Answer
$I_1=\dfrac{11}{13}; \ I_2=\dfrac{6}{13} ; \ I_3=\dfrac{17}{13}$
Work Step by Step
We need to solve the given system of equations as follows: $$I_3=I_1+I_2~~~~(1) \\ 8=3I_3+6I_2~~~~(2)\\8I_1=4+6I_2~~~(3)$$
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:
$$I_3=\dfrac{8-6I_2}{3}=\dfrac{4-3I_2}{2}~~~~(4)\\ I_1=\dfrac{4+6I_2}{8}= \dfrac{2+3I_2}{4}~~~(5)$$
Now, substitute the values of $I_1$ and $I_3$ in the equations $(1)$ to solve $I_2$.
$\dfrac{4-3I_2}{2}=\dfrac{2+3I_2}{4}+I_2\\ 2(4-3I_2) =2+3I_2+4I_2 \\ 8-6I_2=2+7I_2 \\ I_2=\dfrac{6}{13}$
Finally, back substitute the value of $I_2$ in the equation $(4)$ and $(5)$ to solve $I_1, I_3$.
$I_3=\dfrac{4-3\left(\dfrac{6}{13}\right)}{2}=\dfrac{17}{13}$
and $ I_1=\dfrac{2+3\left(\dfrac{6}{13}\right)}{4}=\dfrac{11}{13}$
So, we have: $I_1=\dfrac{11}{13}; \ I_2=\dfrac{6}{13} ; \ I_3=\dfrac{17}{13}$
