Answer
$I_1=\frac{3}{28}$, $I_2=\frac{2}{7}$, and $I_3=\frac{9}{28}$ Amp.
Work Step by Step
Step 1. Multiply the first equation by 4 to get $4I_1+4I_2-4I3=0$ add this to the third equation, we get
$4I_1+12I_2=5$
Step 2. The middle equation can be simplified to $4I_1-2I2=1$. Subtract this equation from the equation obtained in step 1, we have $14I_2=4$ or $I_2=\frac{2}{7}$
Step 3. Thus, $I_1=\frac{1-2I_2}{4}=\frac{3}{28}$ and $I_3=I_1+I_2=\frac{9}{28}$ Amp.
Step 4. We conclude that $I_1=\frac{3}{28}$, $I_2=\frac{2}{7}$, and $I_3=\frac{9}{28}$ Amp.
