Absolute Value Inequality

[student22]

If xyz ≠ 0, is x(y + z) ≥ 0?

1. |y + z| = |y| + |z|
2. |x + y| = |x| + |y|

OA: C

Other than picking numbers, how do you solve this? More importantly, what is this question testing, I can't seem to figure out how to break this question down to a simpler form. Does anybody have an elegant solution? Thanks.

I mean, logically, I can sort of see why it's C, since all 3 variables have to be the same sign for this relationship to work. But hopefully someone can break this down for me.

[liferocks]

The question is whether x and (y+z) are of same sign
Now from 1

|Y+Z|=|Y|+|Z|..this is possible only when y and z are of same sign else the value of y+z will be less that |y|+|z|
but no infor mation about x hence not sufficient

From2
similarly we get x and y are of same sign..but no infor mation aboutz..not sufficient

combining
x,y and z are of same sing..since y and z are of same sign,y+z will also have same sign as x,y and z..hence x(y+z)>0..sufficient
Ans option C

[student22]

liferocks, thanks for the reply. That explanation makes sense. But there's no algebraic way to solve this?

[clock60]

the way i see the problem
(1)|y+z|=|y|+|z|
square both part and left with
yz=|yz|
here we have no info about x, so 1st insufficient, but one thing to notice
both y,z must be +ve, or -ve. for the product of yz to be +ve as |yz|>0
y>0,z>0 or
y<0,z<0

(2) the same is true for st 2 as we have no info about z, but again
y>0,x>0, or
y<0,x<0

together

y>0,z>0 so x also >0 and
x(y+z)>0 as product of two +ve numbers

y<0,z<0 and x<0
x(y+z)>0 as product of two -ve numbers ( i mean x and (y+z))
hope it makes sence

[student22]

Interesting idea squaring both sides...but how did you manage to simplify it down to yz = |yz|

|y+z|=|y|+|z|


The left hand side becomes, (y + z)^2 = y^2 + 2yz + z^2

But how did you square the right hand side?

Won't it be the same thing (|y| + |z|)^2 = y^2 + 2yz + z^2?

[clock60]

Interesting idea squaring both sides...but how did you manage to simplify it down to yz = |yz|

|y+z|=|y|+|z|


The left hand side becomes, (y + z)^2 = y^2 + 2yz + z^2

But how did you square the right hand side?

Won't it be the same thing (|y| + |z|)^2 = y^2 + 2yz + z^2?

it must be y^2+2*|yz|+z^2
from this place you can cancel y^2 and z^2 and left with |yz|
hope it is clear now

[student22]

Yes, got it, now it's clear, thanks.