How do I solve the rational inequality $\frac{x + 10}{3 x - 2} \leq 3$ $(x+10)/(3x-2)<=3$ using a TI-84?

Question

5904 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-28T18:23:11

$x \in (- \infty, \frac{2}{3}) \cup [2, \infty)$ $x\in(-\infty, 2/3)\cup[2, \infty)$

Given inequality

$\frac{x + 10}{3 x - 2} \leq 3$ ${x+10}/{3x-2}\le 3$

$\frac{x + 10}{3 x - 2} - 3 \leq 0$ ${x+10}/{3x-2}-3\le 0$

$\frac{- 8 (x - 2)}{3 x - 2} \leq 0$ $\frac{-8(x-2)}{3x-2}\le 0$

$\frac{x - 2}{3 x - 2} \geq 0$ $\frac{x-2}{3x-2}\ge 0$

Setting $x - 2 = 0 \Rightarrow x = 2$ $x-2=0\implies x=2$ &

$3 x - 2 = 0 \Rightarrow x = \frac{2}{3}$ $3x-2=0\implies x=2/3$

Specifying the critical points on the number line & dividing in the positive & negative intervals. It gives the solution

$x \in (- \infty, \frac{2}{3}) \cup [2, \infty)$ $x\in(-\infty, 2/3)\cup[2, \infty)$

How do I solve the rational inequality x+103x−2≤3(x+10)/(3x-2)<=3 using a TI-84?