Solve y=[(1-x)/x]^2t

$$y = [ ( 1 - x ) / x ] ^ { 2 } t$$

Solve for t

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

Steps for Solving Linear Equation

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

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

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

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

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

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

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

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

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

$$\frac{t-2xt+x^{2}t}{x^{2}}=y$$

Multiply both sides of the equation by x^{2}.

$$t-2xt+x^{2}t=yx^{2}$$

Combine all terms containing t.

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

The equation is in standard form.

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

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

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

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

$$t=\frac{yx^{2}}{x^{2}-2x+1}$$

Divide yx^{2} by 1-2x+x^{2}.

$$t=\frac{yx^{2}}{\left(x-1\right)^{2}}$$

Solve for x

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