Is x^2 equal to xy? (1) x^2 + y^2 = -(x^2 y^2) (2) x = y

the question can be rephrased as \(x^2-xy\) or\(x(x-y)\) So for the inequality to be equal to 0 either \(x=0\) or \(x=y\)
From A:
\(x^2 + y^2 = -(x^2 – y^2)\)
\(2x^2=0\) so \(x=0\)
A is sufficient
From b:
x=y
Y is sufficient

Correct answer is D

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.

Is x^2 equal to xy?

(1) x^2 + y^2 = -(x^2 – y^2)
(2) x = y

If we modify the original condition and the question, we want to know whether x^2-xy=0, which can further be simplified into x(x-y)=0 and whether x=0 or x=y.
There are 2 variables (x,y), meaning we need 2 equations to match the number of variables. The conditions give us 2 equations in total, making it likely that (C) is going to be our answer, but if we look closely into the conditions,
For condition 1, x^2+y^2=-x^2+y^2, x^2=-x^2, 2x^2=0, so x=0, answering the question ‘yes’
For condition 2, x=y, also answering the question ‘yes’. Therefore the 2 conditions are each sufficient alone to answer the question, and the answer becomes (D)

For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

VERITAS PREP OFFICIAL SOLUTION:

Question Type: Yes/No The question asks: “Is x2 = xy?”

For this question it is easier to begin with Statement 2.

Statement 2: x = y. If x = y then x^2 will automatically equal xy. So this statement is sufficient. The answer is either B or D.

Statement 1: x^2 + y^2 = - (x^2 – y^2). Generally with algebra you want to eliminate parentheses, so in this case you’ll want to multiply each term in (x^2 – y^2) by -1 to distribute that negative sign; the statement becomes x^2 + y^2 = -x^2 + y^2. If you then manipulate this statement algebraically by adding x^2 to both sides and by subtracting y^2 from both sides you get 2x^2 = 0. For this to be true x must 0. And if x = 0, then 0^2 = 0(y) by definition—both sides of the equation would equal 0. This statement is also sufficient and the correct answer is D.

How about:

\(x^2+y^2 = -(x^2-y^2)\)

\(x^2+y^2+(x^2-y^2)=0\)

\(x^2+y^2+((x+y)(x-y))=0\)

\(x^2+x^2+y^2-y^2-2xy=0\)

\(2x^2-2xy=0\)

\(2x^2=2xy\)

\(x^2=xy\) same as stem. Suf.

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