If x^2 + 4 y^2 = 4 x y then (x + y)/ (x - y) is
(A) 2/3
(B) 2
(C) 3
(D) 10/3
(E) 4
[spoiler]???????????[/spoiler]
x^2 + 4 y^2 = 4 x y
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- sanju09
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- shovan85
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IMO C
x^2 + 4y^2 - 4xy = 0
=> x^2 - 2 * x * 2y + (2y)^2 = 0
=> (x-2y)^2 = 0
=> x = 2y
=> x/y = 2
Apply Componendo Dividendo:
(x+y)/(x-y) = (2+1)/(2-1) = 3
FYI: https://en.wikipedia.org/wiki/Componendo_and_dividendo
x^2 + 4y^2 - 4xy = 0
=> x^2 - 2 * x * 2y + (2y)^2 = 0
=> (x-2y)^2 = 0
=> x = 2y
=> x/y = 2
Apply Componendo Dividendo:
(x+y)/(x-y) = (2+1)/(2-1) = 3
FYI: https://en.wikipedia.org/wiki/Componendo_and_dividendo
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- fskilnik@GMATH
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Excellent, Shovan.shovan85 wrote: x^2 + 4y^2 - 4xy = 0
=> x^2 - 2 * x * 2y + (2y)^2 = 0
=> (x-2y)^2 = 0
=> x = 2y
=> x/y = 2
Apply Componendo Dividendo:
(x+y)/(x-y) = (2+1)/(2-1) = 3
FYI: https://en.wikipedia.org/wiki/Componendo_and_dividendo
Please note that substituting x by 2y in the expression asked we find the exact same answer, for sure...
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- shovan85
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Thanks. I did that for cross verification but did not post it herefskilnik wrote:Excellent, Shovan.shovan85 wrote: x^2 + 4y^2 - 4xy = 0
=> x^2 - 2 * x * 2y + (2y)^2 = 0
=> (x-2y)^2 = 0
=> x = 2y
=> x/y = 2
Apply Componendo Dividendo:
(x+y)/(x-y) = (2+1)/(2-1) = 3
FYI: https://en.wikipedia.org/wiki/Componendo_and_dividendo
Please note that substituting x by 2y in the expression asked we find the exact same answer, for sure...
If the problem is Easy Respect it, if the problem is tough Attack it
- fskilnik@GMATH
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Sure! I just mentioned it because someone could imagine that this (neat) tool/result you used was essential to get the answer, and of course it was not the case.shovan85 wrote: I did that for cross verification but did not post it here
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- sandeep800
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although at first look i would have tried to solve it by using componendo and dividendo but yeah substitution works well here... !!
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- AtifS
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My explanation:
x^2 + 4y^2 = 4xy
=> x^2 + 4 y^2 - 4xy = 0
=> x^2 - 4 xy + 4 y^2 = 0
=> (x - 2y)^2 = 0
=> x - 2y = 0
=> x = 2y
=> x/2 = y
Now put the value of x/2 into (x + y)/(x - y)
(x + y)/(x - y)
= (x + x/2)/(x - x/2)
= {(2x+x)/2}/{(2x-x)/2}
= (3x/2)/(x/2)
= (3x/2) * (2/x)
= 3x/2 * 2/x ---> 2 will be canceled and x as well (because they are reciprocals of each other)
= 3 --> Answer
So, "C" is the answer for me, which is "3"
x^2 + 4y^2 = 4xy
=> x^2 + 4 y^2 - 4xy = 0
=> x^2 - 4 xy + 4 y^2 = 0
=> (x - 2y)^2 = 0
=> x - 2y = 0
=> x = 2y
=> x/2 = y
Now put the value of x/2 into (x + y)/(x - y)
(x + y)/(x - y)
= (x + x/2)/(x - x/2)
= {(2x+x)/2}/{(2x-x)/2}
= (3x/2)/(x/2)
= (3x/2) * (2/x)
= 3x/2 * 2/x ---> 2 will be canceled and x as well (because they are reciprocals of each other)
= 3 --> Answer
So, "C" is the answer for me, which is "3"
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x^2 + 4 y^2 = 4 x ysanju09 wrote:If x^2 + 4 y^2 = 4 x y then (x + y)/ (x - y) is
(A) 2/3
(B) 2
(C) 3
(D) 10/3
(E) 4
[spoiler]???????????[/spoiler]
x^2 + 4 y^2 - 4 x y=0(it is in form a^2+2ab+b^2 where a=x and b=2y)
so
(x-2y)^2=0
x=2y
now actual question is x+y / x -y =?
substitute x=2y
2y+y / 2y-y =3y/y =3
So C
thanks
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