# How do you solve |\frac { 2x - 3} { 4} | \leq 2?

Sep 27, 2017

See a solution process below:

#### Explanation:

The absolute value function takes any term and transforms it to its non-negative form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.

$- 2 \le \frac{2 x - 3}{4} \le 2$

First, multiply each segment of the system of inequalities by $\textcolor{red}{4}$ to eliminate the fraction while keeping the system balanced:

$\textcolor{red}{4} \times - 2 \le \textcolor{red}{4} \times \frac{2 x - 3}{4} \le \textcolor{red}{4} \times 2$

$- 8 \le \cancel{\textcolor{red}{4}} \times \frac{2 x - 3}{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}}} \le 8$

$- 8 \le 2 x - 3 \le 8$

Next, add $\textcolor{red}{3}$ to each segment to isolate the $x$ term while keeping the system balanced:

$- 8 + \textcolor{red}{3} \le 2 x - 3 + \textcolor{red}{3} \le 8 + \textcolor{red}{3}$

$- 5 \le 2 x - 0 \le 11$

$- 5 \le 2 x \le 11$

Now, divide each segment by $\textcolor{red}{2}$ to solve for $x$ while keeping the system balanced:

$- \frac{5}{\textcolor{red}{2}} \le \frac{2 x}{\textcolor{red}{2}} \le \frac{11}{\textcolor{red}{2}}$

$- \frac{5}{2} \le \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} x}{\cancel{\textcolor{red}{2}}} \le \frac{11}{2}$

$- \frac{5}{2} \le x \le \frac{11}{2}$

Or

$x \ge - \frac{5}{2}$ and $x \le = \frac{11}{2}$

Or, in interval notation:

$\left[- \frac{5}{2} , \frac{11}{2}\right]$