FJ24
gmatopoeia
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Quote:
If y <> = 0 Does x = 0
(1) x^2 y = x^3 y
(2) x^2 / y = x^3 y
Very clear explanation, Bunuel. Thanks!
Initially, I had answer A and realised my mistake after reading Bunuel's solution. My error came from the fact that I cross-multiplied and eliminated as below:
Quote:
From Statement 1
--> x^2y=x^3y --> y/y = x^3/x^2 --> x = 1
Sufficient
Given y does not equal to zero, does x=0?
Statement 1x^2y = x^3y --> x^3=x^2 -->
0 = x^2(x-1)So, x=0 or x=1
Insufficient.
Statement 2x^2/y = x^3y -->
0 = x^2(xy^2-1)So, x=0 or xy^2=1 (i.e. x is not zero)
Insufficient.
Statement 1 + 2x^2(x-1) = x^2(xy^2-1)
x-1 = xy^2-1
x(1-y^2) = 0
So, x = 0 or x does not equals to zero (when y^2=1)
Insufficient.
Answer: E
Thank you for the above explanation. However, I'm still confused as to how we are getting from "x^3=x^2 --> 0 = x^2(x-1)". How are we getting to zero on one side? If we divide both sides by x^2, don't we end up 1. Can someone pls clarify? Same for statement 2. Thank you!
(1) \(x^2=x^3\). (Notice here that we cannot divide this by x^2 to get 1 = x because x can be 0 and we cannot divide by 0. By doing so, we loose the root, namely x = 0, which also satisfies x^2 = x^3)
Rearrange: \(x^3 - x^2 = 0\);
Factor out x^2: \(x^2(x-1)=0\);
Either \(x=0\) or \(x=1\).
(2) \(\frac{x^2}{y}=x^3*y\);
Cross-multiply: \(x^2=x^3y^2\) (the same here: we cannot divide by x^2 to get 1 = xy^2 because x^2 can be zero and we cannot divide by 0.);
Rearrange: \(x^3y^2-x^2=0\);
Factor out x^2: \(x^2(xy^2-1)=0\);
Either \(x=0\) or \(xy^2=1\).