Solve (x+1)(x-5)/(x-1)^2=y

$$\frac { ( x + 1 ) ( x - 5 ) } { ( x - 1 ) ^ { 2 } } = y$$

Solve for y

$$y=\frac{\left(x-5\right)\left(x+1\right)}{\left(x-1\right)^{2}}$$

$$x\neq 1$$

Steps for Solving Linear Equation

Multiply both sides of the equation by \left(x-1\right)^{2}.

$$\left(x+1\right)\left(x-5\right)=y\left(x-1\right)^{2}$$

Use the distributive property to multiply x+1 by x-5 and combine like terms.

$$x^{2}-4x-5=y\left(x-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$x^{2}-4x-5=y\left(x^{2}-2x+1\right)$$

Use the distributive property to multiply y by x^{2}-2x+1.

$$x^{2}-4x-5=yx^{2}-2yx+y$$

Swap sides so that all variable terms are on the left hand side.

$$yx^{2}-2yx+y=x^{2}-4x-5$$

Combine all terms containing y.

$$\left(x^{2}-2x+1\right)y=x^{2}-4x-5$$

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

$$\frac{\left(x^{2}-2x+1\right)y}{x^{2}-2x+1}=\frac{\left(x-5\right)\left(x+1\right)}{x^{2}-2x+1}$$

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

$$y=\frac{\left(x-5\right)\left(x+1\right)}{x^{2}-2x+1}$$

Divide \left(-5+x\right)\left(1+x\right) by x^{2}-2x+1.

$$y=\frac{\left(x-5\right)\left(x+1\right)}{\left(x-1\right)^{2}}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{-y+\sqrt{9-8y}+2}{1-y}\text{; }x=\frac{-y-\sqrt{9-8y}+2}{1-y}\text{, }&y\neq 1\\x=-3\text{, }&y=1\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{-y+\sqrt{9-8y}+2}{1-y}\text{; }x=\frac{-y-\sqrt{9-8y}+2}{1-y}\text{, }&y\neq 1\text{ and }y\leq \frac{9}{8}\\x=-3\text{, }&y=1\end{matrix}\right.$$