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

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

Solve for w (complex solution)

$$\left\{\begin{matrix}w=\left(\frac{x+1}{x-1}\right)^{2}y\text{, }&x\neq 1\text{ and }x\neq -1\\w\in \mathrm{C}\text{, }&x=1\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

To raise \frac{x-1}{x+1} to a power, raise both numerator and denominator to the power and then divide.

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

Express \frac{\left(x-1\right)^{2}}{\left(x+1\right)^{2}}w as a single fraction.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$x^{2}w-2xw+w=yx^{2}+2yx+y$$

Combine all terms containing w.

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

The equation is in standard form.

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

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

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

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

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

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

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

Solve for w

$$\left\{\begin{matrix}w=\left(\frac{x+1}{x-1}\right)^{2}y\text{, }&|x|\neq 1\\w\in \mathrm{R}\text{, }&x=1\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

To raise \frac{x-1}{x+1} to a power, raise both numerator and denominator to the power and then divide.

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

Express \frac{\left(x-1\right)^{2}}{\left(x+1\right)^{2}}w as a single fraction.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$x^{2}w-2xw+w=yx^{2}+2yx+y$$

Combine all terms containing w.

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

The equation is in standard form.

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

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

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

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

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

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

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

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{2\sqrt{wy}+w+y}{w-y}\text{; }x=\frac{-2\sqrt{wy}+w+y}{w-y}\text{, }&w\neq 0\text{ and }y\neq w\\x=0\text{, }&y=w\text{ and }w\neq 0\\x\neq -1\text{, }&y=0\text{ and }w=0\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{2\sqrt{wy}+w+y}{w-y}\text{; }x=\frac{-2\sqrt{wy}+w+y}{w-y}\text{, }&\left(y\neq w\text{ and }y\leq 0\text{ and }w<0\right)\text{ or }\left(y\neq w\text{ and }y\geq 0\text{ and }w>0\right)\\x=0\text{, }&y=w\text{ and }w\neq 0\\x\neq -1\text{, }&y=0\text{ and }w=0\end{matrix}\right.$$