Electronic – 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 – Schematic created using CircuitLab}

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

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

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:

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

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

Either way works.