# How do you solve the compound inequality 3<2x-3<15?

May 21, 2017

See a solution process below:

#### Explanation:

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

$3 + \textcolor{red}{3} < 2 x - 3 + \textcolor{red}{3} < 15 + \textcolor{red}{3}$

$6 < 2 x - 0 < 18$

$6 < 2 x < 18$

Now, divide each segment by $\textcolor{red}{2}$ to solve for $x$ while keeping the system balanced:

$\frac{6}{\textcolor{red}{2}} < \frac{2 x}{\textcolor{red}{2}} < \frac{18}{\textcolor{red}{2}}$

$3 < \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} x}{\cancel{\textcolor{red}{2}}} < 9$

$3 < x < 9$

Or

$x > 3$; $x < 9$

Or, in interval notation:

$\left(3 , 9\right)$

May 21, 2017

Solution: $3 < x < 9$ or in interval notation: $\left(3 , 9\right)$

#### Explanation:

$3 < 2 x - 3 < 15$ Adding $3$ in all sides we get

$6 < 2 x < 18$ Multiplying by $\frac{1}{2}$ in all sides we get

$3 < x < 9$.

Solution: $3 < x < 9$ or in interval notation: $\left(3 , 9\right)$ [Ans]