Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 3 down vote accepted

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:


share|cite|improve this answer

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

share|cite|improve this answer

Say we have a linear fractional equation of the form


We can multiply out:

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

Subtract the right side and recombine:


and then subtract and divide for $x$:


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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.