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Source: — Data Sufficiency |
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by gmatboost » Wed Aug 17, 2011 12:36 pm
Systems of equations in which one equation is

x + y = k or x - y = k
and the other is
xy = p

Always lead to quadratic equations, which almost always have two solutions for each variable.
Don't fall for the "obvious" positive answer for x.
And don't assume that this falls under the "two equations two variables -> sufficient" category.
The equation xy = p is NOT linear, so that idea doesn't apply here.

In this case,
xy = 12
[spoiler](y+1)y = 12[/spoiler]
[spoiler]y^2 + y - 12 = 0[/spoiler]
[spoiler](y + 4)(y - 3) = 0[/spoiler]
[spoiler]y = -4 or y = 3[/spoiler]
[spoiler]x = -3 or x = 4 Insufficient[/spoiler]
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by Anurag@Gurome » Thu Aug 18, 2011 5:37 am
GmatKiss wrote:What is the value of x?

1)x-y=1
2)xy=12
None of the statements individually is sufficient to determine the value of x. So let us consider both of them together. From statement 1, y = (x - 1)

Hence, xy = x(x - 1) = 12
The above quadratic equation will yield two values of x and we cannot determine any unique value of x.

The correct answer is E.
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by gmatboost » Mon Aug 22, 2011 9:51 am
This is generally true but occasionally it isn't:

x(x-6) = -9 and any other equation that is derived from "perfect square = 0" has only one solution.
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