Solve y^2+xy+3x=9

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

Solve for x (complex solution)

$$\left\{\begin{matrix}\\x=3-y\text{, }&\text{unconditionally}\\x\in \mathrm{C}\text{, }&y=-3\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract y^{2} from both sides.

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

Combine all terms containing x.

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

Divide both sides by 3+y.

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

Dividing by 3+y undoes the multiplication by 3+y.

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

Divide -y^{2}+9 by 3+y.

$$x=3-y$$

Solve for x

$$\left\{\begin{matrix}\\x=3-y\text{, }&\text{unconditionally}\\x\in \mathrm{R}\text{, }&y=-3\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract y^{2} from both sides.

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

Combine all terms containing x.

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

Divide both sides by 3+y.

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

Dividing by 3+y undoes the multiplication by 3+y.

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

Divide -y^{2}+9 by 3+y.

$$x=3-y$$

Solve for y

$$y=-3$$

$$y=3-x$$