1.22 Absolute Value Equations and Inequalities

Table of Contents

Equations

In Section 1.5, we learned that
$| x | = c \geq 0 \Leftrightarrow x = \pm c$
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 $Q \geq 0$ .

Example 1

Solve each equation:
(a)

| 2 x - 3 | = 5

(b)

| 5 x - 7 | + 9 = 0.

Solution

(a)

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

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

Example 2

Solve each equation:
(a)

| 3 - 2 x | + 5 x = 18

(b)

| 4 x + 3 | + 3 x = 10

Solution

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

$\begin{matrix} (i) & | 3 - 2 x | = 18 - 5 x \end{matrix}$ which is equivalent to
$Extra close brace or missing open brace$ Solving each equation:
$3 - 2 x = 18 - 5 x \Rightarrow 3 x = 15 \Rightarrow x = 5$ or
$3 - 2 x = - 18 + 5 x \Rightarrow 7 x = 21 \Rightarrow x = 3$ If we substitute $5$ for $x$ in $18 - 5 x$ it becomes $18 - 25 < 0$ . Because the RHS of (i) is negative 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)
$\begin{matrix} (ii) & | 4 x + 3 | = 10 - 3 x \end{matrix}$ $\Leftrightarrow 4 x + 3 = \pm (10 - 3 x)$ 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.

Example 3

Solve

| 2 x + 4 | + 4 | 14 - 3 x | = 30

Solution

Using the definition of the absolute value

$| 2 x + 4 | = {\begin{cases} 2 x + 4 & x \geq - 2 \\ - (2 x + 4) & x < - 2 \end{cases}$ $| 14 - 3 x | = | 3 x - 14 | = {\begin{cases} 3 x - 14 & x \geq \frac{14}{3} \approx 4.67 \\ 14 - 3 x & x < \frac{14}{3} \end{cases}$

When $x \geq 14 / 3 \approx 4.67$ :
$\begin{aligned} | 2 x + 4 | + 4 | 14 - 3 x | = 14 x - 52 & = 30 \\ 14 x - 52 & = 30 \\ 14 x & = 82 \\ x & = \frac{82}{14} = \frac{41}{7} \approx 5.86 \end{aligned}$
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 $- 2 \leq x < \frac{14}{3}$ :
$\begin{aligned} | 2 x + 4 | + 4 | 14 - 3 x | = - 10 x + 60 & = 30 \\ - 10 x + 60 & = 30 \\ - 10 x & = - 30 \\ x & = 3 \end{aligned}$
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 $x < - 2$ :
$\begin{aligned} | 2 x + 4 | + 4 | 14 - 3 x | = - 14 x + 52 & = 30 \\ - 14 x + 52 & = 30 \\ - 14 x & = - 22 \\ x & = \frac{11}{7} \end{aligned}$

Because $x = 11 / 7$ does not lie in the interval $(- \infty, - 2)$ , it is \uline{not} consistent with our assumption that $x < - 2$ , 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
$at x = 14 / 3 - 10 x + 60 = 14 x - 52 = 40 / 3.$

Inequalities

To solve absolute value inequalities, recall (see Section 1.4):

$| x | < c$ is equivalent to $- c < x < c$ .
$| x | > c$ is equivalent $x > c$ or $x < - c$
where $c$ is a positive number.

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

Example 4

Solve each of the following inequalities
(a)

| x - 3 | < 5

(b)

| 5 - 4 x | > 6.

Solution

(a) The inequality

| x - 3 | < 5

is equivalent to

- 5 < x - 3 < 5

Add 3 to each side:
$- 2 < x < 8$ The solution set is the interval $(- 2, 8)$ .
(b) The inequality $| 5 - 4 x | > 6$ is equivalent to
$Extra close brace or missing open brace$

Subtract 5 from each side:
$Extra close brace or missing open brace$

Divide each term by $- 4$ :
$x < - \frac{1}{4} or x > \frac{1}{4}$ [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
${x | x < - \frac{1}{4} or x > \frac{1}{4}} = (- \infty, - \frac{1}{4}) \cup (\frac{1}{4}, \infty) .$