Solving Absolute Value Equations and Inequalities

Absolute value equations and inequalities are solved by removing the absolute value sign and splitting into two cases. The key rule: $| P | = Q$ is equivalent to $P = Q$ or $P = - Q$ , and $| P | < c$ is equivalent to $- c < P < c$ , while $| P | > c$ is equivalent to $P > c$ or $P < - c$ .

Quick Reference

Expression	Equivalent Form	Notes
$\| P \| = c$ , $c > 0$	$P = c$ or $P = - c$	No solution if $c < 0$
$\| P \| = Q$ (variable RHS)	$P = Q$ or $P = - Q$	Check each solution: need $Q \geq 0$
$\| P \| < c$ , $c > 0$	$- c < P < c$	Solution is an interval
$\| P \| > c$ , $c > 0$	$P > c$ or $P < - c$	Solution is a union of intervals
$\leq$ , $\geq$ versions	Same as above with $\leq$ , $\geq$	Endpoints included

Absolute Value Equations

In Section: Absolute Value, we learned that

| x | = c > 0 is equivalent to x = \pm c

provided $c \geq 0$ . So to solve equations involving an absolute value follow these steps:

Isolate the absolute value expression on one side and the rest of terms on the other side. That is, rewrite the equation as

  |
  P
  |
  =
  Q

where 
  P
 and 
  Q
 are two expressions in 
  x
 [to indicate the dependence on 
  x
, we may write them as 
  P
  (
  x
  )
 and 
  Q
  (
  x
  )
].
Equate the expression inside the absolute value notation once with 
  +
 the quantity on the other side and once with 
  −
 the quantity on the other side:

  P
  =
  Q
  
  or
  
  P
  =
  −
  Q

Solve both equations.
Check your answers by substitution in the original equation.

When $Q < 0$ , the equation will not have a solution because always $| P | \geq 0$ . When $Q$ is an expression, we need to substitute the solutions in $Q$ to make sure that $Q \geq 0$ .

Solve each equation:

$| 2 x - 3 | = 5$
$| 5 x - 7 | + 9 = 0$

Solution

(a)

| 2 x - 3 | = 5 \Leftrightarrow 2 x - 3 = \pm 5

\begin{aligned} 2x-3 & =5\Rightarrow x=4\\ 2x-3 & =-5\Rightarrow x=-1 \end{aligned}

So the solutions are $x = 4$ and $x = - 1$ . We can check to see these values satisfy the equation, but it is not necessary because the right hand side is a positive number.

(b)

| 5 x - 7 | = - 9

Because always $| 5 x - 7 | \geq 0$ and $- 9 < 0$ , this equation does not have a solution.

Solve each equation:

$| 3 - 2 x | + 5 x = 18$
$| 4 x + 3 | + 3 x = 10$

Solution

(a) We isolate the absolute value expression on one side:

|3-2x|=18-5x\tag{i}

which is equivalent to

3 - 2 x = 18 - 5 x or 3 - 2 x = - 18 + 5 x .

Solving each equation:

3 - 2 x = 18 - 5 x \Rightarrow 3 x = 15 \Rightarrow x = 5

3 - 2 x = - 18 + 5 x \Rightarrow 7 x = 21 \Rightarrow x = 3

If we substitute $5$ for $x$ in $18 - 5 x$ it becomes negative. Because the RHS of (i) is negative for $x = 5$ , and the LHS is nonnegative, $x = 5$ cannot be a solution. But if we substitute 3 for $x$ in $18 - 5 x$ , it becomes $18 - 15 = 3 > 0$ , so the RHS and the LHS of (i) are both nonnegative and $x = 3$ is the only solution.

Alternatively, we can substitute $x = 5$ and $x = 3$ in the original equation and check if they satisfy the equation.

(b) Similar to (a)

|4x+3|=10-3x\tag{ii}

\Leftrightarrow 4 x + 3 = \pm (10 - 3 x)

We have to solve two equations:

(1) 4 x + 3 = 10 - 3 x \Rightarrow 7 x = 7 \Rightarrow x = 1

(2) 4 x + 3 = - 10 + 3 x \Rightarrow x = - 13

Substituting $1$ for $x$ in RHS of (ii) ( $10 - 3 x$ ) gives $7$. Because the LHS and RHS of (ii) are nonnegative, $x = 1$ is a solution. Substituting $- 13$ for $x$ in the RHS of (ii) gives a positive number, so $x = - 13$ is another solution. Therefore, the solutions are $x = 1$ and $x = - 13$ .

Alternatively, we can substitute $x = 1$ and $x = - 13$ in the original equation and see which one satisfies the equation.

When there are more than one absolute value, for example when we have

| P | + | R | = Q

where $P, Q$ , and $R$ are some expressions, the above technique may not work. In such cases, we need to find where $P$ and $R$ are positive and where they are negative and then solve the equation in the same way that we solve regular equations.

Solve $| 2 x + 4 | + 4 | 14 - 3 x | = 30$ .

