Solve for $x$ in the following inequality: $-5 \leq \frac{2-3x}{4} \leq 9$

(N/A) We have,$-5 \leq \frac{2-3x}{4} \leq 9$.
Multiplying throughout by $4$,we get:
$-20 \leq 2-3x \leq 36$.
Subtracting $2$ from all parts:
$-20-2 \leq -3x \leq 36-2$.
$-22 \leq -3x \leq 34$.
Dividing by $-3$ (and reversing the inequality signs):
$\frac{-22}{-3} \geq x \geq \frac{34}{-3}$.
$\frac{22}{3} \geq x \geq -\frac{34}{3}$.
Thus,$-\frac{34}{3} \leq x \leq \frac{22}{3}$.
In interval notation,$x \in \left[-\frac{34}{3}, \frac{22}{3}\right]$.