Bunuel wrote:

If \(x\neq{y}\) and \(y = \frac{x^2 - y^2}{x-y}\), then what is the value of \(y\)?

(1) \(x+y-3=0\)

(2) \(x(y-3)=0\)

y = x-y * x+y / x-y

y = x + y

from this we can say x =0 but

y can be any value, integer ,fraction or a number(1) \(x+y-3=0\)

From here we get a definite value as y = 3

Sufficient

(2) \(x(y-3)=0\)[/quote]

Now, i didn't realize

this initially

0 * (y-3) = 0

y can be any value > 3

3.1, 6 , 9, these values will satisfy the statement

But will give multiple values for the question what is the value of y

A

NEVER assume things.

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If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.

Many of life's failures happen with people who do not realize how close they were to success when they gave up.