Solve 3a=-b+cquad[c]

$$3 a = - b + c \quad [ c ]$$

Solve for a

$$a=\frac{c^{2}-b}{3}$$

Steps for Solving Linear Equation

Multiply c and c to get c^{2}.

$$3a=-b+c^{2}$$

The equation is in standard form.

$$3a=c^{2}-b$$

Divide both sides by 3.

$$\frac{3a}{3}=\frac{c^{2}-b}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$a=\frac{c^{2}-b}{3}$$

Solve for b

$$b=c^{2}-3a$$

Steps for Solving Linear Equation

Multiply c and c to get c^{2}.

$$3a=-b+c^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$-b+c^{2}=3a$$

Subtract c^{2} from both sides.

$$-b=3a-c^{2}$$

Divide both sides by -1.

$$\frac{-b}{-1}=\frac{3a-c^{2}}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$b=\frac{3a-c^{2}}{-1}$$

Divide -c^{2}+3a by -1.

$$b=c^{2}-3a$$