Is | x + y | - | y + z |

Question

Is | x + y | - | y + z | < 0 ? 1. x < y 2. x < z Looking for an explanation. Please provide details for your answers.

Ravshonbek · Answer

at first look it is easy to say that both statement combined will be suff as the second part of the inequality/modulus is just addition of two greater numbers than x thus it makes the result negative. 

but if let's say x=-5 and y=-1 as well as z=-2 then you will notice that 
6-3 is not less than 0

also, if x=5, y=6 and z=7
then 11-13 is less than 0

so both statements combined or alone not suff.E

bkk145 · Answer

I got E.

(1) x<y
We don't know anything about z here, so can't conclude. 

(2) x<z
Try plugging in some numbers. 
Say x=1, z=2, y=4
5-6 < 0 Yes!
Say x=1, z=2, y=-4
3-2 < 0 No!

Together,
Plug in more numbers:
x=0, z=1, y=2
2-3<0 Yes!
x=-10 y=-1 z=-2
11-3<0 No!

FN · Answer

i would go for E too, it boils down to I dont know the relationship between y and z...thus we cant really know the sign of the (y-z)..if y>z, then +(y-z) however if y<z>0, y>0

and x<y>z then

x-y -y+z; which becomes x-2y+z which is obviously <0

if x<y; y<z> x+z which is obviously >0

I can go on..but suffice to say its E

beckee529 · Answer

thanks guys.. the answer is E

pau.sabria · Answer

I also got E.

|x+y|-|y+z| < 0 then
|x+y| < |y+z|
|x| < |z| (since it is simply adding a number)

1. not sufficient since i says nothing about z
2. not sufficient since x being inferior to z does not assure us that |x| < |z|, check x=-5 and z=-1 for instance. Insufficient.

Is | x + y | - | y + z | < 0 ? 1. x < y 2. x < z

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