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Two non zero real numbers, x and y, satisfy xy = x-y. which of the fol

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Two non zero real numbers, x and y, satisfy xy = x-y. which of the fol  [#permalink]

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New post 14 Jan 2015, 11:52
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Two non zero real numbers, x and y, satisfy \(xy = x-y\). which of the following is a possible value of \(\frac{x}{y}\) + \(\frac{y}{x}\) \(- xy\) ?

A) -2
B) -1/2
C) 1/3
D) 1/2
E) 2
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Two non zero real numbers, x and y, satisfy xy = x-y. which of the fol  [#permalink]

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New post 14 Jan 2015, 15:30
1
manpreetsingh86 wrote:
Two non zero real numbers, x and y, satisfy \(xy = x-y\). which of the following is a possible value of \(\frac{x}{y}\) + \(\frac{y}{x}\) \(- xy\) ?

A) -2
B) -1/2
C) 1/3
D) 1/2
E) 2



\(\frac{x}{y}\) + \(\frac{y}{x}\) \(- xy\)

\(\frac{(x^2+y^2)}{xy}\) \(- (x-y)\)

\(\frac{(x^2+y^2)}{(x-y)}\) \(-\frac{(x-y)^2}{(x-y)}\)

\(\frac{(x^2+y^2)}{(x-y)}\) \(-\frac{(x^2-2xy+y^2)}{(x-y)}\)

\(\frac{2xy}{(x-y)}\)

\(\frac{2xy}{xy}\)

\(2\)

E.
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Re: Two non zero real numbers, x and y, satisfy xy = x-y. which of the fol  [#permalink]

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New post 14 Jan 2015, 17:20
Hi manpreetsingh86,

BassanioGratiano has already presented a great "Algebra" explanation for this question, so I won't rehash that math here. Instead, here's how a combination of TESTing VALUES and some Number Property logic can be used:

We're told that X and Y CANNOT be 0, so the equation XY = X-Y is interesting....What types of numbers have a product that is equal to their DIFFERENCE??? The rest of the prompt and the answer choices strongly *hint* that fractions may be involved.

The first/simplest example I could come up with was...

X = 1
Y = 1/2
1(1/2) = 1 - 1/2
1/2 = 1/2

With those 2 numbers, let's see what result we get in the question...

X/Y + Y/X - XY
1/(1/2) + (1/2)/1 - 1(1/2)
2 + 1/2 - 1/2
2

That answer APPEARS in the answer choices, so we're done.

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Re: Two non zero real numbers, x and y, satisfy xy = x-y. which of the fol  [#permalink]

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New post 15 Jan 2015, 01:55
here is an another solution

xy = x-y --------------------------1)
dividing both sides by xy we have;
1 = 1/y - 1/x ---------------------------2)

now coming to the original equation.

\(\frac{x}{y}\) + \(\frac{y}{x}\) \(- xy\)

substitute the value of xy , from 1 we have

\(\frac{x}{y}\) + \(\frac{y}{x}\) \(-(x-y)\)

\(\frac{x}{y}\) + \(\frac{y}{x}\) \(-x+y\)

\(\frac{x}{y}\)\(-x\) + \(\frac{y}{x}\) \(+y\)

\(x(\frac{1}{y}-1)\) +\(y(\frac{1}{x}+1)\)

from 2) we know that (1/y-1) =1/x and (1/x+1) = 1/y

thus we have

x(1/x)+y(1/y) =2
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Re: Two non zero real numbers, x and y, satisfy xy = x-y. which of the fol  [#permalink]

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New post 07 Jan 2019, 20:00
BassanioGratiano wrote:
manpreetsingh86 wrote:
Two non zero real numbers, x and y, satisfy \(xy = x-y\). which of the following is a possible value of \(\frac{x}{y}\) + \(\frac{y}{x}\) \(- xy\) ?

A) -2
B) -1/2
C) 1/3
D) 1/2
E) 2



\(\frac{x}{y}\) + \(\frac{y}{x}\) \(- xy\)

\(\frac{(x^2+y^2)}{xy}\) \(- (x-y)\)

\(\frac{(x^2+y^2)}{(x-y)}\) \(-\frac{(x-y)^2}{(x-y)}\)

\(\frac{(x^2+y^2)}{(x-y)}\) \(-\frac{(x^2-2xy+y^2)}{(x-y)}\)

\(\frac{2xy}{(x-y)}\)

\(\frac{2xy}{xy}\)

\(2\)

E.



Hello!

Could someone please explain to me why do we have to square the second term?

\(\frac{(x^2+y^2)}{(x-y)}\) \(-\frac{(x-y)^2}{(x-y)}\)

Kind regards!
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Re: Two non zero real numbers, x and y, satisfy xy = x-y. which of the fol   [#permalink] 07 Jan 2019, 20:00
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