
# 1.7: Absolute Value Equations and Inequalities


## Solving an Absolute Value Equation

Recall that the absolute value of a real number $$a$$, denoted $$|a|$$, is defined as the distance between zero (the origin) and the graph of that real number on the number line. For example, $$|−3|=3$$ and $$|3|=3$$.

In addition, the absolute value of a real number can be defined algebraically as a piecewise function.

$$| a | = \left\{ \begin{array} { l } { a \text { if } a \geq 0 } \\ { - a \text { if } a < 0 } \end{array} \right.$$

Given this definition, $$|3| = 3$$ and $$|−3| = − (−3) = 3$$.Therefore, the equation $$|x| = 3$$ has two solutions for $$x$$, namely $$\{±3\}$$.

Next, To solve an equation such as $$|2x−6|=8$$, 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 can be solved independently.

$$|2x−6|=8$$
$$\begin{array} {r|r} 2x-6= 8 & 2x-6= -8 \\ 2x= 14 & 2x= -2 \\ x= 7 & x= -1 \end{array}$$

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.

Definition is a green note

Definition: Absolute Value Equations

The absolute value of any algebraic expression $$u$$, is written as $$|u|$$ and is never negative, although the argument $$u$$ inside the absolute value bars can be either positive or negative.
For any real number $$c$$, the absolute value equation $$| u | = c$$ has the following properties

• If $$c<0$$, then $$|u|=c$$ has no solution.
• If $$c=0$$, then $$|u|=c$$ has one solution.
• If $$c>0$$, then $$|u|=c$$ has two solutions.

Howto: Solve an Absolute Value Equation.

1. Isolate the absolute value expression $$|u|$$  on one side of the equal sign, producing an equation of the form  $$|u|=c$$
2. If $$c>0$$, write and solve two equations: $$u=c$$ and $$u=−c$$.
If $$c=0$$, solve the single equation: $$u=0$$
If  $$c<0$$, the equation has no solution.

Example $$\PageIndex{1}$$: Solving Absolute Value Equations

Solve the following absolute value equations:

1. $$|6x+4|=8$$
2. $$|3x+4| + 7=1$$
3. $$| 2 x + 3 |-1 = 3$$
4. $$2 |5x − 1| − 3 = 9$$.
5. $$|−5x+10| + 3=3$$

Solution:

1. The absolute value expression is isolated already, so rewrite it as two separate equations and solve them.

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

$$\quad$$ The solution set is $$\left\{ \tfrac{2}{3}, -2 \right\}$$

1. Isolate the absolute value expression in the equation.

$$|3x+4| + 7=1$$
$$|3x+4| =-6$$

$$\quad$$ An absolute value cannot be negative. Therefore the equation does not have a solution.
$$\quad$$ The solution set is $$\{ \:\: \}$$

1. Isolate the absolute value expression in the equation.

$$| 2 x + 3 | -1 = 3$$
$$| 2 x + 3 | = 4$$

$$\quad$$ Rewrite the absolute value as two separate equations and solve them.

$$\begin{array} {r|r} 2 x + 3 = - 4 & 2 x + 3 = 4 \\ 2 x = - 7 & 2 x = 1 \\ - \frac { 7 } { 2 } & x = \frac { 1 } { 2 } \\ \end{array}$$

$$\quad$$ The solution set is $$\{ - \frac { 7 } { 2 }, \frac { 1 } { 2 } \}$$

1. Isolate the absolute value expression in the equation.

$$2 |5x − 1| − 3 = 9$$
$$2 |5x − 1| = 12$$
$$|5x − 1| = 6$$

$$\quad$$ Rewrite the absolute value as two separate equations and solve them.

$$\begin{array} {r|r} 5 x - 1 = - 6 & 5 x - 1 = 6 \\ 5 x = - 5 & 5 x = 7 \\ x = - 1 & x = \frac { 7 } { 5 } \\ \end{array}$$

$$\quad$$ The solution set is $$\{ -1, \frac { 7 } { 5 } \}$$

1. Isolate the absolute value expression in $$|−5x+10| + 3=3$$.

$$|−5x+10| + 3=3$$
$$|−5x+10| =0$$

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

\begin{align*} -5x+10&= 0\\ -5x&= -10\\ x&= 2 \end{align*}

$$\quad$$ The solution set is $$\{ 2 \}$$

Try It $$\PageIndex{1}$$

Solve.

