# Solve for $y$: $\frac{y+1}{y-1} = 10^{x^2}$ [closed]

Can someone please show me the steps (all of them… yeah, even the obvious ones) to go from

\begin{align}\frac{y+1}{y-1} = 10^{x^2}\end{align}

to

\begin{align}y=\frac{10^{x^2}+1}{10^{x^2}-1}\end{align}

• This is basically the fact that the function $f(x)=\frac{x+1}{x-1}$ is inverse to itself. This can be seen quite well from the graph: wolframalpha.com/input/… Commented Aug 18, 2015 at 12:33

Solve for $y$:

$$\frac{y+1}{y-1} = 10^{x^2}$$

Multiply both sides by $y-1$: $$y+1=10^{x^2}(y-1)$$ Expand out terms of the right hand side: $$y+1=10^{x^2}y-10^{x^2}$$ Subtract $1+10^{x^2}y$ from both sides: $$y(1-10^{x^2})=-1-10^{x^2}$$ Divide both sides by $1-10^{x^2}$ $$\boxed{\color{blue}{y=\frac{10^{x^2}+1}{10^{x^2}-1}}}$$

$$\frac{y+1}{y-1} = 10^{x^2}$$

$$y+1 = 10^{x^2}(y-1)$$

$$y+1 = 10^{x^2}y-10^{x^2}$$

$$y+1+10^{x^2} = 10^{x^2}y$$

$$1+10^{x^2} = (10^{x^2}-1)y$$

$$y=\frac{10^{x^2}+1}{10^{x^2}-1}$$

For $y\neq1$ $$\frac{y+1}{y-1}=10^{x^2}$$ $$y+1=y\cdot10^{x^2}-10^{x^2}$$ $$1+10^{x^2}=y(10^{x^2}-1 )$$

Then?