# How do you solve (1/4)(3x + 2) ≤ (2/3)(-3 + 3x)?

May 12, 2015

First notice that the right hand side can be rearranged as follows:

$\left(\frac{2}{3}\right) \left(- 3 + 3 x\right)$

$= \left(\frac{2}{3}\right) \left(3 \cdot \left(- 1\right) + 3 \cdot x\right)$

$= \left(\frac{2}{3}\right) \cdot 3 \left(- 1 + x\right)$

$= 2 \left(x - 1\right)$

$= 2 x - 2$

So we have:

$\left(\frac{1}{4}\right) \left(3 x + 2\right) \le 2 x - 2$

Multiply both sides by 4 to get:

$3 x + 2 \le 8 x - 8$

Add 8 to both sides to get:

$3 x + 10 \le 8 x$

Subtract $3 x$ from both sides to get:

$10 \le 5 x$

Divide both sides by 5 to get:

$2 \le x$

Or to put it another way:

$x \ge 2$