What is the error in my analysis?

So, I am to analyze this circuit. 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?

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.

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

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