Does x = y ? (1) x^2 = y^2 (2) |x| = |y|

Does x = y ?

(1) x^2 = y^2
If x=1 and y=1 [So, (1)^2=(1)^2] , x=y
If x=1 and y=-1 [So, (1)^2=(-1)^2 ] , x is NOT equal to Y
INSUFFICIENT.

(2) |x| = |y|
If x=1 and y=1 [So, |1| = |1| ] , x=y
If x=1 and y=-1 [So, |1| = |-1| ] , x is NOT equal to Y
INSUFFICIENT.

Combining:
Since same values make both statements INSUFFICIENT, no need to combine.
E is the answer

The answer is E.

x=y?

1) X^2=y^2
Remember to beware of even exponents! They are dangerous. This simply means that |x|=|y|.
2)|x|=|y| Says the same thing as the above.

None of these tell us whether x and y are negative or postive. The answer is E.

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Since we have 2 variables (\(x\) and \(y\)) and 0 equations, C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)

The condition 1) is equivalent to \(x = y\) or \(x = -y\) for the following reason.
\(x^2 = y^2\)
\(⇔ x^2 - y^2 = 0\)
\(⇔ (x-y)(x+y)=0\)
\(⇔ x = y\) or \(x = -y\)


The condition 2) is equivalent to \(x = y\) or \(x = -y\) for the following reason.
\(|x|=|y|\)
\(⇔ |x|^2 = |y|^2\), we can square both sides since they are greater than or equal to
\(⇔ x^2 = y^2\), we can eliminate | | notations from both sides
\(⇔ x^2 - y^2 = 0\)
\(⇔ (x-y)(x+y)=0\)
\(⇔ x = y\) or \(x = -y\)

"\(x=y\)" means the answer is 'yes'.
"\(x=-y\)" means the answer is 'no'.

Since both conditions together do not yield a unique solution, they are not sufficient.

Therefore, E is the answer.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B, or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D, or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.

x = y?

St 1: x^\({2}\) = y^\({2}\) ==> (x-y)(x+y) = 0 ==> x = y or -y ==> insufficient

St 2: |x| = |y| ==> x = y or x = -y ==> insufficient

Sts 1 & 2 ==> we have no new info since sts 1 and 2 are the same ==> answer is E

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.