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(1) Given \(y\) is non negative value and \(|x - 3|\geq{y}\), so \(|x - 3|\) is more than 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) \(|x-3|\leq{-y}\), \(y\geq{0}\) --> \(0\leq{-y}\), equation says that \(|x-3|\) less or equals to zero, but \(|x-3|\) never negative (\(|x-3|\geq{0}\)), so only solution is if \(|x-3|=0=y\) --> \(x-3=0\) --> \(x=3\). 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 \(0\).

stmt1: |x-3| >= y in case y>= 0 if we consider y = 0, |x-3| >= 0 but y can be 0, 1,2,3 anything so insufficient.

stmt2: |x-3| <= -y since y >= 0 => -y <= 0 so we can say |x-3| <= -y <= 0 but |x-3| has to be positive which is only possible if y = 0 and |x-3| = 0 only one condition suffice this if x=3, hence sufficient. so B is the answer _________________

I know that the value of an abs expression cannot be less than 0. so does that imply that x is 0 units from 3?

As \(y\) is some non-negative value (0, 7, 1.4, ...) then \(-y\) is some non-positive value (0, -9, -5.6, ...). Statement 2 says that \(|x-3|\leq{-y}\) (\(|x-3|\leq{non-positive}\)) - absolute value is less than or equal to some non-positive value, BUT absolute value can not be negative, least value of it is zero, thus \(|x-3|={0}=y\) --> \(x=3\).

I know that the value of an abs expression cannot be less than 0. so does that imply that x is 0 units from 3?

As \(y\) is some non-negative value (0, 7, 1.4, ...) then \(-y\) is some non-positive value (0, -9, -5.6, ...). Statement 2 says that \(|x-3|\leq{-y}\) (\(|x-3|\leq{non-positive}\)) - absolute value is less than or equal to some non-positive value, BUT absolute value can not be negative, least value of it is zero, thus \(|x-3|={0}=y\) --> \(x=3\).

Refer for the full solution to my previous post.

Hope it's clear.

I trust your explanations - very clear. Thanks _________________

KUDOS me if you feel my contribution has helped you.

The inequality represents the regino between the x-axis and the blue line. Knowing y>=0 is not enough to know x

(2)-y>=|x-3| OR y<=-|x-3|

The inequality represents the region below the blue line. We know y>=0, this can only represent the point where, the blue line meets the x-axis or x=3. Sufficient

(2) \(|x-3|\leq{-y}\), \(y\geq{0}\) --> [highlight]\(0\leq{-y}\)[/highlight], equation says that \(|x-3|\) less or equals to zero, but \(|x-3|\) never negative (\(|x-3|\geq{0}\)), so only solution is if \(|x-3|=0=y\) --> \(x-3=0\) --> \(x=3\). SUFFICIENT It should be [highlight]\(0\geq{-y}\)[/highlight] in my opinion

No, it should be as it is.

(2) \(|x-3|\leq{-y}\) --> now, as LHS is absolute value, which is always non-negative, then we have that \(-y\) is more than or equal to some non-negative value: \(0\leq{-y}\) --> \(y\leq{0}\). As also is given that \(y\geq{0}\) then \(y=0\) --> \(|x-3|\leq{0}\) --> absolute value can not be negative, so \(|x-3|=0\) --> \(x=3\). _________________

(2) \(|x-3|\leq{-y}\), \(y\geq{0}\)-->[highlight]\(0\leq{-y}\)[/highlight], equation says that \(|x-3|\) less or equals to zero, but \(|x-3|\) never negative (\(|x-3|\geq{0}\)), so only solution is if \(|x-3|=0=y\) --> \(x-3=0\) --> \(x=3\). SUFFICIENT It should be [highlight]\(0\geq{-y}\)[/highlight] in my opinion

No, it should be as it is.

(2) \(|x-3|\leq{-y}\) --> now, as LHS is absolute value, which is always non-negative, then we have that \(-y\) is more than or equal to some non-negative value: \(0\leq{-y}\) --> \(y\leq{0}\). As also is given that \(y\geq{0}\) then \(y=0\) --> \(|x-3|\leq{0}\) --> absolute value can not be negative, so \(|x-3|=0\) --> \(x=3\).

Does this symbol ---> means "implies that"?. Because I interpreted it that way in the statement. Rest of the explanation is clear to me.

Ahhhhh i hate absolute value questions. Gets me every time! _________________

******************************************************************* ~ PickyTooth - Eat Like a Local Foodie // www.pickytooth.com ~ *******************************************************************

Re: If y>=0, What is the value of x? (1) |x-3|>=y (2) |x-3|<=-y [#permalink]

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12 Jan 2014, 07:38

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Re: If y>=0, What is the value of x? (1) |x-3|>=y (2) |x-3|<=-y [#permalink]

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29 Jun 2015, 05:13

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