# 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?

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@Justin: Assuming this is a physics problem involving Newton's law of gravity, the units do not match. If you calculate a force, you need a distance^2 in the denominator. – Ross Millikan Jul 25 '11 at 19:26
I neglected to mention that F = 0 in this scenario. Will I still need to use the quadratic formula? – Justin Jul 25 '11 at 19:28
If you put $f=0$ in Arturo Magidin's answer, the quadratic term disappears, so not in that case. And if the denominators are really $r_2^2$ and $(d-r_2)^2$ you will just have the ratio of squares with no linear term. – Ross Millikan Jul 25 '11 at 19:39

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 $$(M-m){r_2}^2 - 2Mdr_2 + Md^2 = 0.$$

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Or (in the denominator squared case) you can write $M(d-r_2)^2=mr_2^2$, take square roots (masses are positive), and have a linear equation. – Ross Millikan Jul 25 '11 at 20:08
@Ross: You have to note that you need to have either $r_2,d-r_2\gt 0$ or $r_2,d-r_1\lt 0$ in that case, so that the original equation holds, but I agree that once you notice that your simplification works. – Arturo Magidin Jul 25 '11 at 21:06
true, I am hypnotized by the appearance of Newton's law as opposed to an arbitrary equation. These conditions will then be satisified. – Ross Millikan Jul 25 '11 at 23:47

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.

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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.

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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.

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