# How do you solve (2x-1)/(3x-2) <=1?

May 10, 2015

Solve $\frac{2 x - 1}{3 x - 2} \le 1$

Multiply both sides by $3 x - 2$

$\frac{\cancel{3 x - 2} \times 2 x - 1}{\cancel{3 x - 2}} \le 1 \times \left(3 x - 2\right)$ =

$2 x - 1 \le 3 x - 2$

Subtract $3 x$ from both sides.

$2 x - 1 - 3 x \le - 2$

Add $1$ to both sides.

$2 x - 3 x \le - 2 + 1$ =

$- x \le - 1$

Multiply by $- 1$

$x \ge 1$

May 10, 2015

$\frac{2 x - 1}{3 x - 2} \le 1$

Case 1:
If $3 x - 2 < 0$
which implies $\textcolor{red}{x < \frac{2}{3}}$
then
$2 x - 1 \ge 3 x - 2$
( since multiplying by a negative reverses the inequality)
and
$\textcolor{red}{x \le 1}$
Combining the Case 1 restrictions on $x$
$\textcolor{red}{x < \frac{2}{3}}$ and $\textcolor{red}{x \le 1}$
gives
$\textcolor{red}{x < \frac{2}{3}}$

Case2
If $3 x - 2 > 0$
which implies $\textcolor{b l u e}{x > \frac{2}{3}}$
then
$2 x - 1 \le 3 x - 2$
$\textcolor{b l u e}{x \ge 1}$
Combining the Case 2 restrictions on $x$
$\textcolor{b l u e}{x > \frac{2}{3}}$ and $\textcolor{b l u e}{x \ge 1}$
gives
$\textcolor{b l u e}{x \ge 1}$

Solution
$x < \frac{2}{3}$ or $x \ge 1$