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

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

Solve for y

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

Steps for Solving Linear Equation

Divide each term of x^{2}+z+y by 3 to get \frac{1}{3}x^{2}+\frac{1}{3}z+\frac{1}{3}y.

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

Subtract \frac{1}{3}x^{2} from both sides.

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

Subtract \frac{1}{3}z from both sides.

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

The equation is in standard form.

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

Multiply both sides by 3.

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

Dividing by \frac{1}{3} undoes the multiplication by \frac{1}{3}.

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

Divide x-\frac{x^{2}}{3}-\frac{z}{3} by \frac{1}{3} by multiplying x-\frac{x^{2}}{3}-\frac{z}{3} by the reciprocal of \frac{1}{3}.

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

Solve for x

$$x=\frac{\sqrt{9-4z-4y}+3}{2}$$

$$x=\frac{-\sqrt{9-4z-4y}+3}{2}\text{, }z\leq \frac{9}{4}-y$$