# How do you solve Solve: 1/2x - 3 <=-1/1.5x - .5 <=1/2x + 2?

May 26, 2015

Given $\frac{1}{2} x - 3 \le - \frac{1}{1.5} x = .5 \le \frac{1}{2} x + 2$

Simplify by multiplying each expression by $6$ to clear the fractions (Remember that you can multiply by any value greater than zero without effecting the orientation of the inequalities).
$3 x - 9 \le - 4 x - 3 \le 3 x + 12$

Break this up into two compound inequalities:
$\text{[1] } 3 x - 9 \le - 4 x - 3$
and
$\text{[2] } - 4 x - 3 \le 3 x + 12$

Evaluate Inequality [1]
$3 x - 9 \le - 4 x - 3$
$\textcolor{w h i t e}{\text{MMMMMM}}$Add $\left(4 x + 9\right)$ to both sides:
$7 x \le 6$
$\textcolor{w h i t e}{\text{MMMMMM}}$Divide by $7$ (which does not change the inequality orientation)
$x \le \frac{6}{7}$

Evaluate Inequality [2]
$- 4 x - 3 \le 3 x + 12$
$\textcolor{w h i t e}{\text{MMMMMM}}$Add $\left(4 x - 12\right)$ to both sides
$- 15 \le 7 x$
$\textcolor{w h i t e}{\text{MMMMMM}}$Divide both sides by 7
$\left(- \frac{15}{7}\right) \le x$

Combine the Compound Inequalities (with and)
$x \le \frac{6}{7}$ and $x \ge \left(- \frac{15}{7}\right)$
$\textcolor{w h i t e}{\text{MMMMMM}}$Re-write as a single compound statement
$\left(- \frac{15}{7}\right) \le x \le \frac{6}{7}$