@elecman, there are multiple ways in solving this circuit. One is with Mesh Analysis and the other way is Nodal Analysis. I really wish there was a way you and I can chat while we work on this problem, but let's still solve this circuit for both ways.
FOR MESH ANALYSIS:
You've have simplified the circuit by combining the resistors in parallel. Resistors in parallel have the same voltage. So now once we get the voltage across the 2 ohms resistor, it will be the same voltage across the 3 ohms resistor. Then from Ohm's Law, we can calculate the current going through the 3 ohms resistor.
In order to get the voltage across the 2 ohms resistor, we need to get the current entering it. Let's call this current I2 (since it is in the second loop). Let's call the current in the first loop I1. By Looking at the circuit, I1 = 2A. So there is no need to do a mesh equation for the first loop since we know what I1 is. However, to find I2, let's do a mesh loop for loop 2:
The current going through the 12 ohms resistor can go up or down, depending on what current (between I1 and I2) is stronger, however since we want the current to go in the direction of I2, we will make this current I2 - I1.
-36v + 12 * (I2 - I1) + 2 * I2 = 0
From there you can solve for I2. With I2, you can get the voltage across the 2 ohms resistor which would be 2 * I2 (which is the same voltage across the 3 ohms resistor) and from there you should get the current going through the 3 ohms resistor which should be = (2 ohms * I2)/3 ohms.
FOR NODAL ANALYSIS:
All we have to do is find the voltage across the 3 ohms and from there get the current across the 3 ohms using Ohm's Law. There is no need to combine anything, just write your nodal equations. Let's call the voltage across the 3 ohms resistor, V3. So assuming currents that are entering the top V3 node are -ve, while currents leaving the top node are +ve, we have
-2amps + (V3 - 36v)/12 + V3/6 + V3/3 = 0.
From there you can find V3, and use V3/3 to get the current going through the 3 ohm resistor.
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.
Best Answer
Yes, any SPICE-like simulator can solve this problem. LTSpice, ngspice, Microcap, and TINA-TI are four free (as in beer) ones that I know of. You can even use Circuitlab (the built in circuit editor attached to this site) to solve this problem.
For the general case, no. What they're basically teaching you is how SPICE (and related simulators) solve this kind of problem. In the real world, you'll probably never solve a circuit with more than 3 or 4 resistors by hand --- you'll input it to a simulator and let the computer do all the matrix manipulation. But you should learn how the simulator works so you understand what kinds of problems it can and can't solve, what situations make it more or less accurate, etc.