# 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.

$\mid x + 3\mid = 5$

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}

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

• ## Absolute Value (in Equations)

Find the value(s) of $x$:

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