1. ## solve absolute value

|2x-3/9|=5

2. $\displaystyle |2x-3/9|=5$ Is equivalent to say that $\displaystyle -5<2x-\frac{3}{9}<5$, or $\displaystyle -5+\frac{3}{9}<2x<5+\frac{3}{9}$, or $\displaystyle \frac{{-5}+{\frac{3}{9}}}{2}<x<\frac{{5}+{\frac{3}{9}}}{2}$. Almost done, I let you simplify the fractions.

3. Hello, super9xman!

Solve: .$\displaystyle \left| \frac{2x-3}{9}\right|\:=\:5$

We have: .$\displaystyle \begin{Bmatrix}\dfrac{2x-3}{9} & = & \text{-}5 \\ \\ \dfrac{2x-3}{9} &=& 5 \end{Bmatrix} \quad\Rightarrow\quad \begin{Bmatrix}2x-3&=&\text{-}45 \\ \\ 2x-3&=&45 \end{Bmatrix}$ . $\displaystyle \Rightarrow\quad\begin{Bmatrix}2x &=&\text{-}42 \\ \\ 2x &=&48 \end{Bmatrix}\quad\Rightarrow\quad \begin{Bmatrix}x &=&\text{-}21 \\ \\ x&=& 24\end{Bmatrix}$