if y>=0, what is the value of x?
1) lx-3l >= y
2) lx-3l <= -y
absolute
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From 1 we have 2 possibilities
|x| = x or -x so two values
From 2 also the same thing
Combining we have -x + 3 >=y
i.e. x + y >=3 ... so i think E
|x| = x or -x so two values
From 2 also the same thing
Combining we have -x + 3 >=y
i.e. x + y >=3 ... so i think E
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Answer is B.
|x-3| <y>= 0
if y = 0, then x MUST be 3.
if y is say 4 then |x-3| <= -4 there is no value for x. So 2) can't be true. But it is given that |x-3| <= -y.. hence y must be 0 . so x must be 3.
Calista.
|x-3| <y>= 0
if y = 0, then x MUST be 3.
if y is say 4 then |x-3| <= -4 there is no value for x. So 2) can't be true. But it is given that |x-3| <= -y.. hence y must be 0 . so x must be 3.
Calista.
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Correcting some typos in my earlier post.
Answer is B.
From 2) |x-3| <y>= 0
if y = 0, then x MUST be 3.
if y is say 4 then |x-3| <= -4 there is no value for x. So 2) can't be true. But it is given that |x-3| <= -y.. hence y must be 0 . so x must be 3.
Calista.
Answer is B.
From 2) |x-3| <y>= 0
if y = 0, then x MUST be 3.
if y is say 4 then |x-3| <= -4 there is no value for x. So 2) can't be true. But it is given that |x-3| <= -y.. hence y must be 0 . so x must be 3.
Calista.
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Hi GMATDUD, I would solve for stmt 2 like this
if y>=0, what is the value of x?
B) lx-3l <y>=0 so -y will be 0 or -ve
now |x-3| is an absolute value so it will be always +ve
so in the given inequality the only real world case that will arise will be
when y=0 as for other values of y -y will be -ve which cannot be more than a +ve value (|x-3|)
so we have |x-3| = 0
or x =3
B
if y>=0, what is the value of x?
B) lx-3l <y>=0 so -y will be 0 or -ve
now |x-3| is an absolute value so it will be always +ve
so in the given inequality the only real world case that will arise will be
when y=0 as for other values of y -y will be -ve which cannot be more than a +ve value (|x-3|)
so we have |x-3| = 0
or x =3
B
Regards
Samir
Samir
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- jayhawk2001
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magical cook wrote:if y>=0, what is the value of x?
1) lx-3l >= y
2) lx-3l <y>= 0.
Since y >=0, only value of y that satisfies this is y=0.
So |x-3| <= 0. So, we can find x.
Hence B
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