# How do you solve 9-2x \le 3 or 3x+10 \le 6-x?

May 27, 2018

See explanation

#### Explanation:

We have two conditions that are combined to define the limit of values that may be assigned to $x$

Condition 1: $9 - 2 x \le 3$
Condition 2: $2 x + 10 \le 6 - x$

Consider condition 1

Add $2 x$ to both sides

$9 \textcolor{w h i t e}{\text{dd}} \underbrace{- 2 x + 2 x} \le 3 + 2 x$

$9 \textcolor{w h i t e}{\text{ddd")+ 0color(white)("dddd}} \le 2 x + 3$

Subtract 3 from both sides

$9 - 3 \le 2 x + 0$

Divide both sides by 2

$\frac{9 - 3}{2} \le \frac{2}{2} \textcolor{w h i t e}{.} x$

$3 \le x$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider condition 2

Add $x$ to both sides

$3 x + 10 \le 6$

Subtract 10 from both sides

$3 x \le - 4$

Divide both sides by 3

$x \le - \frac{4}{3}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Combining these we have:

$- \frac{4}{3} \ge x \ge 3$

In other words $x$ does not take on any values between and excluding $- \frac{4}{3} \mathmr{and} 3$