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16 Sep 2014, 01:28
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If \(x\neq{y}\) and \(y = \frac{x^2 - y^2}{x-y}\), what is the value of \(y\)?
(1) \(x+y-3=0\)
(2) \(x(y-3)=0\)
(1) \(x+y-3=0\)
(2) \(x(y-3)=0\)
16 Sep 2014, 01:28
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Official Solution:
If \(x\neq{y}\) and \(y = \frac{x^2 - y^2}{x-y}\), what is the value of \(y\)?
We are given \(y = \frac{x^2 - y^2}{x-y}\). Simplify: \(y = \frac{(x - y)(x + y)}{x - y}\). Cancel out the \((x - y)\) terms: \(y = x + y\). From this, we get that \(x = 0\).
(1) \(x+y-3=0\).
Substituting \(x = 0\) into the equation above gives us \(y = 3\). Sufficient.
(2) \(x(y-3)=0\).
Since we know that \(x = 0\), the product \(x(y - 3)\) will be 0, regardless of the value of \(y\). This means \(y\) can take any value. Not sufficient.
Answer: A
If \(x\neq{y}\) and \(y = \frac{x^2 - y^2}{x-y}\), what is the value of \(y\)?
We are given \(y = \frac{x^2 - y^2}{x-y}\). Simplify: \(y = \frac{(x - y)(x + y)}{x - y}\). Cancel out the \((x - y)\) terms: \(y = x + y\). From this, we get that \(x = 0\).
(1) \(x+y-3=0\).
Substituting \(x = 0\) into the equation above gives us \(y = 3\). Sufficient.
(2) \(x(y-3)=0\).
Since we know that \(x = 0\), the product \(x(y - 3)\) will be 0, regardless of the value of \(y\). This means \(y\) can take any value. Not sufficient.
Answer: A
RenanBragion
11 Jun 2020, 06:32
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I think this is a high-quality question and I agree with explanation.
