Answer
$v_x=17.5V
Work Step by Step
Ohm's law dictates $v_2=5V$ and since all three resistors are in parallel we can find that $i_2=v_2\div 10\Omega=.5A$ and $i_3=v_2\div5\Omega=1A$
Then, using KCL in conjunction with Ohm's law, $v_1=5\Omega\times(1A+1A+0.5A)=12.5V$
Finally, with KVL we can find $v_x-v_1-v_2=0, v_x=12.5V+5V=17.5V$
