I have this system of equations: \begin{cases} I_1 = I_2 + I_3 \\ \epsilon_1 - I_1(R_1 + R_2) - I_2 R_3 = 0 \\ \epsilon_1 - I_1(R_1 + R_2) - I_3(R_4 + R_5) + \epsilon_2 = 0 \end{cases} I want to solve it for $I_1, I_2$ and $I_3$ (so that neither is expressed in function of the other).
This is what I did so far:
From the first equation we have that $I_2 = I_1 - I_3$. Substitute that into the second equation for $I_2$ to get \begin{align*} \epsilon_1 - I_1(R_1 + R_2) - (I_1 - I_3) R_3 = 0. \end{align*} Substracting the third equation from that one gives \begin{align*} -(I_1 - I_3) R_3 + I_3(R_4 + R_5) - \epsilon_2 = 0. \end{align*} Then I'm not sure what to do, I still have that $I_1$ which I want to remove.
Any help?
Edit: Another try, making use of a given hint. Substituting $I_1 = I_2 + I_3$ into equation two and three gives \begin{cases} \epsilon_1 - (I_2 + I_3)(R_1 + R_2) - I_2 R_3 = 0 \\ \epsilon_1 - (I_2 + I_3)(R_1 + R_2) - I_3(R_4+R_5) + \epsilon_2 = 0. \end{cases} After distribution this becomes \begin{cases} \epsilon_1 - I_2 R_1 - I_2 R_2 - I_3 R_1 - I_3 R_2 - I_2 R_3 = 0 \\ \epsilon_1 - I_2 R_1 - I_2 R_2 - I_3 R_1 - I_3 R_2 - I_3 R_4 - I_3 R_5 + \epsilon_2 = 0. \end{cases} Substracting the second from the first gives \begin{align*} -I_2 R_3 + I_3 R_4 + I_3 R_5 + \epsilon_2 = 0, \end{align*} which has still the $I_2$ factor in it =(