{3} \amp\gt \divideunder{32}{3}\\ x \amp\lt \frac{4}{3} \amp\amp\text{or}\amp x \amp\gt \frac{32}{3} \end{align*}

The solution set is $\left(-\infty,\frac{4}{3}\right) \cup \left(\frac{32}{3},\infty\right)\text{.}$