Is x^2-y^2>x+y?

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Is x^2-y^2>x+y?

by Max@Math Revolution » Mon Jun 13, 2016 9:00 am
Is x^2-y^2>x+y?
1) x<y
2) x+y<0

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by flsh » Wed Jun 15, 2016 7:25 am
Max@Math Revolution wrote:Is x^2-y^2>x+y?
1) x<y
2) x+y<0
x^2 - y^2 = (x - y)(x + y)
So, question is: (x - y)(x + y) > x + y ?

(1) x < y => x - y < 0, no info about x + y. Insufficient
(2) x + y < 0, no info about x - y. Insufficient
(Both) (x - y)(x + y) > 0, x + y < 0 => x^2 - y^2 > x + y. Sufficient

C is the answer.
Last edited by flsh on Thu Jun 16, 2016 7:19 am, edited 1 time in total.

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by Max@Math Revolution » Wed Jun 15, 2016 6:41 pm
We can modify the original condition and the question. Then, the question becomes x^2-y^2>x+y? (x-y)(x+y)-(x+y)>0?, (x-y-1)(x+y)>0?. Then, since there are 2 variables, there is high chance that C is the correct answer. Using the condition 1) and the condition 2), the answer is always yes and the conditions are sufficient. Hence, the correct answer is C.

- Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.