How to solve for an unknown in two different denominators? How to solve for an unknown in two different denominators?
In this equation:
$$F = \frac{GMm_3}{r_2} - \frac{Gmm_3}{d-r_2}.$$
Everything is given except for $r_2$.  How do I solve for $r_2$?  Is it possible?
 A: First, rewrite the right hand side as a single fraction by adding the fractions; factoring out $Gm_3$ simplifies matters a bit:
$$\begin{align*}
f &= \frac{GMm_3}{r_2} - \frac{Gmm_3}{d-r_2}\\
  &= Gm_3\left(\frac{M}{r_2} - \frac{m}{d-r_2}\right)\\
  &= Gm_3\left(\frac{M(d-r_2) - mr_2}{r_2(d-r_2)}\right).
\end{align*}$$
Then clear denominators by cross multiplying, and collect appropriate powers of $r_2$; you'll end up with a quadratic equation in $r_2$ that can be solved using the quadratic formula or other methods:
$$\begin{align*}
f & = Gm_3\left(\frac{Md - (M+m)r_2}{r_2(d-r_2)}\right)\\
fr_2(d-r_2) &= Gm_3\left(Md - (M+m)r_2\right)\\
fdr_2 - f(r_2)^2 &= Gm_3Md - Gm_3(M+m)r_2\\
0&= GMm_3d - \Bigl(Gm_3(M+m)+fd\Bigr)r_2 + f(r_2)^2.
\end{align*}$$
Added. If $f=0$, as you now write, then the equation becomes
$$0 = Gm_3\left(\frac {Md - (M+m)r_2}{r_2(d-r_2)}\right).$$
If $G\neq 0$ and $m_3\neq 0$, then this holds if and only if the numerator is $0$, if and only if
$$0 = Md - (M+m)r_2,$$
which is easy to solve.
If, as Ross suggests, the denominators should be squared and $f=0$, then you would instead get
$$0 = M(d-r_2)^2 - m{r_2}^2$$
which yields a quadratic equation in $r_2$ again, namely
$$(M-m){r_2}^2 - 2Mdr_2 + Md^2 = 0.$$
A: Try multiplying each side by both denominators, i.e. both $r_2$ and $d-r_2$, so that there are no fractions left. Combine like terms - in other words, group the terms with $r_2^2$ in them, the terms with $r_2$ in them, and the terms with no $r_2$'s in them. Then, treating $r_2$ as a variable we are solving for, we can apply the quadratic formula.
A: You can, but you will probably use the quadratic formula.
$$f = \frac{GMm_3}{r_2} - \frac{Gmm_3}{d-r_2}.$$
$$f \cdot (r_2)(d-r_2) = GMm_3 \cdot (d-r_2) - Gmm_3 \cdot r_2$$
$$ -f \cdot r_2^2  + (fd + GMm_3 + Gmm_3) \cdot r_2 - GMm_3 d = 0$$
And this is quadratic in $r_2$, so solve using the quadratic formula. Aha - just as I am ready to post, this tells me that Zev has posted a similar answer. Well, this is his, except Texed up.
A: A brief plea for symmetry!
The problem probably came from a situation in which there are two distances, $r_1$ and $r_2$, whose sum is $d$.  It could make life easier to use $r_1$ instead of $d-r_2$.  We then need to specify additionally that $r_1+r_2=d$.  
So instead of a single equation, we have a system of two equations.  But there is much more symmetry. Calculations often are more pleasant, and the calculation is more likely to reveal structural information.
