# How do you solve 4 > -3x + 3 or 8 ≤ -2x + 5?

Jun 28, 2015

$4 > - 3 x + 3 \rightarrow x > - \frac{1}{3}$

$8 \le - 2 x + 5 \rightarrow x \le - \frac{3}{2}$

#### Explanation:

Solving an inequation is the same as solving an equation, except that the orientation of the inequality will change when you multiply or divide by a negative number.

Let's start with $4 > - 3 x + 3$

$\rightarrow 4 - 3 > - 3 x + \cancel{3} - \cancel{3}$
$\rightarrow \frac{1}{3} > - \frac{\cancel{3} x}{\cancel{3}}$
$\rightarrow - \frac{1}{3} < - \left(- x\right) \rightarrow x > - \frac{1}{3}$

Now $8 \le - 2 x + 5$

$\rightarrow 8 - 5 \le - 2 x + \cancel{5} - \cancel{5}$
$\rightarrow \frac{3}{2} \le - \frac{\cancel{2} x}{\cancel{2}}$
$\rightarrow - \frac{3}{2} \ge - \left(- x\right) \rightarrow x \le - \frac{3}{2}$