1. $$|1−4x|+8=13$$.
2. $$2 - 7 | x + 4 | = - 12$$.

a, $$x=−1, x=\dfrac{3}{2}$$ $$\qquad$$ b. $$-6, -2$$

If two absolute value expressions are equal, then the arguments can either be the exact same value, or be the exact same value but with opposite signs.

Example $$\PageIndex{2}$$:

Solve: $$| 2 x - 5 | = | x - 4 |$$.

Solution

Set $$2x-5$$ equal to $$\pm ( x - 4 )$$ and then solve each linear equation.

$$\begin{array} { c } { | 2 x - 5 | = | x - 4 | } \\ { 2 x - 5 = - ( x - 4 ) \:\: \text { or }\:\: 2 x - 5 = + ( x - 4 ) } \\ { 2 x - 5 = - x + 4 }\quad\quad\quad 2x-5=x-4 \\ { 3 x = 9 }\quad\quad\quad\quad\quad\quad \quad\quad x=1 \\ { x = 3 \quad\quad\quad\quad\quad\quad\quad\quad\quad\:\:\:\:} \end{array}$$

The solution set is $$\{ 1, 3 \}$$

Try It $$\PageIndex{2}$$

Solve: $$| x + 10 | = | 3 x - 2 |$$.

$$-2, 6$$

## Absolute Value Inequalities

Consider the solutions to the inequality $$| x | \leq 3$$. The absolute value of a number represents the distance from the origin. Therefore, this inequality describes all numbers whose distance from zero is less than or equal to $$3$$. We can graph this solution set by shading all such numbers.

This solution set is expressed using set notation or interval notation as follows:

$$\begin{array} { c } { \{ x | - 3 \leq x \leq 3 \} \color{Cerulean} { Set\: Notation } } \\ { [ - 3,3 ] \quad \color{Cerulean}{ Interval \:Notation } } \end{array}$$

In contrast, examine solutions to the inequality $$| x | \geq 3$$. This inequality describes all numbers whose distance from the origin is greater than or equal to $$3$$. On a graph, we can shade all such numbers.

This solution set is expressed using using set notation and interval notation as follows:

$$\begin{array} { l } { \{ x | x \leq - 3 \text { or } x \geq 3 \} \:\:\color{Cerulean} { Set\: Notation } } \\ { ( - \infty , - 3 ] \cup [ 3 , \infty ) \:\:\color{Cerulean} { Interval\: Notation } } \end{array}$$

Howto: Solve a Linear Absolute Value Inequality.

Given an algebraic expression $$u$$ and a positive number $$c$$,

1. Isolate the absolute value expression $$|u|$$  on left side of the inequality symbol
2. The solution of $$| u | < c$$ is a single interval composed of the numbers that satisfy the compound inequality $$-c < u < c$$.
3. The solution of $$| u | > c$$ is a combination of two separate intervals that satisfy the two separate inequalities $$u < -c$$ or $$u > c$$. NOTE that  $$| u | > c$$ CANNOT be written as $$-c < u > c$$.
4. These rules are valid if $$<$$ is replaced with $$\le$$ and $$>$$ is replaced with $$\ge$$.

Example $$\PageIndex{3}$$:

Solve and graph the solution set: $$|x+2|<3$$.

Solution

The absolute value expression is already isolated. Bound the argument $$x+2$$ by $$−3$$ and $$3$$ in a compound inequality and solve.

$$\begin{array} { c } { | x + 2 | < 3 } \\ { - 3 < x + 2 < 3 } \\ { - 3 \color{Cerulean}{- 2}\color{Black}{ <} x + 2 \color{Cerulean}{- 2}\color{Black}{ <} 3 \color{Cerulean}{- 2} } \\ { - 5 < x < 1 } \end{array}$$

Here we use open dots to indicate strict inequalities on the graph as follows.

Using interval notation, $$(−5,1)$$.

Example $$\PageIndex{4}$$:

Solve: $$4 |x + 3| − 7 ≤ 5$$.

