applying nodal inspection technique to supernodes with dependent sources
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If the numbers you've added in your PS are indeed correct then the assignment contains an error. "\$ 5 \text{ }i\$" has the dimension of current, if it should have been a voltage it should have said "\$ 5 \Omega \text{ }i\$". My answer assumes what the schematic says, not what may have been intended. I've added the answer for the corrected assignment below.
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Like I said in my answer to your other question, draw the chosen current directions so that you don't get confused and switch one of the directions halfway the calculation. I'll use the resistor values as current index, so the current though the 3 &Omega; resistor is \$I_{R3}\$.



I'm not sure what this supernode is, but I presume you want to apply KCL to a region instead of a single node. You'll have to clarify what this region is, because with the current equations there must be a mistake. 

Anyway, we don't really need a supernode, you can set up a set of simultaneous equations as follows:  

For the node \$v_1\$:  

\$ I_{10V} = I_{R2} + I_{R6}  \$  

For the node \$v_2\$:  

\$  I_{10V} + 5 I_{R2} + I_{R4} = 0  \$

For the node \$v_3\$:  

\$ 5 I_{R2} + I_{R6} = I_{R3} \$

and

\$ v_1 = 2\Omega \cdot I_{R2} \$

\$ v_2 = 4\Omega \cdot I_{R4} \$

\$ v_3 = 3\Omega \cdot I_{R3} \$

\$ v_1 - v_3 = 6\Omega \cdot I_{R6} \$

We have 7 equations for 8 variables, so we need another independent equation:  

\$ v_1 = v_2 + 10 V \$

This set of linear equations is easy to solve with some substitutions, and the result is:

\$ \begin{cases}  v_1 = 0.989 V \\ v_2 = -9.01 V \\ v_3 = 5.27 V \\ I_{R2} = 0.495 A \\ I_{R3} = 1.76 A \\ I_{R4} = -2.25 A \\ I_{R6} = -0.714 A \\ I_{10V} = -0.220 A \end{cases} \$



The solution for the corrected assignment:  



\$ v_1 = v_2 + 10 V \$

\$ v_3 = v_2 + 5 \Omega \cdot \dfrac{v_1}{2 \Omega} = v_2 + 2.5 v_1 \$  

and applying KCL to the ground node:  

\$ \dfrac{v_1}{2 \Omega} + \dfrac{v_2}{4 \Omega} + \dfrac{v_3}{3 \Omega} = 0 \$

That gives us a set of 3 simultaneous linear equations in 3 unknowns, which indeed gives us 

\$ \begin{cases} v_1 = 3.04348 V \\ v_2 = -6.95652 V \\ v_3 = 0.652174 V \end{cases}\$