Find the numerical value of x/y

**DivideBy0** · October 20th 2007, 05:22 PM

If I try this question in different ways I get a different answer:

For positive $x,y,z$ ,

$\frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}$

Find the numerical value of $\frac{x}{y}$

(Using the fact that $\frac{a}{b}=\frac{c}{d}=\frac{a+kc}{b+kd}$ ),

Solution 1:

$\frac{y+(x+y)}{x-z+(z)}=\frac{x}{y}$

$\implies \frac{2y+x}{x}=\frac{x}{y}$

$\implies 2y+x=y$

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

Solution 2:

$\frac{x}{y}=\frac{y+(x+y)+(x)}{x-z+(z)+(y)}=\frac{2(x+y)}{x+y}=2$

**DivideBy0** · October 20th 2007, 05:28 PM

Nope, it works for all k, supposedly.

$\frac{a+kc}{b+kd}=\frac{a(1+k(\frac{c}{a}))}{b(1+k (\frac{d}{b}))}=\frac{a(1+k(\frac{c}{a}))}{b(1+k(\ frac{c}{a}))}=\frac{a}{b}$

**Jhevon** · October 20th 2007, 05:28 PM

Originally Posted by topsquark

Where did you get this identity from? It is not true for all k:
$\frac{1}{2} = \frac{3}{6} = \frac{1 + 3k}{2 + 6k}$
is only valid for k = 0.

-Dan

actually, it is true otherwise. did you try other values for k?

**DivideBy0** · October 20th 2007, 05:30 PM

Ooops, look how I canceled the x's in the first solution. Happens way to much.

**ticbol** · October 20th 2007, 05:32 PM

And

if (2y +x)/x = x/y

Then (2y +x) is not equal to y.

**topsquark** · October 20th 2007, 05:33 PM

Originally Posted by DivideBy0

For positive $x,y,z$ ,

$\frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}$

Find the numerical value of $\frac{x}{y}$

Take the equations like so:
$\frac{y}{x-z} = \frac{x+y}{z}$
and
$\frac{y}{x-z} = \frac{x}{y}$
(The third equation is implied by these two.)

The first one says:
$yz = (x + y)(x - z)$

$\frac{z}{y} = \left ( \frac{x}{y} + 1 \right ) \left ( \frac{x}{y} - \frac{z}{y} \right )$

and the second says
$y^2 = x(x - z)$

$1 = \left ( \frac{x}{y} \right ) \left ( \frac{x}{y} - \frac{z}{y} \right )$

Now define $a = \frac{x}{y}$ and $b = \frac{z}{y}$ and you get the system:
$b = (a + 1)(a - b)$
and
$1 = a(a - b)$

You are after the value of a.

-Dan

**DivideBy0** · October 20th 2007, 05:33 PM

thanks, lol

Interesting approach @ topsquark

geeze, you guys are replying so faast today

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