absolute

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

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.

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.

can someone confirm oa?

Sorry for the late reply - OA is B.

I still did not get the explanation .

B) lx-3l <= -y

|x|<= -Y+3

All this says is x is negative and 3 more than y. Absolute val is my weak area. Can someone help by explaining in more detail ?

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

Could someone explain ..i still dont get it

makes sense thanks ..