If x > y, x² – 2xy + y² – 9 = 0, and x + y = 15, what is x?
A. -3
B. 0
C. 3
D. 6
E. 9
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find the value of x
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parallel_chase
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if you look closely x² – 2xy + y² – 9 = 0 can be simplified as (x-y)^2 =9
Rest is just a cake walk.
x-y = 3
x+y=15
2x=18
x=9
Rest is just a cake walk.
x-y = 3
x+y=15
2x=18
x=9
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sibbineni
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If x > y, x² – 2xy + y² – 9 = 0 and x + y = 15, what is x?
(x² – 2xy + y²)-9=0
(x² – 2xy + y²) is of the form (x-y)^2
i.e (a-b)^2=a^2-2ab+b^2
so (x-y)^2=9
x-y=+/- 3
but as x>y
x-y=3--(1)
x+y=15---(2)
solve for x
x+x-3=15
2x=18
x=9
(x² – 2xy + y²)-9=0
(x² – 2xy + y²) is of the form (x-y)^2
i.e (a-b)^2=a^2-2ab+b^2
so (x-y)^2=9
x-y=+/- 3
but as x>y
x-y=3--(1)
x+y=15---(2)
solve for x
x+x-3=15
2x=18
x=9
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jimmytwoshoes
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Couldn't you also say that because x>y and x+y=15 that x has to be at least 7.5, and e is the only one greater than 7.5?
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*arch*
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An even simpler way would be to look at the answer choices.
We have a condition x>y that must be satisfied in the equation x+y = 15.
Only option (E) where x = 9 satisfies this condition. In all other options (A), (B), (C) & (D), x will be less than y which violates the condition stated in the question.
We have a condition x>y that must be satisfied in the equation x+y = 15.
Only option (E) where x = 9 satisfies this condition. In all other options (A), (B), (C) & (D), x will be less than y which violates the condition stated in the question.
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gmat_novice
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luiscarlos59
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I have a question regarding the point where you get
(x-y) = +/- 3
While using the crtieria of x>y , i dont understand why to chose the
x-y = +3 and not x-y= -3 ?
If x is greater than y , you could get -3 as an answer also.
(x-y) = +/- 3
While using the crtieria of x>y , i dont understand why to chose the
x-y = +3 and not x-y= -3 ?
If x is greater than y , you could get -3 as an answer also.
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cans
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x>y means x-y>0 but -3 is not greater than 0. Thus we can't get -3 as an answer.luiscarlos59 wrote:I have a question regarding the point where you get
(x-y) = +/- 3
While using the crtieria of x>y , i dont understand why to chose the
x-y = +3 and not x-y= -3 ?
If x is greater than y , you could get -3 as an answer also.
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Cans!!
Contact me about long distance tutoring!
[email protected]
Cans!!
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aftableo2006
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smackmartine
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If you take x-y= -3 , and solve x-y = -3 & x+y=15 , you will get:luiscarlos59 wrote:I have a question regarding the point where you get
(x-y) = +/- 3
While using the crtieria of x>y , i dont understand why to chose the
x-y = +3 and not x-y= -3 ?
If x is greater than y , you could get -3 as an answer also.
x=6 , and y = 9 ,but its given that x>y , So you cannot take x-y = -3
Solving x-y = 3 & x+y=15, we get x=9 , y= 6
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krishnasty
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excellent answer arch..*arch* wrote:An even simpler way would be to look at the answer choices.
We have a condition x>y that must be satisfied in the equation x+y = 15.
Only option (E) where x = 9 satisfies this condition. In all other options (A), (B), (C) & (D), x will be less than y which violates the condition stated in the question.
