Okay I think I'm just having a major brain block, but I need help solving this proportion for my physics class.

$$\frac {6.0\times 10^{-6}}{ x^2} = \frac {2.0\times 10^{-6}}{ (x-20)^2}$$

What's confusing me is the solution manual to this problem lists writing the proportion as,

$$\frac {(x-20)^2} { x^2} = \frac {2.0\times 10^{-6}}{ 6.0\times 10^{-6}}$$

and then proceeds to solve the problem from there... but that doesn't seem right to me. Usually you would cross multiply a proportion and solve, but they seemed to do some illegal math or something. Could you guys work me through how to solve this? This answer is 47 by the way.

  • $\begingroup$ What do they do from here on? Please write the next few "illegal" steps so that the judges can deliver the verdict. $\endgroup$ – Teresa Lisbon Jan 26 '17 at 1:05
  • $\begingroup$ Cross-multiplying and solving works just as fine. Try it and see if it matches the result. $\endgroup$ – Simply Beautiful Art Jan 26 '17 at 1:05
  • $\begingroup$ The answer is not exactly $47$, that should be an approximation. $\endgroup$ – user160738 Jan 26 '17 at 1:06
  • $\begingroup$ What's wrong with cross-multiplying? Looks valid to me so far. $\endgroup$ – lulu Jan 26 '17 at 1:06
  • $\begingroup$ Assuming $a,b,c,d$ are nonzero, $$a/b = c/d \iff ad = bc \iff a/c = b/d$$ so the solution manual just chose a valid equivalent form (presumably to allow an immediate simplification of the scientific fractions). $\endgroup$ – quasi Jan 26 '17 at 1:08

It's only $47$ in physics. In mathematics, it would be

$$ \frac{20\sqrt{3}}{\sqrt{3}-1} = 47.320\ldots $$


Anyway: You start off with a proportion that can be written, generally, as

$$ \frac{a}{b} = \frac{c}{d} $$

The book then proceeds to rewrite this as

$$ \frac{d}{b} = \frac{c}{a} $$

That the two are equivalent (provided $a \not= 0$) can be seen by multiplying both sides of the first equation by $d$, and then dividing by $a$. You can also see that both equations yield the same result after cross-multiplication.

  • $\begingroup$ I guess I'm not understanding how you can rewrite it like that. Say for instance you have 3/5 = 3/5, if you rearrange it like you did above with the corresponding a,b,c,and d, then you'd get 5/5 = 3/3... which is 1 = 1. So how is that the same when one says 3/5 = 3/5, and the other says 1 = 1? $\endgroup$ – Luke Jan 26 '17 at 1:23
  • $\begingroup$ Multiply by d, divide by a. $\endgroup$ – Kaynex Jan 26 '17 at 1:24
  • $\begingroup$ @Luke: The issue isn't whether the equations are identical; it's whether they're logically equivalent. The first is true if and only if the second is true (provided $a \not= 0$), so they are logically equivalent. For instance, if I have $a = 2$, then it is also true that $2a = 4$ (and vice versa); the two are logically equivalent, even though the quantity of equality is $2$ in one case and $4$ in the other. $\endgroup$ – Brian Tung Jan 26 '17 at 2:47
  • $\begingroup$ @Luke: Speaking more generally, I'd say that if you aren't comfortable multiplying both sides of an equation by the same value and getting a logically valid result, you probably should work with it until you are comfortable with it, because it is used all the time in algebra (and therefore physics!), and getting stuck on this will obstruct further progress. $\endgroup$ – Brian Tung Jan 26 '17 at 2:48

The proportion they gave is correct. Simply divide both sides of the equation you started with by $6\cdot10^{-6}$ and multiply both sides by $(x-20)^2$.

So we have $$ \frac{(x-20)^2}{x^2} = \frac{2.0\cdot10^{-6}}{6.0\cdot10^{-6}}=\frac{1}{3}.$$ The left-hand side becomes $$\frac{(x-20)^2}{x^2}=\frac{x^2-40x+400}{x^2}=\frac{1}{3}.$$ Multiplying both sides by $x^2$, we get $$x^2-40x+400=\frac{x^2}{3} \quad\Rightarrow\quad\frac{2}{3}x^2-40x+400=0.$$ From here you use the quadratic equation to give you $x=10(3±\sqrt{3})$. It's physics, so you probably want the positive one, so $x=10(3+\sqrt{3})\approx47$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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