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schematic

simulate this circuit – Schematic created using CircuitLab

At node 1, we have:$$0.001+i_1-i_4-i_5=0$$. At node 2, we have: $$i_4+i_5+i_2=0$$.

$$i_1=\frac{0-V_1}{1000}=\frac{-V_x}{1000}$$(Vx is the voltage across R2) $$i_2=\frac{0-V_2}{6000}$$ $$i_3=0.001A$$ $$i_4=i_1+i_3=-i_2-i_5$$ $$i_5=\frac{V_1-V_2}{1000}=\frac{V_x}{1000}$$

Use the equation at node 1, to find an expression for i4 and plug that into the equation at node 2: $$i_1+0.001+i_2=0$$ $$\frac{-V_x}{1000}+0.001-\frac{V_2}{6000}=0$$

But there are V2 and Vx to solve in one equation.

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  • \$\begingroup\$ Your first equation(at node 1) should also include \$i_5\$ (going out of the node). Also, your expression for \$i_1\$ should be \$-\frac{V_x}{1000}\$ look at the polarity of \$V_x\$ with respect to \$i_1\$ \$\endgroup\$ – Big6 Sep 24 '17 at 23:41
  • \$\begingroup\$ It still not solvable. there is one equation at the end, but Vx and V2 \$\endgroup\$ – most venerable sir Sep 25 '17 at 0:01
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You have too much variables and arrows (not wrong, but unnecessary). I redrawed the circuit:

enter image description here

You obviously missed that the voltage over R4 can be expressed in terms of Vx. This is Kirchoffs voltage law.

I have written the current equations for nodes 1 and 2. There are 2 unknowns and 2 equations.

Hopefully clear now.

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