Solution

Using the definition of the absolute value

|2x+4|=\begin{cases} 2x+4 & x\ge-2\\ -(2x+4) & x<-2 \end{cases} |14-3x|=|3x-14|=\begin{cases} 3x-14 & x\ge\frac{14}{3}\approx4.67\\ 14-3x & x<\frac{14}{3} \end{cases}

$x$	$- \infty < x \leq - 2$	$- 2 \leq x \leq \frac{14}{3}$	$\frac{14}{3} \leq x < \infty$
$\| 2 x + 4 \| =$	$- 2 x - 4$	$2 x + 4$	$2 x + 4$
$4\|14-3x\|=$	$4 (14 - 3 x)$	$4 (14 - 3 x)$	$4 (3 x - 14)$
$\| 2 x + 4 \| + 4 \| 14 - 3 x \| =$	$- 14 x + 52$	$- 10 x + 60$	$14 x - 52$

When $x \geq 14 / 3 \approx 4.67$ :

\begin{aligned} |2x+4|+4|14-3x|=14x-52 & =30\\ 14x-52 & =30\\ 14x & =82\\ x & =\frac{82}{14}=\frac{41}{7}\approx5.86 \end{aligned}

Because $x = 41 / 7$ lies in the interval $[14 / 3, \infty)$ , it is consistent with our assumption that $x \geq 14 / 3$ and thus $x = 41 / 7$ is a solution.

When $- 2 \leq x \leq \frac{14}{3}$ :

\begin{aligned} |2x+4|+4|14-3x|=-10x+60 & =30\\ -10x+60 & =30\\ -10x & =-30\\ x & =3 \end{aligned}

Because $x = 3$ lies in the interval $[- 2, \frac{14}{3}]$ , it is consistent with our assumption that $- 2 \leq x < 14 / 3$ and thus $x = 3$ is a solution.

When $x < - 2$ :

\begin{aligned} |2x+4|+4|14-3x|=-14x+52 & =30\\ -14x+52 & =30\\ -14x & =-22\\ x & =\frac{11}{7} \end{aligned}

Because $x = 11 / 7$ does not lie in the interval $(- \infty, - 2]$ , it is not consistent with our assumption that $x \leq - 2$ , and thus $x = 11 / 7$ cannot be a solution. Therefore, the solution set is

{3, \frac{41}{7}} .

It does not matter in which interval we consider the endpoints because

at x = - 2 - 14 x + 52 = - 10 x + 60 = 80

and

at x = 14 / 3 - 10 x + 60 = 14 x - 52 = 40 / 3.

Absolute Value Inequalities

To solve absolute value inequalities, recall (see here)

$| x | < c$ is equivalent to $- c < x < c$ .
$| x | > c$ is equivalent to $x > c$ or $x < - c$

where $c$ is a positive number.

The above equivalent statements hold true if we replace $<$ by $\leq$ and $>$ by $\geq$ .

Solve each of the following inequalities:

$| x - 3 | < 5$
$| 5 - 4 x | > 6$

Solution

(a) The inequality $| x - 3 | < 5$ is equivalent to

- 5 < x - 3 < 5

-2

The solution set is the interval $(- 2, 8)$ .

(b) The inequality $| 5 - 4 x | > 6$ is equivalent to

5 - 4 x > 6 or 5 - 4 x < - 6

-4x>1\qquad\text{or}\qquad-4x<-11\tag{subtract 5} x<-\frac{1}{4}\qquad\text{or}\qquad x>\frac{11}{4}\tag{divide by $-4$}

For the last step, recall that when we divide both sides of an inequality by a negative number, the direction of the inequality changes.

The solution set is

{x ∣ x < - \frac{1}{4} or x > \frac{11}{4}} = (- \infty, - \frac{1}{4}) \cup (\frac{11}{4}, \infty) .

Frequently Asked Questions

What does

| P | = Q

mean and how do you solve it?

The equation

| P | = Q

means that

P

is a distance of

Q

from zero on the number line. It splits into two cases:

P = Q

P = - Q

. If

Q < 0

, there is no solution because absolute values are always non-negative. If

Q

contains a variable, substitute each solution back to verify

Q \geq 0

Why do you get two equations when solving an absolute value equation?

Because

| P | = Q

means

P

is either

+ Q

away from zero or

- Q

away from zero. Both cases satisfy the original equation, so both must be considered and solved. After solving, check that each solution makes

Q \geq 0

in the original equation.

What is the difference between

| P | < c

and

| P | > c

For

| P | < c

(with

c > 0

): the solution is a single interval

- c < P < c

, meaning

P

is within distance

c

of zero. For

| P | > c

: the solution is a union of two intervals

P > c

P < - c

, meaning

P

is farther than distance

c

from zero. The same rules apply with

\leq

and

\geq

Why must you check solutions when the right-hand side contains a variable?

When solving

| P | = Q

where

Q

is an expression in

x

, splitting gives

P = Q

P = - Q

. But the original equation also requires

Q \geq 0

(since the left side is non-negative). A solution that makes

Q

negative is extraneous and must be discarded.

How do you handle an equation with more than one absolute value?

When two absolute values appear, such as

| P | + | R | = Q

, determine the critical points where each expression inside the absolute values equals zero. These points divide the number line into intervals. In each interval, both expressions have fixed signs, so you can replace the absolute values with signed expressions and solve an ordinary equation. Check that each solution lies in the interval from which it came.

Solving Absolute Value Equations and Inequalities

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Absolute Value Equations

Absolute Value Inequalities

Frequently Asked Questions

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