tarek99 wrote:

If zy < xy < 0, is |x-z| + |x| = |z|?

(1) z<x

(2) y>0

I usually have problems interpreting the absolute value expressions such as |x-z| + |x| = |z|. I do know at least that this expression can be rearranged into |x-z| = |z| - |x|

What does this expression mean??

Thanks

zy < xy < 0

--> x and y must have oppostive sigins

z and y must have oppostive sigins

--> x and z have same signs.

1) if z<x State 1 is not sufficient alone

Assume x and z are +ve numbers

--> x-z>0

|x-z| + |x| = |z|

--> x-z+x=z --> 2x-z =z

-- alwasy false if x, z are +ve numbers

Assume x and z are -ve numbers

-- > z<x --> |z|>|x| ( E.G -2 >-1 |-2| >|-1|)

( x=-1 z= -2)

|x-z| + |x| = |z| -->

--> |x-z| =|z| -|x| when other z and x are -ve and z<x

--> always true if x ,z are negative sign

2) if y>0 x and z must be -ve numbers.

then |x-z| + |x| = |z|.

assume that z<x --> |z|>|x|

|x-z| + |x| = |z| -- always true

assume that z>x --> |z|<|x|

|x-z| + |x| = |z| -- alwasy false

Statement 2 alone not sufficient.

Combine two..

z<x and y>0 --> both are negative numbers |z|>|x|

|x-z| + |x| = |z| -- alwasy true.

Both Statements combined can answer the question.

My answer is C

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