Let's Learn Basic Absolute Value Equations

Section 2.2 Basic Absolute Value Equations

A graph of the function $y=\abs{x}$ is shown in Figure 2.2.1. The line $y=3$ is also shown. Note that the line intersects the absolute value at two points: $(-3,3)$ and $(3,3)\text{.}$ These points of intersection indicate that the equation $\abs{x}=3$ has two solutions: $-3$ and $3\text{.}$

$A graph of $y=\abs(x)\text{.}$ The graph resembles the printed letter vee. Starting at the origin there is an arrow pointing upward through the points $(-1,1)\text{,}$ $(-2,2)\text{,}$ $(-3,3)\text{,}$ etc. There is a second that starts at the origin and points upward through the points $(1,1)\text{,}$ $(2,2)\text{,}$ $(3,3)\text{,}$ etc. The line $y=3$ is also graphed and it is indicated that this line intersects $y=\abs(x)$ at the points $(-3,3)$ and $(3,3)\text{.}$$

Figure 2.2.1. $\abs{x}=3$

In general, if $k$ is a positive number, an equation of form $\abs{ax+b}=k$ will have two solutions. Specifically:

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

is equivalent to the pair of equations

\begin{equation*} ax+b=-k \text{ or } ax+b=k\text{.} \end{equation*}

Example 2.2.2.

Determine the solution set to the equation $\abs{4x-1}=29\text{.}$

Solution

We begin by writing an equivalent pair of equations that do not include absolute value expressions. We then solve that pair of equations.

\begin{align*} 4x-1\amp=-29 \amp\amp\text{or}\amp 4x-1\amp=29\\ 4x-1\addright{1}\amp=-29\addright{1} \amp\amp\text{or}\amp 4x-1\addright{1}\amp=29\addright{1}\\ 4x\amp=-28 \amp\amp\text{or}\amp 4x\amp=30\\ \divideunder{4x}{4}\amp=\divideunder{-28}{4} \amp\amp\text{or}\amp \divideunder{4x}{4}\amp=\divideunder{30}{4}\\ x\amp=-7 \amp\amp\text{or}\amp x\amp=\frac{15}{2} \end{align*}

The solution set to the given equation is $\left\{-7,\frac{15}{2}\right\}\text{.}$

Example 2.2.3.

Determine the solution set to the equation $\abs{\frac{3-x}{7}}=4\text{.}$

Solution

We begin by writing a pair of equations (without absolute value expressions) that are collectively equivalent to the given equation. We then solve those equations.

\begin{align*} \frac{3-x}{7}\amp=-4 \amp\amp\text{or}\amp \frac{3-x}{7}\amp=4\\ \multiplyleft{7}\frac{3-x}{7}\amp=\multiplyleft{7}-4 \amp\amp\text{or}\amp \multiplyleft{7}\frac{3-x}{7}\amp=\multiplyleft{7}4\\ 3-x\amp=-28 \amp\amp\text{or}\amp 3-x\amp=28\\ 3-x\subtractright{3}\amp=-28\subtractright{3} \amp\amp\text{or}\amp 3-x\subtractright{3}\amp=28\subtractright{3}\\ -x\amp=-31 \amp\amp\text{or}\amp -x\amp=25\\ \multiplyleft{-1}-x \amp=\multiplyleft{-1}-31 \amp\amp\text{or}\amp \multiplyleft{-1}-x\amp=\multiplyleft{-1}25\\ x\amp=31 \amp\amp\text{or}\amp x\amp=-25 \end{align*}

The solutions set is $\{-25,31\}\text{.}$

Example 2.2.4.

Determine the solution set to the equation $\abs{6x-29}=0\text{.}$

Solution

We need to first make note that we do not have a positive constant on the non-absolute value side of the equation, so we need to proceed with caution. There is only one number whose absolute value is zero, and that's zero itself. So when we write an equivalent statement without absolute value bars, there will still be only one equation.

\begin{align*} \abs{6x-29}\amp=0\\ 6x-29\amp=0\\ 6x-29\addright{29}\amp=0\addright{29}\\ 6x\amp=29\\ \divideunder{6x}{6}\amp=\divideunder{29}{6}\\ x\amp=\frac{29}{6} \end{align*}

The solution set to the given equation is $\left\{\frac{29}{6}\right\}\text{.}$

Exercises Exercises

Find the solution set for each of the following equations.

