If \(y\geq{0}\), what is the value of x?(1) \(|x - 3|\geq{y}\). As given that \(y\) is non negative value then \(|x - 3|\) is more than (or equal to) some non negative value, (we could say the same ourselves as absolute value in our case (\(|x - 3|\)) is never negative). So we can not determine single numerical value of \(x\). Not sufficient.
Or another way: to check \(|x - 3|\geq{y}\geq{0}\) is sufficient or not just plug numbers:
A. \(x=5\), \(y=1>0\), and B. \(x=8\), \(y=2>0\): you'll see that both fits in \(|x - 3|>=y\), \(y\geq{0}\).
Or another way:\(|x - 3|\geq{y}\) means that:
\(x - 3\geq{y}\geq{0}\) when \(x-3>0\) --> \(x>3\)
OR (not and)
\(-x+3\geq{y}\geq{0}\) when \(x-3<0\) --> \(x<3\)
Generally speaking \(|x - 3|\geq{y}\geq{0}\) means that \(|x - 3|\), an absolute value, is not negative. So, there's no way you'll get a unique value for \(x\). INSUFFICIENT.
(2) \(|x-3|\leq{-y}\). Now, as \(|x-3|\) is never negative (\(0\leq{|x-3|}\)) then \(0\leq{-y}\) --> \(y\leq{0}\) BUT stem says that \(y\geq{0}\) thus \(y=0\). \(|x-3|\leq{0}\) --> \(|x-3|=0=y\) (as absolute value, in our case |x-3|, can not be less than zero) --> \(x-3=0\) --> \(x=3\). SUFFICIENT
In other words:\(-y\) is zero or less, and the absolute value (\(|x-3|\)) must be at zero or below this value. But absolute value (in this case \(|x-3|\)) can not be less than zero, so it must be \(0\).
Answer: B.
Hope it helps..
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