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If 0 < x < y, what is the value of \(\frac{(x + y)^2}{(x- y)^2}\)?
(1) \(x^2 + y^2 = 3xy\)
(2) \(xy = 3\)
15.gif [ 19.21 KiB | Viewed 16838 times ]
(1) \(x^2 + y^2 = 3xy\)
(2) \(xy = 3\)
Attachment:
15.gif [ 19.21 KiB | Viewed 16838 times ]
24 Mar 2013, 02:40
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If 0 < x < y, what is the value of (x + y)^2/( x- y)^2?
\(\frac{(x + y)^2}{( x- y)^2}=\frac{x^2+2xy+y^2}{x^2-2xy+y^2}\)
(1) x^2 + y^2 = 3xy. Substitute: what is the value of \(\frac{x^2+2xy+y^2}{x^2-2xy+y^2}=\frac{3xy+2xy}{3xy-2xy}=\frac{5xy}{xy}=5\). Sufficient.
(2) xy = 3. Not sufficient.
Answer: A.
\(\frac{(x + y)^2}{( x- y)^2}=\frac{x^2+2xy+y^2}{x^2-2xy+y^2}\)
(1) x^2 + y^2 = 3xy. Substitute: what is the value of \(\frac{x^2+2xy+y^2}{x^2-2xy+y^2}=\frac{3xy+2xy}{3xy-2xy}=\frac{5xy}{xy}=5\). Sufficient.
(2) xy = 3. Not sufficient.
Answer: A.
24 Mar 2013, 00:29
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1)
x^2+y^2=3xy => x^2+y^2-2xy=xy => (x-y)^2=xy
So you can replace : (x+y)^2/xy
And then just finish the work : (x+y)^2/xy => (x^2+y^2+2xy)/xy => (3xy+2xy)/xy => 5
1 is enough
2) not enough. (x+y)^2/(x-y)^2 => (x^2+y^2+2xy)/(x^2+y^2-2xy) => (x^2+y^2+6)/(x^2+y^2-6) => you can't know the value of x^2 or y^2
Hope it helps.
x^2+y^2=3xy => x^2+y^2-2xy=xy => (x-y)^2=xy
So you can replace : (x+y)^2/xy
And then just finish the work : (x+y)^2/xy => (x^2+y^2+2xy)/xy => (3xy+2xy)/xy => 5
1 is enough
2) not enough. (x+y)^2/(x-y)^2 => (x^2+y^2+2xy)/(x^2+y^2-2xy) => (x^2+y^2+6)/(x^2+y^2-6) => you can't know the value of x^2 or y^2
Hope it helps.
