Determining voltage and current in a circuit
This problem is, I think, a good case for using superposition with dependent sources.

Zero all sources (including dependent sources) except for the 50V DC source.

Then, by voltage division and Ohm's Law:

\$V_x = -50V \dfrac{20}{10 + 20} =  -33.33V\$

\$I_y = \dfrac{50}{20 + 25} = 1.111A \$

Now, zero that source and enable the 5A current source.

Then, by inspection and current division:

\$V_x = 5A \cdot 20\Omega || 10 \Omega = 33.33V\$

\$I_y = 5A \cdot \dfrac{20}{20 + 25} =  2.222A\$

Now, zero that source and enable the 5Iy source.

Then, by inspection:

\$V_x = 0V \$

\$I_y = -\dfrac{5I_y}{20 + 25} =  -0.1111 \cdot I_y\$

Finally, zero that source and enable the 0.1Vx source.

Then, by inspection:

\$V_x = 0.1V_x \cdot 20\Omega || 10 \Omega = 0.6667 V_x \$

\$I_y = 0A \$

The sums of the associated four terms give the answers:

\$V_x = -33.33V + 33.33V + 0V + 0.6667V_x \rightarrow V_x = 0V  \$

\$I_y = 1.111A + 2.222A +0A - 0.1111 \cdot I_y \rightarrow I_y =  3A\$