Absolute Value (in equations with extra terms)

You know that, in order to solve for a variable in an absolute value equation, you set the equation inside the absolute value to both the positive and negative versions of the answer.

What do you do when there are additional terms in the equation, outside of the absolute value expression?

Whenever an absolute value equation has additional terms, outside of the absolute value expression, you use algebra to get rid of those extra terms BEFORE you set the equation equal to both the positive and negative forms of the answer.

For any terms NOT inside the absolute value expression, you know exactly what the value is.  There is no need to try two different answers.  So, what you want to do, essentially, is treat the entire absolute value expression as a term that you are solving for.  Once you get that absolute value expression alone, you set it equal to the positive and negative version of the answer (see Absolute Value (in equations) for more details) and solve for the two possible answers to your equation. 

Think about this:

$\mid x + 3\mid -7= 5$

Before we can deal with the absolute value, we need to get rid of the $^-7$.

$$\eqalign{\mid x + 3\mid -7&= 5\\+7 \quad &\quad +7\\ \mid x+3 \mid & = 12}$$

Now, you have a basic absolute value equation, with an absolute value expression isolated on one side, and a value on the other.  Now, you can write two different equations, to solve for $x$.

$$\eqalign {\mid x + 3 \mid &= 12\\x+3&=12 \text{ or } ^-12}$$

$$\begin {array}{cc}\eqalign{\mathbf{x+3}&\mathbf{=12} \\ -3&=-3 \\x&=9 }\end {array}\qquad \qquad \begin{array}{cc}\eqalign{\mathbf{x+3}&\mathbf{=^-12}\\ -3&=-3\\x&=-15}\end{array}$$

$x=9 \text{ or } ^-15$

No matter how your additional terms are attached to your absolute value expression, you need to use algebra to get rid of them, and isolate your absolute value expression, before you write your two equations and solve for the variable. 

An example with an additional multiplication term:

$$\require{cancel}\eqalign {3 \mid 2x + 5 \mid &= 9  && \text{The absolute value expression is multiplied by 3}\\ \div 3 \quad & \quad \div 3 && \text{Divide by sides by 3 to isolate absolute value expression}\\ \mid 2x+5 \mid &= 3\\ \text{So,} \qquad 2x+5&=3 \text{ or } ^-3 && \text{Now, write your two equations}}$$

$$\begin {array}{cc}\eqalign{\mathbf{2x+5}&\mathbf{=3} \\ -5 & \text {}-5 \\2x&=-2\\\div 2 & \text{} \div 2\\x&=-1 }\end {array}\qquad \qquad \begin{array}{cc}\eqalign{\mathbf{2x+5}&\mathbf{=^-3}\\ -5&\text{}-5\\2x&=^-8\\\div 2 & \text{} \div 2\\x&=-4}\end{array}$$

$x=^-1 \text{ or } ^-4$

An example with subtraction and division:

$$\eqalign { \dfrac{\mid x-2 \mid}{3}-5 &= 12\\+5\quad&\quad +5 &&\text {Add 5 to each side to get rid of -5}\\\dfrac{\mid x-2 \mid}{3} &= 17\\\times 3 \quad & \quad \times 3 && \text {Multiply both sides by 3 to get rid of denominator}\\\mid x-2 \mid & = 51 && \text {Absolute value expression is isolated!}\\ x-2 &= 51 \text{ or } -51&&\text {Now, write your two equations}}$$

$$\begin {array}{cc}\eqalign{\mathbf{x-2}&\mathbf{=51} \\ +2&\text {}+2 \\2x&=53\\\div 2 & \text{} \div 2\\x&=\dfrac{53}{2} }\end {array}\qquad \qquad \begin{array}{cc}\eqalign{\mathbf{x-2}&\mathbf{=^-51}\\+2&\text{}+2\\2x&=^-49\\\div 2 & \text{} \div 2\\x&=-\dfrac{49}{2}}\end{array}$$

$x=\dfrac{53}{2} \text{ or } ^-\dfrac{49}{2}$

The rule with any equation that contains an absolute value: isolate the absolute value expression (just like it was a variable) and then, once you find the value of the absolute value expression, write your two (one positive, and one negative) equations, and solve.