Solution

Isolate the absolute value expression.

$$\begin{array} { c } { 4 | x + 3 | - 7 \leq 5 } \\ { 4 | x + 3 | \leq 12 } \\ { | x + 3 | \leq 3 } \end{array}$$

Rewrite the absolute value inequality as a compound inequality and solve.

$$\begin{array} { c } { - 3 \leq x + 3 \leq 3 } \end{array}$$

\begin{aligned} - 3 \color{Cerulean}{- 3} \color{Black}{ \leq} x + 3 \color{Cerulean}{- 3} & \color{Black}{ \leq} 3 \color{Cerulean}{- 3} \\ - 6 \leq x \leq 0 \end{aligned}

Shade the solutions on a number line and present the answer in interval notation. Here we use closed dots to indicate inclusive inequalities on the graph as follows:

Using interval notation, $$[−6,0]$$

Try It $$\PageIndex{4}$$

Solve and graph the solution set: $$3 + | 4 x - 5 | < 8$$.

Interval notation: $$(0, \frac{5}{2})$$

Example $$\PageIndex{5}$$:

Solve and graph the solution set: $$|x+2|>3$$.

Solution

The absolute value expression is already isolated. The argument $$x+2$$ must be less than $$−3$$ or greater than $$3$$.

$$\begin{array} { c } { | x + 2 | > 3 } \\ { x + 2 < - 3 \quad \text { or } \quad x + 2 > 3 } \\ { x < - 5 }\quad\quad\quad\quad\quad\: x>1 \end{array}$$

Using interval notation, $$(−∞,−5)∪(1,∞)$$.

Example $$\PageIndex{6}$$:

Solve: $$13 - 2 |4x − 7| ≤ 3$$.

Solution

Isolate the absolute value expression. Notice in the last step that division by $$-2$$ changes the inequality symbol.

$$\begin{array} { r } { 13 - 2 | 4 x - 7 | \leq 3 } \\ { -2 | 4 x - 7 | \leq -10 } \\ { | 4 x - 7 | \geq 5 } \end{array}$$

Next, apply the theorem and rewrite the absolute value inequality as a compound inequality with two separate intervals and solve.

$$\begin{array} 4 x - 7 \leq - 5 \quad \text { or } \quad 4 x - 7 \geq 5 \end{array}$$

$$\begin{array} { l } \quad\:\:{ 4 x \leq 2 } \quad\quad\quad\:\:\: 4x\geq 12\\ \quad\:\:{ x \leq \frac { 2 } { 4 } } \quad\quad\quad\quad x\geq \frac { 12 } { 4 } \\ \quad\:\:{ x \leq \frac { 1 } { 2 } } \quad\quad\quad\quad x\geq 3 \end{array}$$

Shade the solutions on a number line and present the answer using interval notation.

Using interval notation, $$(−∞, \tfrac { 1 } { 2 }]∪[3,∞)$$

Try It $$\PageIndex{6}$$

Solve and graph: $$3 | 6 x + 5 | - 2 > 13$$.

Using interval notation, $$\left( - \infty , - \frac { 5 } { 3 } \right) \cup ( 0 , \infty )$$

Up to this point, the solution sets of linear absolute value inequalities have consisted of a single bounded interval or two unbounded intervals. This is not always the case.

Example $$\PageIndex{7}$$:

Solve and graph: $$|2x−1|+5>2$$.

Solution

Begin by isolating the absolute value.

$$\begin{array} { c } { | 2 x - 1 | + 5 > 2 } \\ { | 2 x - 1 | > - 3 } \end{array}$$

Notice that we have an absolute value greater than a negative number. For any real number x the absolute value of the argument will always be positive. Hence, any real number will solve this inequality.

All real numbers, $$ℝ$$.

Example $$\PageIndex{8}$$:

Solve and graph: $$|x+1|+4≤3$$.

Solution

Begin by isolating the absolute value.

$$\begin{array} { l } { | x + 1 | + 4 \leq 3 } \\ { | x + 1 | \leq - 1 } \end{array}$$

In this case, we can see that the isolated absolute value is to be less than or equal to a negative number. Again, the absolute value will always be positive; hence, we can conclude that there is no solution.

Answer: $$Ø$$ or $$\{ \:\: \}$$

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