solve for x. |4x-1|=4

I am ging to finsh up in a minute so I'll show you a little more to make sure you can at least start

$$\left({\frac{{\mathtt{2}}}{{\mathtt{z}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right){\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{4}}}} = {\frac{{\mathtt{3}}}{\left({\mathtt{z}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}$$

$$\\\frac{2}{z}-1 -\frac{ 1}{4}\right = \frac{3}{(z+1)}\\\\
\frac{2}{z}-\frac{ 5}{4}\right = \frac{3}{(z+1)}\\\\
4z(z+1)\left[\frac{2}{z}-\frac{ 5}{4}\right] = 4z(z+1)\frac{3}{(z+1)}\\\\
4z(z+1)\frac{2}{z}-4z(z+1)\frac{ 5}{4} = 4z(z+1)\frac{3}{(z+1)}\\\\
4\not{z}(z+1)\times \frac{2}{\not{z}}-\not{4}z(z+1)\frac{ 5}{\not{4}} = 4z\times \frac{3}{1}\\\\
8(z+1)-5z(z+1) = 12z\\\\$$

etc

(2/z-1) - 1/4 = 3/(z+1)

this is what you have asked for is it what you want?

$$\left({\frac{{\mathtt{2}}}{{\mathtt{z}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right){\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{4}}}} = {\frac{{\mathtt{3}}}{\left({\mathtt{z}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}$$

anyway if it is firstly recognise that z cannot be 0 or -1

Simplify the LHS 

then

multiply both sides by   4z(z+1)

that will get rid of all the fractions and make it easy to work with.   

give it a go and if you get stuck or need more help then show us what you've done or ask us to help more.

$$\left({\frac{{\mathtt{2}}}{{\mathtt{z}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right){\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{4}}}} = {\frac{{\mathtt{3}}}{\left({\mathtt{z}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}$$. yes

I have added to my answer above :)