# How do you solve and write the following in interval notation:  -4<=( -2x+5)/3<=3?

Aug 1, 2016

$x \in \left[- 2 , \frac{17}{2}\right]$

#### Explanation:

Your goal here is to isolate $x$ in the middle of this compound inequality.

Start by multiplying all sides by $3$

$- 4 \cdot 3 \le \frac{- 2 x + 5}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} \le 3 \cdot 3$

$\textcolor{w h i t e}{a} - 12 \le \textcolor{w h i t e}{a} - 2 x + 5 \textcolor{w h i t e}{a a} \le 9$

Next, subtract $5$ from all sides

$- 12 - 5 \le - 2 x + \textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} \le 9 - 5$

$\textcolor{w h i t e}{a a a} - 17 \le \textcolor{w h i t e}{a a a} - 2 x \textcolor{w h i t e}{a a a a} \le 4$

Finally, divide all sides by $- 2$. Do not forget that when you're multiplying or dividing an inequality by a negative number, the sign of the inequality is flipped!

$\frac{- 17}{- 2} \textcolor{b l u e}{\ge} \textcolor{w h i t e}{a} x \textcolor{w h i t e}{a} \textcolor{b l u e}{\ge} \frac{4}{- 2} \to$ the signs are flipped !

will get you

$\textcolor{w h i t e}{-} \frac{17}{2} \ge \textcolor{w h i t e}{a} x \textcolor{w h i t e}{a} \ge - 2$

This tells you that in order to be a solution to this compound inequality, a value of $x$ must be greater than or equal to $- 2$ and smaller than or equal to $\frac{17}{2}$.

In interval notation, this is expressed like this

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{x \in \left[- 2 , \frac{17}{2}\right]} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

You can find this solution interval by breaking up the compound inequality into two simple inequalities

$x \le \frac{17}{2} \text{ }$ and $\text{ } x \ge - 2$

In interval notation, these solution intervals will get you

$x \in \left(- \infty , \frac{17}{2}\right] \cap \left[- 2 , + \infty\right) \implies x \in \left[- 2 , \frac{17}{2}\right]$