How to solve for node voltages using mesh analysis?
It's often best to redraw the schematic. Sometimes, this works well in your favor. This is an unbalanced wheatstone bridge:







simulate this circuit &ndash; Schematic created using CircuitLab

This would be very easy with nodal. But you want mesh. So:

$$\begin{align*}
20\:\text{V}-\left(I_1+I_3\right)\cdot R_8-\left(I_1+I_2\right)\cdot R_7-I_1\cdot R_6&amp;=20\:\text{V}\\\\
0\:\text{V}-\left(I_2-I_3\right)\cdot R_9-\left(I_1+I_2\right)\cdot R_7-I_2\cdot R_{10}&amp;=0\:\text{V}\\\\
20\:\text{V}-\left(I_1+I_3\right)\cdot R_8-\left(I_3-I_2\right)\cdot R_9&amp;=0\:\text{V}
\end{align*}$$

You should be able to solve for \$I_1\$, \$I_2\$, and \$I_3\$ with ease.

Once you have the currents, it's very easy to solve for \$V_x\$ and \$V_y\$. For example, \$V_x=20\:\text{V}-\left(I_1+I_3\right)\cdot R_8\$.



Should you have wanted nodal, then it would be these:

$$\begin{align*}
\frac{V_x}{R_7}+\frac{V_x}{R_8}+\frac{V_x}{R_9}&amp;=\frac{V_y}{R_7}+\frac{20\:\text{V}}{R_8}+\frac{0\:\text{V}}{R_9}\\\\
\frac{V_y}{R_6}+\frac{V_y}{R_7}+\frac{V_y}{R_{10}}&amp;=\frac{20\:\text{V}}{R_6}+\frac{V_x}{R_7}+\frac{0\:\text{V}}{R_{10}}
\end{align*}$$

Here, you'd solve for just two voltages. And from there, you could get the currents if you wanted them.

Either way works.