Solve x^2y+3x-y+xy=0

$$x ^ { 2 } y + 3 x - y + x y = 0$$

Solve for y

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

$$x\neq \frac{\sqrt{5}-1}{2}\text{ and }x\neq \frac{-\sqrt{5}-1}{2}$$

Steps for Solving Linear Equation

Subtract 3x from both sides. Anything subtracted from zero gives its negation.

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

Combine all terms containing y.

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

The equation is in standard form.

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

Divide both sides by x^{2}-1+x.

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

Dividing by x^{2}-1+x undoes the multiplication by x^{2}-1+x.

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

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{5y^{2}+6y+9}-y-3}{2y}\text{; }x=-\frac{\sqrt{5y^{2}+6y+9}+y+3}{2y}\text{, }&y\neq 0\\x=0\text{, }&y=0\end{matrix}\right.$$