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


The top equation is the problem. But I think I've solved it correctly for x
$\frac{1}{3y} = \frac{x}{x}-\frac{y}{x}$

$\frac{1}{3y} = 1-\frac{y}{x}$

$\frac{y}{x} = 1-\frac{1}{3y}$

$\frac{X}{Y} = 1-3y$

$x = y-3y^2$
Put the X equal to eachother and solve for Y?
$y-3y^2 = 4y$

That gives Y values $-1$ and $0$, which is not correct. So do I have to solve for Y first and find X-values or did I do a mistake somewhere?

share|cite|improve this question
As a matter of solving strategy, it seems better first to use the second equation to conclude that $x=4y$, and then to substitute for $x$ in the first equation. – André Nicolas Oct 4 '11 at 15:03
Is the recreational-math tag really appropriate? Wouldn't algebra-precalculus be better? – Gerry Myerson Oct 5 '11 at 5:43
up vote 2 down vote accepted

$\frac{y}{x} = 1-\frac{1}{3y}$ implies $\frac{x}{y} = \frac{1}{1-\frac{1}{3y}}$ , not $\frac{x}{y} = 1-3y$.

share|cite|improve this answer
That explains a bit, thanks. – Algific Oct 4 '11 at 14:54

$3y=\frac{x}{x-y} /(x-y) $ multiply by $(x-y) $ both sides so we get


From the second equation we may conclude that $x=4y$ ,If we substitute $x$ with $4y$ at the first equation we may write

$3y(4y-y)=4y \Rightarrow 9y^2-4y=0 \Rightarrow y(9y-4)=0$ , so: $y_1=0$ and $y_2=\frac{4}{9}$ which means that $x_1=0$ and $x_2=\frac{16}{9}$ ,Since $x-y\neq0$ only solution is $(x_2,y_2)=(\frac{16}{9},\frac{4}{9})$

share|cite|improve this answer

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.