1.

$\abs{x-6}=8$

Solution

We begin by rewriting a pair of equations that do not include absolute value expressions. We then solve those two equations.

\begin{align*} x-6\amp=-8 \amp\amp\text{or}\amp x-6\amp=8\\ x-6\addright{6}\amp=8\addright{6} \amp\amp\text{or}\amp x-6\addright{6}\amp=8\addright{6}\\ x\amp=-2 \amp\amp\text{or}\amp x\amp=14 \end{align*}

The solution set to the given equation is $\{-2,14\}$

2.

$\abs{2x+7}=21$

Solution

We begin by rewriting a pair of equations that do not include absolute value expressions. We then solve those two equations.

\begin{align*} 2x+7\amp=-21 \amp\amp\text{or}\amp 2x+7\amp=21\\ 2x+7\subtractright{7}\amp=-21\subtractright{7} \amp\amp\text{or}\amp 2x+7\subtractright{7}\amp=21\subtractright{7}\\ 2x\amp=-28 \amp\amp\text{or}\amp 2x\amp=14\\ \divideunder{2x}2\amp=\divideunder{-28}{2} \amp\amp\text{or}\amp \divideunder{2x}{2}\amp=\divideunder{14}{2}\\ x\amp=-14 \amp\amp\text{or}\amp x\amp=7 \end{align*}

The solution set to the given equation is $\{-14,7\}\text{.}$

3.

$\abs{3x-2}=-11$

Solution

We begin by observing that no number has an absolute value equal to $-11\text{.}$ We can therefore conclude that there are no solutions to the given equation and that the solution set is $\emptyset\text{.}$

4.

$\abs{3-5x}=30$

Solution

We begin by rewriting a pair of equations that do not include absolute value expressions. We then solve those two equations.

\begin{align*} 3-5x\amp=-30 \amp\amp\text{or}\amp 3-5x\amp=30\\ 3-5x\subtractright{3}\amp=-30\subtractright{3} \amp\amp\text{or}\amp 3-5x\subtractright{3}\amp=30\subtractright{3}\\ -5x\amp=-33 \amp\amp\text{or}\amp -5x\amp=27\\ \divideunder{-5x}{-5}\amp=\divideunder{-33}{-5} \amp\amp\text{or}\amp \divideunder{-5x}{-5}\amp=\divideunder{27}{-5}\\ x\amp=\frac{33}{5} \amp\amp\text{or}\amp x\amp=-\frac{27}{5} \end{align*}

The solution set is $\left\{\frac{33}{5},-\frac{27}{5}\right\}\text{.}$

5.

$\abs{9-x}=0$

Solution

We begin by observing that the only number whose absolute value is $0$ is $0$ itself. So the given equation is equivalent to the equation $9-x=0\text{.}$ Let's solve the new equation.

\begin{align*} 9-x\amp=0\\ 9-x\subtractright{9}\amp=0\subtractright{9}\\ -x\amp=-9\\ \multiplyleft{-1}-x\amp=\multiplyleft{-1}-9\\ x\amp=9 \end{align*}

The solution set is $\{9\}\text{.}$

6.

$\abs{-4x-9}=99$

Solution

We begin by rewriting a pair of equations that do not include absolute value expressions. We then solve those two equations.

\begin{align*} -4x-9\amp=-99 \amp\amp\text{or}\amp -4x-9\amp=99\\ -4x-9\addright{9}\amp=-99\addright{9} \amp\amp\text{or}\amp -4x-9\addright{9}\amp=99\addright{9}\\ -4x\amp=-90 \amp\amp\text{or}\amp -4x\amp=108\\ \divideunder{-4x}{-4}\amp=\divideunder{-90}{-4} \amp\amp\text{or}\amp \divideunder{-4x}{-4}\amp=\divideunder{108}{-4}\\ x\amp=\frac{45}{2} \amp\amp\text{or}\amp x\amp=-27 \end{align*}

The solution set is $\left\{\frac{45}{2},-27\right\}\text{.}$