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if x and y are integers

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if x and y are integers

by navalpike » Fri Jul 17, 2009 9:00 pm
X and y are integers such that x<y<0. What is x?

1. (x+y) (x-y) = 7
2. xy = 12

For 1. After FOILing the left side and isolating x, we reach
X^2 = y^2 + 7
And then we take the Sq. root of both sides, resulting in
X = Sq. root (y^2+7)

After this GMATfocus simply says that “trial and error will show that the only value of y for which this equation results in an integer is -3, which makes x = -4”.

Trial and error is fine, but I wasn't able to pick the right numbers. Can you think of another method to solve the problem?

Thanks,
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by tohellandback » Fri Jul 17, 2009 10:51 pm
I think you better not do calculations for this.
given,
both x and y negative and they are integers
1)(x+y) (x-y) = 7 , product of two numbers is 7
possibilities(-1,-7), (1,7)
(1,7) not possible because sum of two negative integers cannot be positive
so (-1,-7). from this you can find X because we know that x+y=-7 and x-y=-1.

1- SUFFICIENT

2)xy=12
not sufficient
answer IMO A
The powers of two are bloody impolite!!
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by navalpike » Sat Jul 18, 2009 9:23 am
Nice solution, thanks.
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by PussInBoots » Sun Jul 19, 2009 11:34 pm
possibilities are (-1,-7), (1,7) and possibilities(-7,-1), (7,1). 2nd set does not matter in this problem, but it definitely might in one of similar type of problems.
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by Domnu » Mon Jul 20, 2009 7:04 pm
The answer should be A. Since 7 is prime,

x+y = -7
x-y = -1

OR

x+y = -1
x-y = -7

are the only possibilities. You will find that the former works while the latter doesn't as far as the x<y<0 constraint goes.
Have you wondered how you could have found such a treasure? -T
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by adityanarula » Tue Jul 21, 2009 8:46 am
Perhaps I am missing something, but doesnt Smtnt 1 imply x^2-y^2 = 7

Thus, x & y could be 4 & 3 respectively?
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by Domnu » Tue Jul 21, 2009 8:49 am
But x, y < 0 according to the constraints.
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by adityanarula » Tue Jul 21, 2009 9:14 am
I am sorry, I meant x -4 and y = -3

In this case x<y<0

and x^2 = 16
y^2 = 9

and x^2-y^2 = 7!
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