Finding the linear system of equations of a circuit using Kirchoff's rules I have the following circuit and I am trying to find the linear system of equations using Kirchoff's rules

For the 2 nodes, I can determine that the equations for both are $I_1 + I_2 - I_3 = 0$
For the left loop, I can determine that the equation is $-R_1*I_2 = E_0$, but according to my textbook, the overall loop is $R_2 *I_3= E_0$. I don't understand this because I thought that the overall loop would be $-R_2 *I_3= E_0$ since the current flow is clockwise. And the right loop would be $-R_3*I_3 + R_1*I_2 = 0$. Can someone help me understand this? Thanks!
 A: 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:
$$ \sum\Delta V = 0 $$
$$ \Rightarrow \Delta V_1 + \Delta V_2 = 0 $$
$$ \Rightarrow E_0 - I_3R_2 = 0 $$
$$ \Rightarrow E_0 = I_3R_2 .$$
A: The system of equations will be:
$$\begin{aligned}
I_1+I_2-I_3&=0\\ R_1 I_2 &= -E_0 \\R_2 I_3 &= E_0
\end{aligned}
$$
