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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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.