# How do you solve and write the following in interval notation: -3<3-2x<= 11?

Jul 18, 2017

See a solution process below:

#### Explanation:

First, subtract $\textcolor{red}{3}$ from each segment of the system of inequalities to isolate the $x$ term while keeping the system balanced:

$- \textcolor{red}{3} - 3 < - \textcolor{red}{3} + 3 - 2 x \le - \textcolor{red}{3} + 11$

$- 6 < 0 - 2 x \le 8$

$- 6 < - 2 x \le 8$

Next, divide each segment by $\textcolor{b l u e}{- 2}$ to solve for $x$ while keeping the system balanced. However, because we are multiplying or dividing inequalities by a negative number we need to reverse the inequality operators:

$\frac{- 6}{\textcolor{b l u e}{- 2}} \textcolor{red}{>} \frac{- 2 x}{\textcolor{b l u e}{- 2}} \textcolor{red}{\ge} \frac{8}{\textcolor{b l u e}{- 2}}$

$3 \textcolor{red}{>} \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 2}}} x}{\cancel{\textcolor{b l u e}{- 2}}} \textcolor{red}{\ge} - 4$

$3 \textcolor{red}{>} x \textcolor{red}{\ge} - 4$

Or

$x \ge - 4$ and $x < 3$

Or, in interval notation:

$\left[- 4 , 3\right)$

Jul 18, 2017

$\left[- 4 , 3\right)$

#### Explanation:

First we solve the inequality, then we put it into the notation we’d like.
−3 < 3 − 2x ≤ 11 Subtract the 3, divide by the -2 ( reverses the direction of the inequality)
(-3 – 3)/(-2) > x >= (11 – 3)/(-2)

$3 > x \ge - 4$

CHECK!: −3 < 3 − 2x ≤ 11 ; −3 < 3 − 2(2) ≤ 11 ; −3 < 3 − 4 ≤ 11

−3 < − 1 ≤ 11 CORRECT!

In interval notation our result is written with an open bracket on the left, and a closed bracket on the right.
$\left(3 , - 4\right]$, or in increasing numberline order, $\left[- 4 , 3\right)$