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So, I am to analyze this circuit. enter image description here

So, this seems easy enough:
Node 3:
\$v_3=20 \space \mathrm{v}\$
Node 2:
\$\frac{1}{110}(v_1-v_2)+\frac{1}{100}v_1+2(\frac{1}{10})(v_3-v_2)=0\\\frac{28}{55}v_2-\frac{1}{110}v_1=6\$
Node 1:
\$\frac{1}{110}(v_1-v_2)+\frac{1}{100}v_1+2\frac{1}{10}(v_3-v_2)=-4\$

However, when I use, say, Cramer's rule (or wolframalpha) to solve for the equation, we get \$v_1=-67\$, which is unlikely. The other numbers don't look any better. Would someone be so kind as to tell me what I am doing wrong here?

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Somebody already gave you the answer but I can answer your question "what I am doing wrong here?"

For Node 2, you forgot to include the current \$i_x\$ and you also forgot to include the current from the \$5 \Omega\$ branch. Additionally, you don't need to add the current from the \$100 \Omega\$ branch because that current is going into Node 1.

For Node 1, you set the equation equal to -4, which implies that there is a constant 4 Amp current going into Node 1, but that is not the case. You should set that equation to 0.

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Your equations are wrong, they should look like this, denoting currents as being positive when running into the nodes:

Node 3:

\$ v_3 = 20 \$

Node 2:

\$ \frac{v_3 - v_2}{10} + \frac{v_1 - v_2}{110} - \frac{v_2}{5} + 2 \cdot \frac{v_3-v_2}{10} = 0 \$

Node 1:

\$-2 \cdot \frac{v_3-v_2}{10} - \frac{v_1}{100} + \frac{v_2 - v_1}{110} = 0 \$

Calculating it using Wolframalpha will give you the result

\$ v_1 = -100, v_2=10, v_3 = 20 \$

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