Answer
$I_1=\dfrac{10}{71}; \ I_2=\dfrac{65}{71} ; \ I_3=\dfrac{55}{71}$
Work Step by Step
We need to solve the given system of equations as follows: $$I_2=I_1+I_3~~~~(1) \\ 5-3I_1-5I_2=0~~~~(2) \\10-5I_2-7I_3=0~~~(3)$$
Re-write the equation $(2)$ in terms of $I_2$ and equation (3) in terms $I_2$.Thus, we get the new system of equations as: $$I_2=I_1+I_3~~~~(4)\\ 5-5I_2=3I_1~~~~(5) \\10-5I_2=7I_3~~~(6)$$
From equations $(5)$ and $(6)$, we have: $$I_1=\dfrac{5-5I_2}{3}~~~~(7) \\ I_3=\dfrac{10-5I_2}{7}~~~(8)$$
Now, substitute the values of $I_1$ and $I_3$ in the equations (4) to solve $I_2$.
$I_2=I_1+I_3= \dfrac{5-5I_2}{3}+\dfrac{10-5I_2}{7} \\
21 \cdot I_2=7(5-5I_2) +3(10-5I_2) \\ 21\cdot I_2 =65-50I_2 \\ 71\cdot I_2=65 \\ I_2 =\dfrac{65}{71}$
Finally, back substitute the value of $I_2$ in the equation $(7)$ and $(8)$ to solve $I_1, I_3$.
$I_1=\dfrac{5-5(\dfrac{65}{71})}{3}=\dfrac{10}{71}$
and $ I_3=\dfrac{10-5(\dfrac{65}{71})}{7}=\dfrac{55}{71}$
So, we have: $I_1=\dfrac{10}{71}; \ I_2=\dfrac{65}{71} ; \ I_3=\dfrac{55}{71}$
