MA001: College Algebra

Next, we will learn how to solve an absolute value equation. To solve an equation such as $|2 x-6|=8$, we notice that the absolute value will be equal to 8 if the quantity inside the absolute value bars is 8 or $-8$. This leads to two different equations we can solve independently.

$\begin{array}{rllr}2 x-6 & =8 & \text { or } & 2x-6&=-8 \\2 x & =14 && 2x&=-2 \\x & =7 && x&=-1\end{array}$

Knowing how to solve problems involving absolute value functions is useful. For example, we may need to identify numbers or points on a line that are at a specified distance from a given reference point.

ABSOLUTE VALUE EQUATIONS

The absolute value of $x$ is written as $|x|$. It has the following properties:

$\text{If} \, x \geq 0, \text{then} |x|=x$.

$\text{If} \, x < 0, \text{then} |x|=-x$.

For real numbers $A$ and $B$, an equation of the form $|A|=B$, with $B \geq 0$, will have solutions when $A=B$ or $A=-B$. If $B < 0$, the equation $|A|=B$ has no solution.

An absolute value equation in the form $|a x+b|=c$ has the following properties:

$\begin{array}{l} \text{If c < 0,|a x+b|=c has no solution.} \\ \text{If c = 0,|a x+b|=c has one solution.} \\ \text{If c > 0,|a x+b|=c has two solutions.} \end{array}$

HOW TO

Given an absolute value equation, solve it.

1. Isolate the absolute value expression on one side of the equal sign.

2. If $c > 0$, write and solve two equations: $a x+b=c$ and $a x+b=-c$.

EXAMPLE 8

Solving Absolute Value Equations

Solve the following absolute value equations:

(a) $|6 x+4|=8$

(b) $|3 x+4|=-9$

(c) $|3 x-5|-4=6$

(d) $|-5 x+10|=0$

Solution

(a)

$|6 x+4|=8$

Write two equations and solve each:

$\begin{array}{rlrl}6 x+4 & =8 & 6 x+4 & =-8 \\6 x & =4 & 6 x & =-12 \\x & =\dfrac{2}{3} & x & =-2\end{array}$

The two solutions are $\dfrac{2}{3}$ and $-2$.

b)

$|3 x+4|=-9$

There is no solution as an absolute value cannot be negative.

(c)

$|3 x-5|-4=6$

Isolate the absolute value expression and then write two equations.

$\begin{array}{rlrlr} && {|3 x-5|-4} =6 \\&& |3 x-5|=10 & \\3 x-5 & =10 & & 3 x-5=-10 \\3 x & =15 & & 3 x=-5 \\x & =5 & & x=-\dfrac{5}{3}\end{array}$

There are two solutions: $5$, and $-\dfrac{5}{3}$.

(d)

(d) $|-5 x+10|=0$

The equation is set equal to zero, so we have to write only one equation.

$\begin{aligned} -5 x+10 &=0 \\ -5 x &=-10 \\ x &=2 \end{aligned}$

There is one solution: $2$.

TRY IT #7

Solve the absolute value equation: $|1-4 x|+8=13$.

Source: Rice University, https://openstax.org/books/college-algebra/pages/2-6-other-types-of-equations
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