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

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Senior Manager
Joined: 13 Jun 2013
Posts: 271
Two non zero real numbers, x and y, satisfy xy = x-y. which of the fol  [#permalink]

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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
Manager
Joined: 23 Oct 2014
Posts: 86
Concentration: Marketing
Two non zero real numbers, x and y, satisfy xy = x-y. which of the fol  [#permalink]

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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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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

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Senior Manager
Joined: 13 Jun 2013
Posts: 271
Re: Two non zero real numbers, x and y, satisfy xy = x-y. which of the fol  [#permalink]

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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
Senior Manager
Joined: 12 Sep 2017
Posts: 295
Re: Two non zero real numbers, x and y, satisfy xy = x-y. which of the fol  [#permalink]

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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!
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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