# Absolute Value (in equations)

You have already learned that the absolute value of a variable tells you the value of the variable, but not whether the variable is positive or negative (and, without more information, you have to assume that it could be either).

**Any time you solve for a variable whose absolute value is given, your answer will be two possible answers**, as in the following equation (to review, go to the Absolute Value (with variables) lesson).

$$\eqalign{\mid x \mid &= 6\\x&=6 \text{ or } ^-6}$$

The same general rules apply when you are working with the absolute value of a variable expression. You know what the expression equals, but because absolute values always have a positive answer, **you don't know if the non-absolute value answer is actually positive or negative**. Thus, you have to solve the equation twice: once as if the answer was positive and once as if it was negative. That will give you two possible answers for your variable.

Think about this:

$\mid x + 3\mid = 5$

We know that $x+3$ equals either $5$ or $^-5$, but we don't know which. In order to solve for $x$, we will write **two equations** (one with the positive answer and one with the negative answer) and then solve both equations.

$$\eqalign {\mid x + 3 \mid &= 5\\x+3&=5 \text{ or } ^-5}$$

$$\begin {array}{cc}\eqalign{\mathbf{x+3}&\mathbf{=5} \\ -3&=-3 \\x&=2 }\end {array}\qquad \qquad \begin{array}{cc}\eqalign{\mathbf{x+3}&\mathbf{=^-5}\\ -3&=-3\\x&=-8}\end{array}$$

$x=2 \text{ or } ^-8$

The two-equation strategy works whenever you have to solve for a variable that is between absolute value signs. We like to write the two equations side by side so that we don't forget to do one of them!

*An example with addition and multiplication:*

$$\eqalign {\mid 2x + 5 \mid &= 9\\2x+5&=9 \text{ or } ^-9}$$

$$\begin {array}{cc}\eqalign{\mathbf{2x+5}&\mathbf{=9} \\ -5&\text {}-5 \\2x&=4\\\div 2 & \text{} \div 2\\x&=2 }\end {array}\qquad \qquad \begin{array}{cc}\eqalign{\mathbf{2x+5}&\mathbf{=^-9}\\ -5&\text{}-5\\2x&=^-14\\\div 2 & \text{} \div 2\\x&=-7}\end{array}$$

*$x=2 \text{ or } ^-7$*

*An example with subtraction and division:*

$$\eqalign {\mid \dfrac{x}{3}-5 \mid &= 7\\\dfrac{x}{3}-5&=7 \text{ or } ^-7}$$

$$\begin {array}{cc}\eqalign{\mathbf{\dfrac{x}{3}-5}&\mathbf{=7} \\ +5&\text {}+5 \\\dfrac{x}{3}&=12\\\times 3 & \text{} \times 3\\x&=36 }\end {array}\qquad \qquad \begin{array}{cc}\eqalign{\mathbf{\dfrac{x}{3}-5}&\mathbf{=^-7}\\ +5&\text{}+5\\\dfrac{x}{3}&=^-2\\\times 3 & \text{} \times 3\\x&=-6}\end{array}$$

$x=36 \text{ or } ^-6$

#### Practice Problems:

## Absolute Value (in Equations)

Find the value(s) of $x$:

- $|x+1|=2$
- $|x-5|=-5$
- $|x-5|=5$
- $|5x|=10$
- $|2x|=3$
- $|\dfrac{x}{2}|=6$
- $|\dfrac{x}{3}|=10$
- $|\dfrac{1}{x}|=2$
- $|3x+4|=12$
- $|12x-7|=20$
- $|\dfrac{2x+1}{2}|=10$
- $|\dfrac{9x-10}{5}|=3$

#### Answer Key: