Solve -3I_{A}+2I_{B}+I_{C}=-9

$$- 3 I _ { A } + 2 I _ { B } + I _ { C } = - 9$$

Solve for I_A

$$I_{A}=\frac{2I_{B}+I_{C}+9}{3}$$

Steps for Solving Linear Equation

Subtract 2I_{B} from both sides.

$$-3I_{A}+I_{C}=-9-2I_{B}$$

Subtract I_{C} from both sides.

$$-3I_{A}=-9-2I_{B}-I_{C}$$

The equation is in standard form.

$$-3I_{A}=-2I_{B}-I_{C}-9$$

Divide both sides by -3.

$$\frac{-3I_{A}}{-3}=\frac{-2I_{B}-I_{C}-9}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$I_{A}=\frac{-2I_{B}-I_{C}-9}{-3}$$

Divide -9-2I_{B}-I_{C} by -3.

$$I_{A}=\frac{I_{C}}{3}+\frac{2I_{B}}{3}+3$$

Solve for I_B

$$I_{B}=\frac{3I_{A}-I_{C}-9}{2}$$

Steps for Solving Linear Equation

Add 3I_{A} to both sides.

$$2I_{B}+I_{C}=-9+3I_{A}$$

Subtract I_{C} from both sides.

$$2I_{B}=-9+3I_{A}-I_{C}$$

The equation is in standard form.

$$2I_{B}=3I_{A}-I_{C}-9$$

Divide both sides by 2.

$$\frac{2I_{B}}{2}=\frac{3I_{A}-I_{C}-9}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$I_{B}=\frac{3I_{A}-I_{C}-9}{2}$$