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# If (x - y)^2/(x^2 - y^2) = 9, then (x + y)/(x - y) =

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Joined: 02 Sep 2009
Posts: 58427
If (x - y)^2/(x^2 - y^2) = 9, then (x + y)/(x - y) =  [#permalink]

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04 Feb 2018, 21:37
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Difficulty:

5% (low)

Question Stats:

90% (00:52) correct 10% (00:44) wrong based on 61 sessions

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If $$\frac{(x - y)^2}{x^2 - y^2} = 9$$, then $$\frac{x + y}{x - y} =$$

(A) 1/9
(B) 1/3
(C) 1
(D) 3
(E) 9

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Re: If (x - y)^2/(x^2 - y^2) = 9, then (x + y)/(x - y) =  [#permalink]

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04 Feb 2018, 21:44
A

(x-y)²/(x²-y²)=9

(x-y)(x-y)/(x+y)(x-y)=9

(x-y)/(x+y)=9

Inverse of it,

(x+y)/(x-y)=1/9
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Re: If (x - y)^2/(x^2 - y^2) = 9, then (x + y)/(x - y) =  [#permalink]

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19 Mar 2018, 09:40
Bunuel wrote:
If $$\frac{(x - y)^2}{x^2 - y^2} = 9$$, then $$\frac{x + y}{x - y} =$$

(A) 1/9
(B) 1/3
(C) 1
(D) 3
(E) 9

$$\frac{(x - y)^2}{x^2-y^2}$$ $$= 9$$

$$\frac{(x-y)(x-y)}{(x-y)(x+y)}$$ $$= 9$$

$$\frac{(x-y)}{(x+y)}$$ $$= 9$$

Therefore, $$\frac{x + y}{x - y}$$$$=$$$$\frac{1}{9}$$

(A)
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Re: If (x - y)^2/(x^2 - y^2) = 9, then (x + y)/(x - y) =  [#permalink]

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19 Mar 2018, 09:49
Bunuel wrote:
If $$\frac{(x - y)^2}{x^2 - y^2} = 9$$, then $$\frac{x + y}{x - y} =$$

(A) 1/9
(B) 1/3
(C) 1
(D) 3
(E) 9

$$\frac{(x - y)^2}{x^2 - y^2} = 9$$

Or, $$\frac{(x - y)(x - y)}{(x + y)(x - y)} = 9$$

Or, $$\frac{(x - y)}{(x + y)} = 9$$

So, $$\frac{x + y}{x - y} =\frac{1}{9}$$, Answer must be (A)
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Re: If (x - y)^2/(x^2 - y^2) = 9, then (x + y)/(x - y) =   [#permalink] 19 Mar 2018, 09:49
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