Q:

Nodal Analysis Help

Please see attached image. Please can you check why my nodal analysis of for node 'x' is incorrect (according to solutions)
Also could please explain to me how one gets from the second line to third line in the solutions.
Thanks

A:

The only issue I see with the way you set up your problem is the right hand side. It should be a negative, but I'll set it up for you:
$$ \frac{V_1-V_x}{R_2}+\frac{V_2-V_x}{R_1}+\frac{0-V_x}{R_p}+\frac{V_{out}-V_x}{R_f}=0$$
By the previous equation (Kirchhoff's law) you are saying that all the currents that go into that node add up to zero. By your convention, you are saying that the currents that go into the node are positive, that's why the fractions have a positive sign in front of them. One way to keep track, at least I would do it, is to set up your equation on one side of the equal sign that why you won't mess up the signs (the summation of all currents add up to zero). You could assume all your currents are going into the node and by your convention those are positive. That's what I did.
Breaking up the previous fractions you get:
$$\frac{V_1}{R_2}+\frac{V_2}{R_1}+\frac{V_{out}}{R_f}-\frac{V_x}{R_2}-\frac{V_x}{R_1}-\frac{V_x}{R_p}-\frac{V_x}{R_f}=0$$
Now all you have to do is collect \$V_x\$ terms and solve for \$V_x\$
$$\frac{V_1}{R_2}+\frac{V_2}{R_1}+\frac{V_{out}}{R_f}-V_x\bigg(\frac{1}{R_2}+\frac{1}{R_1}+\frac{1}{R_p}+\frac{1}{R_f}\bigg)=0$$
From there, you can isolate \$V_x\$
$$ V_x=\bigg(\frac{V_1}{R_2}+\frac{V_2}{R_1}+\frac{V_{out}}{R_f}\bigg)\cdot\frac{1}{\bigg(\frac{1}{R_2}+\frac{1}{R_1}+\frac{1}{R_p}+\frac{1}{R_f}\bigg)}$$
We've arrived to the final solution you have there. If you look that last term in the right hand side (the inverse of the inverse sum) is the same as finding the parallel equivalent of \$R_1\$, \$R_2\$, \$R_p\$, and \$R_f\$, which is the way the final solution is expressed.
Hope it helps!

