What is the thevenin voltage and resistance of the following circuit?
The equivalent resistance, short-circuiting the voltage source is:
$$R_{eq}=(5||20+8)||12=(5(1||4)+8)||12=(5\frac{4}{5}+8)||12=12||12=12(1||1)=6\Omega$$
where \$||\$ is the parallel resistance operator:
$$r=\frac{r_1r_2}{r_1+r_2}$$

The equivalent voltage source is obtained by calculating the open-circuit voltage. There are two loops, one for the power source (\$i_1\$) and other for the \$12\Omega\$ resistor (\$i_2\$). Hence the equations are:
$$
5(i_1-i_2)+20i_1=72\\
12i_2+8i_2+5(i_2-i_1)=0
$$
In matrix notation:
$$
\left[ 
{\begin{array}{cc}
25 &amp; -5 \\ -5 &amp; 25 
\end{array} }
\right]
\left[ 
{\begin{array}{cc}
i_1 \\ i_2 
\end{array} }
\right]=
\left[ 
{\begin{array}{cc}
72 \\ 0
\end{array} }
\right]
$$
Hence:
$$
\left[ 
{\begin{array}{cc}
i_1 \\ i_2 
\end{array} }
\right]=
\left[ 
{\begin{array}{cc}
3 \\ 3/5
\end{array} }
\right]
$$
And thus, the voltage is:
$$v_{eq}=20i_1=60V$$