# How do I solve the rational inequality (3x-2)/(x+2)<=1/3 using a TI-83?

Mar 8, 2016

$x \le 1$
We start with $\frac{3 x - 2}{x + 2} \le \frac{1}{3}$
The first thing I am going to do is get rid of the $x + 2$ in the denominator by multiplying both sides by $x + 2$ to give me $3 x - 2 \le \frac{1}{3} x + \frac{2}{3}$. Now I just add $2$ on both sides to arrive at $3 x \le \frac{1}{3} x + \frac{2}{3} + 2$, or $3 x \le \frac{1}{3} x + \frac{8}{3}$. From there I subtract $\frac{1}{3} x$ on both sides, which gives us $3 x - \frac{1}{3} x \le \frac{8}{3}$, which can be rewritten as $\frac{8}{3} x \le \frac{8}{3}$. Divide both sides by $\frac{8}{3}$, and we get $x \le 1$.