It is easier to write the three equations and use Gaussian Elimination, the equations being
$$ 2i_1 + 8i_2 + 0 i_3 = 10 \ \text{V}\\ 0 i_1 - 8i_2 +4i_3 = 6 \ \text{V} \\ i_1 - i_2 - i_3 = 0 \text{V}\\ $$
Method 1: We have the augmented matrix
$$ \left(
\begin{array}{ccc|c}
2 & 8 & 0 & 10 \\
0 & -8 & 4 & 6 \\
1 & -1 & -1 & 0 \\
\end{array}
\right)$$
Using RREF, we reduce to
$$\left(
\begin{array}{ccc|c}
1 & 0 & 0 & 3 \\
0 & 1 & 0 & \dfrac{1}{2} \\
0 & 0 & 1 & \dfrac{5}{2} \\
\end{array}
\right)$$
From this result, we have
$$i_3 = \dfrac{5}{2}, i_2 = \dfrac{1}{2}, i_1 = 3$$
Method 2: We can also use elimination from the augmented system, lets add equation one and two
$$2 i_1 + 4 i_3 = 16$$
Next, subtract $8$ times equation three from equation two
$$-8 i_1 + 12 i_3 = 6$$
Now we have two equations in two unknowns and we can add four times equation one to equation two
$$28 i_3 = 70 \implies i_3 = \dfrac{5}{2}$$
It is easy to move forward from here.
Method 3: Your substitution approach is also perfectly fine, we have
$$i_2 = \dfrac{5-i_1}{4}$$
Substituting that into the second equation
$$-6 - 8\left(\dfrac{5-i_1}{4} \right) + 4 i_3 = 0$$
This reduces to
$$2i_1 + 4 i_3 = 16$$
Next, substitute the same into equation three
$$i_1 - \left(\dfrac{5-i_1}{4} \right) -i_3 = 0$$
this reduces to
$$5 i_1 - 4 i_3 = 5$$
Now you have two equation in two unknowns, adding them
$$7i_1 = 21 \implies i_1 = 3$$
Best Answer
It's better to think about the changes in potentials as a sum, it'll help you keep your signs straight.
Think of a battery as going uphill, while a resistor is like going downhill (assuming the current is moving through them in the same direction, clockwise or counterclockwise). When you make a complete loop, and add up all the changes in potential, you need to end up at the same altitude you were at when you started - with no net change in potential (Kirchoff's law for voltage).
In this case, as you move clockwise around the outside loop, you've got an increase in potential over the battery of $\Delta V_1 = E_0$, and a decrease in potential over the resistor of $\Delta V_2 = -I_3R_2$. If you add these up, you have to have no net change in potential:
$$ \sum\Delta V = 0 $$ $$ \Rightarrow \Delta V_1 + \Delta V_2 = 0 $$ $$ \Rightarrow E_0 - I_3R_2 = 0 $$ $$ \Rightarrow E_0 = I_3R_2 .$$