# Solve for $y$ in $x=\sqrt{(y-1)/(y+1)}$

I always struggle with this:

Express $y$ in terms of $x$ where $$x = \sqrt\frac{y-1}{y+1}$$

I know to square both sides and get $x^2 = \frac{y-1}{y+1}$

Then I'm thinking multiply both sides by $y+1?$ So we get $(y+1)x^2 = y-1$

But where do I go from here? If there was only one y I would be fine.

• Distribute the $x^2$ into $(y+1)$. i.e. $(y+1)x^2=yx^2+x^2$. Then move all terms involving $y$ to the left-hand-side and the rest to the right-hand-side. Commented Jan 20, 2015 at 21:13
• Regroup the terms on $y$ in one side and factor by $y$.
– user63181
Commented Jan 20, 2015 at 21:14

You got to this point correctly:$$(y+1)x^2=y-1$$Now use the distribution rule, i.e.:$$(a+b)c=ac+bc$$to expand this to:$$yx^2+x^2=y-1$$Then move all terms involving $y$ to the left-hand-side and the rest of the terms to the right-hand-side, i.e.:$$yx^2-y=-1-x^2$$Then factor the $y$ on the left-hand-side and solve from here.
$$x=\sqrt\frac{y-1}{y+1}\implies x^2y+x^2=y-1\implies y(x^2-1)=-x^2-1\implies ??$$
• Sorry, I'm not seeing how you got $y(x^2 - 1) = -x^2 -1??$ Commented Jan 20, 2015 at 21:18
• $x^2y+x^2-y=-1$ Commented Jan 20, 2015 at 21:19
• Yes yes, sorry I just got it. So our answer shall be $y =\frac{-1-x^2}{x^2-1}$ yes? thanks Commented Jan 20, 2015 at 21:21
• @moony Yes that the result, or "nicer":$$y=\frac{1+x^2}{1-x^2}$$ Commented Jan 20, 2015 at 21:21
Here is what I would do: \begin{align} x=\sqrt{\frac{y-1}{y+1}} &\Longleftrightarrow x^2 = \frac{y-1}{y+1}\\[1em] &\Longleftrightarrow (y+1)x^2 = y-1\\[1em] &\Longleftrightarrow yx^2+x^2=y-1\\[1em] &\Longleftrightarrow yx^2-y=-(x^2+1)\\[1em] &\Longleftrightarrow y(x^2-1)=-(x^2+1)\\[1em] &\Longleftrightarrow y = \frac{-1}{-1}\cdot \frac{x^2+1}{1-x^2}\\[1em] &\Longleftrightarrow y = \frac{x^2+1}{1-x^2} \end{align} Make sure you do not forget about the value restricts though (i.e., $x$ cannot be negative, $y\neq -1$, etc.).