Solve xp^2+(y-x)p-y=0

$$x p ^ { 2 } + ( y - x ) p - y = 0$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=-\frac{y}{p}\text{, }&p\neq 0\\x\in \mathrm{C}\text{, }&p=1\text{ or }\left(y=0\text{ and }p=0\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply y-x by p.

$$xp^{2}+yp-xp-y=0$$

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

$$xp^{2}-xp-y=-yp$$

Add y to both sides.

$$xp^{2}-xp=-yp+y$$

Reorder the terms.

$$xp^{2}-px=-py+y$$

Combine all terms containing x.

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

The equation is in standard form.

$$\left(p^{2}-p\right)x=y-py$$

Divide both sides by p^{2}-p.

$$\frac{\left(p^{2}-p\right)x}{p^{2}-p}=\frac{y-py}{p^{2}-p}$$

Dividing by p^{2}-p undoes the multiplication by p^{2}-p.

$$x=\frac{y-py}{p^{2}-p}$$

Divide -yp+y by p^{2}-p.

$$x=-\frac{y}{p}$$

Solve for x

$$\left\{\begin{matrix}x=-\frac{y}{p}\text{, }&p\neq 0\\x\in \mathrm{R}\text{, }&p=1\text{ or }\left(y=0\text{ and }p=0\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply y-x by p.

$$xp^{2}+yp-xp-y=0$$

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

$$xp^{2}-xp-y=-yp$$

Add y to both sides.

$$xp^{2}-xp=-yp+y$$

Reorder the terms.

$$xp^{2}-px=-py+y$$

Combine all terms containing x.

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

The equation is in standard form.

$$\left(p^{2}-p\right)x=y-py$$

Divide both sides by p^{2}-p.

$$\frac{\left(p^{2}-p\right)x}{p^{2}-p}=\frac{y-py}{p^{2}-p}$$

Dividing by p^{2}-p undoes the multiplication by p^{2}-p.

$$x=\frac{y-py}{p^{2}-p}$$

Divide -yp+y by p^{2}-p.

$$x=-\frac{y}{p}$$

Solve for p (complex solution)

$$\left\{\begin{matrix}\\p=1\text{, }&\text{unconditionally}\\p=-\frac{y}{x}\text{, }&x\neq 0\\p\in \mathrm{C}\text{, }&y=0\text{ and }x=0\end{matrix}\right.$$

Solve for p

$$\left\{\begin{matrix}\\p=1\text{, }&\text{unconditionally}\\p=-\frac{y}{x}\text{, }&x\neq 0\\p\in \mathrm{R}\text{, }&y=0\text{ and }x=0\end{matrix}\right.$$