Solve 3x^2=x/3-y

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

Solve for y

$$\left\{\begin{matrix}y=3-\frac{1}{3x}\text{, }&x\neq 0\\y\neq 3\text{, }&x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Variable y cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by -y+3.

$$3x^{2}\left(-y+3\right)=x$$

Use the distributive property to multiply 3x^{2} by -y+3.

$$-3x^{2}y+9x^{2}=x$$

Subtract 9x^{2} from both sides.

$$-3x^{2}y=x-9x^{2}$$

The equation is in standard form.

$$\left(-3x^{2}\right)y=x-9x^{2}$$

Divide both sides by -3x^{2}.

$$\frac{\left(-3x^{2}\right)y}{-3x^{2}}=\frac{x\left(1-9x\right)}{-3x^{2}}$$

Dividing by -3x^{2} undoes the multiplication by -3x^{2}.

$$y=\frac{x\left(1-9x\right)}{-3x^{2}}$$

Divide x\left(1-9x\right) by -3x^{2}.

$$y=3-\frac{1}{3x}$$

Variable y cannot be equal to 3.

$$y=3-\frac{1}{3x}\text{, }y\neq 3$$

Solve for x

$$x=\frac{1}{3\left(3-y\right)}$$

$$x=0\text{, }y\neq 3$$