How to solve the given system of equations for $I_1, I_2, I_3$? 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 =(
 A: 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 $-R_3\epsilon_1$.
A: This is a circuit system of equations.  Be sure to utilize Kirchhoff's Current Law and Ohm's Law to obtain additional equations to allow you to solve more easily. If you want to solve for just the I's in terms of R and E, then organize equations with I's as the coefficients.
Then solve system of 3 equations for the 3 unknowns.
