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 =(
If you substitue $I_2+I_3$ for $I_1$ in the second and third equations, then you have two equations in two variables, $I_2$ and $I_3$. Do the simplifications, then do a similar thing to reduce it to one unknown quantity.
Later update in response to comments: You have $$ \begin{cases} \epsilon_1 - (I_2 + I_3)(R_1 + R_2) - I_2 R_3 = 0 \\ \epsilon_1 - (I_2 + I_3)(R_1 + R_2) - I_3(R_4+R_5) + \epsilon_2 = 0. \end{cases} $$ Collecting all of the instances of $I_2$ into one term and all of those of $I_3$ into one term, we get $$ \begin{cases} \epsilon_1 - I_2(R_1+R_2+R_3) - I_3(R_1+R_2) = 0 \\ \epsilon_1 - I_2(R_1+R_2) - I_3(R_1+R_2+R_4+R_5) +\epsilon_2 = 0 \end{cases} $$ If you multiply both sides of the first of these by $R_1+R_2$ and both sides of the second by $R_1+R_2+R_3$ and then subtract, then you eliminate $I_2$ from the system, and you're left with only one variable to solve for, namely $I_3$. When you've done that, you can plug in the solution for $R_3$ in place of $I_3$ in one of the two equations above and solve it for $I_2$. Then you can find $I_1$ since you've shown it's the sum $I_1+I_2$. Notice that $(R_1+R_2)\epsilon_1 - (R_1+R_2+R_3)\epsilon_1$, which you get when you subtract, simplifies to $-R_3\epsilon_1$.