Solve 1+x^2y+cxy=0

$$1 + x ^ { 2 } y + c x y = 0$$

Solve for c

$$c=\frac{-yx^{2}-1}{xy}$$

$$y\neq 0\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}y+cxy=-1$$

Subtract x^{2}y from both sides.

$$cxy=-1-x^{2}y$$

Reorder the terms.

$$cxy=-yx^{2}-1$$

The equation is in standard form.

$$xyc=-yx^{2}-1$$

Divide both sides by xy.

$$\frac{xyc}{xy}=\frac{-yx^{2}-1}{xy}$$

Dividing by xy undoes the multiplication by xy.

$$c=\frac{-yx^{2}-1}{xy}$$

Divide -yx^{2}-1 by xy.

$$c=-x-\frac{1}{xy}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{\left(cy\right)^{2}-4y}}{2y}-\frac{c}{2}$$

$$x=-\frac{\sqrt{\left(cy\right)^{2}-4y}}{2y}-\frac{c}{2}\text{, }y\neq 0$$

Solve for x

$$x=\frac{\sqrt{\left(cy\right)^{2}-4y}}{2y}-\frac{c}{2}$$

$$x=-\frac{\sqrt{\left(cy\right)^{2}-4y}}{2y}-\frac{c}{2}\text{, }y\neq 0\text{ and }\left(c=0\text{ or }y\geq \frac{4}{c^{2}}\text{ or }y<0\right)$$