Solve 2x+3y+z^2=3

$$2 x + 3 y + z ^ { 2 } = 3$$

Solve for x

$$x=\frac{3-z^{2}-3y}{2}$$

Steps for Solving Linear Equation

Subtract 3y from both sides.

$$2x+z^{2}=3-3y$$

Subtract z^{2} from both sides.

$$2x=3-3y-z^{2}$$

The equation is in standard form.

$$2x=3-z^{2}-3y$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{3-z^{2}-3y}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{3-z^{2}-3y}{2}$$

Solve for y

$$y=-\frac{z^{2}}{3}-\frac{2x}{3}+1$$

Steps for Solving Linear Equation

Subtract 2x from both sides.

$$3y+z^{2}=3-2x$$

Subtract z^{2} from both sides.

$$3y=3-2x-z^{2}$$

The equation is in standard form.

$$3y=3-z^{2}-2x$$

Divide both sides by 3.

$$\frac{3y}{3}=\frac{3-z^{2}-2x}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$y=\frac{3-z^{2}-2x}{3}$$

Divide 3-2x-z^{2} by 3.

$$y=-\frac{z^{2}}{3}-\frac{2x}{3}+1$$