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# If (x + y)/(x − y) = 1/2, then (xy + x^2)/(xy − x^2) =

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Re: If (x + y)/(x − y) = 1/2, then (xy + x^2)/(xy − x^2) = [#permalink]
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Bunuel wrote:
If $$\frac{x + y}{x − y} = \frac{1}{2}$$, then $$\frac{xy + x^2}{xy − x^2}=$$

(A) –4.2

(B) –1/2

(C) 1.1

(D) 3

(E) 5.3

Taking x as common from the fraction; $$\frac{xy + x^2}{xy − x^2}$$
$$\frac{x(y + x)}{x(y - x)}$$
Cancelling out "x" we get = $$\frac{x+y}{-(x-y)}$$

Given; $$\frac{x + y}{x − y} = \frac{1}{2}$$
Therefore ; $$\frac{x+y}{-(x-y)}$$ = - $$\frac{1}{2}$$
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Re: If (x + y)/(x − y) = 1/2, then (xy + x^2)/(xy − x^2) = [#permalink]
(xy + x^2)/(xy-x^2)
= x(y+x)/x(y-x)
= (y+x)/(y-x)
= -(x+y)/(x-y) = -1/2

Option B should be the correct answer.
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Re: If (x + y)/(x − y) = 1/2, then (xy + x^2)/(xy − x^2) = [#permalink]
(xy+x^2)/(xy-x^2)= (y+x)/(y-x)= -((x+y)/(x-y))= -1/2
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Re: If (x + y)/(x − y) = 1/2, then (xy + x^2)/(xy − x^2) = [#permalink]
Bunuel wrote:
If $$\frac{x + y}{x − y} = \frac{1}{2}$$, then $$\frac{xy + x^2}{xy − x^2}=$$

(A) –4.2

(B) –1/2

(C) 1.1

(D) 3

(E) 5.3

Let’s simplify (xy + x^2)/(xy - x^2):

x(y + x)/[x(y - x)]

We know x cannot equal zero (because (x + y)/(x - y) will not equal 1/2); therefore we can cancel the x:

(x + y)/(y - x)

(x + y)/(x - y) * -1

½ * -1 = -½

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Re: If (x + y)/(x − y) = 1/2, then (xy + x^2)/(xy − x^2) = [#permalink]
Take the x from the numerator and denominator and -1 from the denominator to get the given condition.

-1*(x+y)/(x-y)= -1/2

Hence Ans:B
Re: If (x + y)/(x − y) = 1/2, then (xy + x^2)/(xy − x^2) = [#permalink]
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