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Equation Systems

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Equation Systems

by yellowho » Sat Feb 26, 2011 7:17 pm
If you are given an two equations one LINEAR and one NOT-LINEAR can you solve for it? I know it gives multiple answers but in the context of DS. When you see it do you still have to solve and check or is that sufficient to say the choice is insufficient?

Ex: (Totally made up by me)

If x+y=-5, What is the value of X and Y?

1) xy=6

2) x-y=4
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Source: — Data Sufficiency |
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by Night reader » Sat Feb 26, 2011 7:34 pm
I prefer killing DS till end :(
my strategy here would be to start from st(2): (x-y=4)+(x+y=-5) --> 2x=-1, x=-1/2; y=-4.5 Sufficient
then st(1) (x+y)^2=25, 2xy=12, x^2+y^2=25-12, x^2+y^2=13 Not Sufficient as there can be mod x and mod y answers i.e. -ve /+ve
yellowho wrote:If you are given an two equations one LINEAR and one NOT-LINEAR can you solve for it? I know it gives multiple answers but in the context of DS. When you see it do you still have to solve and check or is that sufficient to say the choice is insufficient?

Ex: (Totally made up by me)

If x+y=-5, What is the value of X and Y?

1) xy=6

2) x-y=4
in general quadratic equations have more than one answer unless it's just (x+c)^2=0, then only x=-c
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by Anurag@Gurome » Sat Feb 26, 2011 8:24 pm
(1) x(-x - 5) = 6 or x^2 + 5x + 6 = 0 or (x + 2)(x + 3) = 0 gives 2 values of x and hence two values of y. Since we don't get a unique answer, so (1) is NOT SUFFICIENT.

(2) We have two linear equations and two variables. So, they can be solved to get values of x and y. Hence, (2) is SUFFICIENT.

The correct answer is B.
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