Solve 3xy+36+x^2=3

$$3xy+36+ { x }^{ 2 } =3$$

Solve for y

$$y=-\frac{x}{3}-\frac{11}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Subtract 36 from both sides.

$$3xy+x^{2}=3-36$$

Subtract 36 from 3 to get -33.

$$3xy+x^{2}=-33$$

Subtract x^{2} from both sides.

$$3xy=-33-x^{2}$$

The equation is in standard form.

$$3xy=-x^{2}-33$$

Divide both sides by 3x.

$$\frac{3xy}{3x}=\frac{-x^{2}-33}{3x}$$

Dividing by 3x undoes the multiplication by 3x.

$$y=\frac{-x^{2}-33}{3x}$$

Divide -33-x^{2} by 3x.

$$y=-\frac{x}{3}-\frac{11}{x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{9y^{2}-132}-3y}{2}$$

$$x=\frac{-\sqrt{9y^{2}-132}-3y}{2}$$

Solve for x

$$x=\frac{\sqrt{9y^{2}-132}-3y}{2}$$

$$x=\frac{-\sqrt{9y^{2}-132}-3y}{2}\text{, }|y|\geq \frac{2\sqrt{33}}{3}$$