# How do you solve 4 > 3x + 3 or 9 ≤ - 5x + 3?

Jul 1, 2015

#### Explanation:

(1) 4 > 3x + 3 --> 3x < 4 - 3 --> $x < \frac{1}{3}$

(2) $9 \le - 5 x + 3 \to - 5 x \ge 6$ --> $x \le - \frac{6}{5}$

Jul 1, 2015

$x < \frac{1}{3}$

#### Explanation:

Part 1: $4 > 3 x + 3$
$4 > 3 x + 3$
$\textcolor{w h i t e}{\text{XXXX}}$subtract 3 from both sides
$1 > 3 x$
$\textcolor{w h i t e}{\text{XXXX}}$divide both sides by 3
$\frac{1}{3} > x$

Part 2: 9<=−5x+3
$9 \le - 5 x + 3$
$\textcolor{w h i t e}{\text{XXXX}}$subtract 3 from both sides
$6 \le - 5 x$
$\textcolor{w h i t e}{\text{XXXX}}$divide both sides by $\left(- 5\right)$ remembering to reverse the inequality orientation because of the negative division.
$- \frac{6}{5} \ge x$

Part 3: combining with or
If it is only necessary for
$\textcolor{w h i t e}{\text{XXXX}}$$x < \frac{1}{3}$
or
$\textcolor{w h i t e}{\text{XXXX}}$$x \le - \frac{6}{5}$
then
it is only necessary for
$\textcolor{w h i t e}{\text{XXXX}}$$x < \frac{1}{3}$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$since $\left(x \le - \frac{6}{5}\right)$ is a subset of $\left(x < \frac{1}{3}\right)$)