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Bunuel
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Bunuel

Tough and Tricky questions: Algebra.



If \(\frac{\sqrt{x} + \sqrt{y}}{x - y} = \frac{2\sqrt{x} + 2\sqrt{y}}{x + 2\sqrt{xy} + y}\), what is the ratio of x to y ?

A. 1/2
B. 2
C. 4
D. 7
E. 9

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x-y= (√x)^2-(√y)^2
=(√x +√y)(√x - √y)

thus (√x + √y)/(x – y) = (√x + √y)/ (√x +√y)(√x - √y)
= 1/(√x - √y)
also (√x + √y)^2 = x + 2√(xy) + y

thus (2√x + 2√y)/[x + 2√(xy) + y] = 2 (√x +√y)/ [(√x + √y)^2]
= 2/(√x + √y)
thus, we have
1/(√x - √y) = 2/(√x + √y)
or,
(√x + √y)= 2√x-2√y
√x = 3√y
squaring both sides we have
x=9y
or x/y=9
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Bunuel

Tough and Tricky questions: Algebra.



If \(\frac{\sqrt{x} + \sqrt{y}}{x - y} = \frac{2\sqrt{x} + 2\sqrt{y}}{x + 2\sqrt{xy} + y}\), what is the ratio of x to y ?

A. 1/2
B. 2
C. 4
D. 7
E. 9

Kudos for a correct solution.

Another Excellent Bunuel

Here After Factoring i got (√x + √y)^2 = 2 (x-y)
without writing x-y = (√x)^2 - (√y)^2 this question couldnt be solved..

Kudos ...!!
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Set a=x^1/2 and b=y^1/2

Substituting

(a+b)/(a^2-b^2)=

2(a+b)/(a+b)^2


Factoring both sides

1/(a-b) = 2/(a+b)

So a+b = 2a-2b and a=3b, so

a/b=3=(x/y)^1/2

So x/y = 9

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x√+y√x−y=2x√+2y√x+2xy−−√+yx+yx−y=2x+2yx+2xy+y

x√+y√(x√−y√)(x√+y√)=2(x√+y√)(x√+y√)2x+y(x−y)(x+y)=2(x+y)(x+y)2

x√+y√x√−y√=2x+yx−y=2

Using componendo / dividend

2x√x√−y√=2+11=32xx−y=2+11=3 ....................... (1)

2y√x√−y√=2−11=12yx−y=2−11=1 ........................ (2)

Divide (1) by (2)

x√y√=31xy=31

xy=9xy=9

Answer = E
out here instead of compenendo nd dividendo just substitute under root x as a and underoot y as b
so a+b/a-b = 2
a+b = 2(a-b)
3b = a
now resubstitute 3rooty = rootx
square both sides hence youll get x = 9y
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