# Trying to solve for X. Memory fails me.

Hopefully, I am in the right place. It's been a while since I did any math, so this might be embarrasing...

I have an equation, where I am trying to solve for X.

$$Reach = \frac{X * AQH * CUME}{X * AQH + CUME - AQH}$$

How do I solve for X?

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Multiply both sides by the denominator:

$$Reach(X \cdot AQH + CUME - AQH)=X \cdot AQH \cdot CUME$$

Group the X's:

$$X(Reach \cdot AQH-AQH \cdot CUME)=-Reach(CUME-AQH)$$

Divide by what's next to X:

$$X=\frac{Reach(AQH-CUME)}{AQH(Reach-CUME)}$$

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$$x\cdot Reach\cdot AQH + Reach\cdot(CUME-AQH)=x\cdot AQH\cdot CUME$$ $$x(Reach\cdot AQH -AQH\cdot CUME)=-Reach\cdot(CUME-AQH)$$ $$x=\frac{-Reach\cdot(CUME-AQH)}{Reach\cdot AQH -AQH\cdot CUME}=\frac{Reach\cdot(AQH-CUME)}{AQH(Reach-CUME)}$$

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Say we have a linear fractional equation of the form

$$y=\frac{ax+b}{cx+d}.$$

We can multiply out:

$$y(cx+d) = ax+b$$

Subtract the right side and recombine:

$$(cy-a)x+(dy-b)=0$$

and then subtract and divide for $x$:

$$x=\frac{-dy+b}{cy-a}.$$

Your particular equation has $y=\mathrm{reach}, a=\mathrm{AQH}\cdot\mathrm{CUME}, b=0, c=\mathrm{AQH}, d=\mathrm{CUME}-\mathrm{AQH}.$

More generally (you might not have the background for this), we can define an action of $\rm SL$ by

$$\begin{pmatrix}a & b \\ c & d \end{pmatrix}x=\frac{ax+b}{cx+d}.$$

Lo and behold, it is well-defined, i.e. $A(Bx)=(AB)x$ and so $y=Ax$ implies $x=A^{-1}y$ when $A$ is an invertible matrix. Note that we speak of $\rm SL$ instead of $\rm GL$ because $A$ and $cA$ define the same transformation for any $c\ne0$, hence the quotient.

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It's better to introduce symbols and abbreviations the first time you use them. In particular, I would say "we can define an action of the special linear group $\mathrm{SL}(2)$" and similarly for $\mathrm{GL}(2)$. That way, even if someone doesn't have the background for what you're saying, they know how to look it up if they want. – Rahul Apr 11 '12 at 23:24