The question asks to determine V_x and I_y.
I tried to find it by using KVL in to lower LHS mesh and KCL in the middle node. But, my final value is a contradictory statement(such as I_x equal all real numbers). Any help would be appreciated.
Using KCL in the central node, the current that passes through the 10 ohm resistor is -5-0.1V_x+y(where y is the current that goes through V_x/20 ohms resistor). Then I used KVL in lower LHS mesh, so -50-10(-5-0.1V_x+y)-V_x=0 As you can see V_x cancel and 10y=0 is left. However V_x=20y.
Best Answer
KCL at the center node gives:
$$ \frac{-V_x-50}{10} + 5 + 0.1V_x + \frac{-V_x}{20} = 0 $$
Which is only an equation of \$V_x\$ and can be immediately solved to get \$V_x = 0\$.
KCL at the top node (with voltage \$V_T\$):
$$ \frac{V_T - 50}{20} - 5 + I_Y = 0 $$
Ohm's law on the 25 ohm resistor gives \$ I_Y = \frac{V_T - 5I_Y}{25} \implies V_T = 30I_Y\$. Using this in the preceeding equation gives:
$$ \frac{30I_Y - 50}{20} - 5 + I_Y = 0 \implies I_Y = 3 \text{A}$$
So, along with your own answer and Alfred Centauri's answer, we can conclude that your book's answer of \$V_X = 70 V\$ is incorrect.