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$3y=\frac{x}{x-y}$
$x+4y=2x$

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?
$y-3y^2 = 4y$

$-3y^2-3y=0$
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?

http://www.wolframalpha.com/input/?i=3y%3Dx%2F%28x-y%29%2C+x%2B4y%3D2x

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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
1  
Is the recreational-math tag really appropriate? Wouldn't algebra-precalculus be better? –  Gerry Myerson Oct 5 '11 at 5:43

2 Answers 2

up vote 2 down vote accepted

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

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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

$3y(x-y)=x$

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 $(x_2,y_2)=(\frac{16}{9},\frac{4}{9})$

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