How do I simplify $\frac{(bS)^{2}}{(bS)^{2} + y}$ to $\frac{1}{1+\left[\frac{y}{bS}\right]^{2}}$?

As the above mentions I have the fraction $$\frac{(bS)^{2}}{(bS)^{2} + y}$$ and the next step in the equation I am following simply states "it works out to equal" $$\frac{1}{1+\left[\frac{y}{bS}\right]^{2}}$$.

I'm sure its simple but I am very rusty on my algebra, do I need to multiply by the conjugate?

I'd be very grateful for a worked example

edit: I accidentally wrote - when I meant to put + in both denominators. Below is a clarification of terms.

The initial expression was $$\frac{M^{2}}{M^{2}+y}$$, where M = bS

edit2: Having checked the paper the text I'm following is based on I have found that the initial expression should have been $$\frac{M^{2}}{M^{2}+y^{2}}$$, solving the issues I was having.

• I do not thonk it is true unless some extra conditions are mentioned – user665856 Apr 30 at 9:48
• $y^2$ is problematic. $$\frac{(bS)^{2}}{(bS)^{2} + y} \cdot \frac{(bS)^{2}}{(bS)^{2}}$$ $$\frac{1}{1 + \frac{y}{(bS)^{2}}}$$ – kelalaka Apr 30 at 9:48
• I would divide both the numerator and denominator by $bS^2$ ... but there's a problem, the sign of the term involving $y$ would be opposite, yet everything else matches. Like Shamim mentioned, unless there's other conditions/info, I think there's a typo. EDIT: Oh, and the $y$ itself is a problem ... – Eevee Trainer Apr 30 at 9:49
• Even more than one typo. – Claude Leibovici Apr 30 at 9:50
• Having checked the paper the text is based on (Hanski's Incidence Function Model for anyone interested) I have seen that the typo in the text im following is that the initial expression should indeed be $\frac{M^{2}}{M^{2}+y^{2}}$. Sorry for wasting peoples time! – tom91 Apr 30 at 10:09

$$\frac{(bS)^{2}}{(bS)^{2} + y}$$ should have been $$\frac{(bS)^{2}}{(bS)^{2} + y^{2}}$$
Simplification is therefore a process of dividing both the numerator and denominator by $$(bS)^{2}$$ to end up with $$\frac{1}{1 + [\frac{y}{bS}]^{2}}$$