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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

Thanks in advance

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.

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    $\begingroup$ I do not thonk it is true unless some extra conditions are mentioned $\endgroup$ – user665856 Apr 30 at 9:48
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    $\begingroup$ $y^2$ is problematic. $$\frac{(bS)^{2}}{(bS)^{2} + y} \cdot \frac{(bS)^{2}}{(bS)^{2}}$$ $$\frac{1}{1 + \frac{y}{(bS)^{2}}} $$ $\endgroup$ – kelalaka Apr 30 at 9:48
  • $\begingroup$ 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 ... $\endgroup$ – Eevee Trainer Apr 30 at 9:49
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    $\begingroup$ Even more than one typo. $\endgroup$ – Claude Leibovici Apr 30 at 9:50
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    $\begingroup$ 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! $\endgroup$ – tom91 Apr 30 at 10:09
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As the good people in the comments suspected, it turns out that the reason I was struggling was because of a typo in the text I was following.

$\frac{(bS)^{2}}{(bS)^{2} + y}$ 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 $\frac{1}{1 + [\frac{y}{bS}]^{2}}$

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