If zy < xy < 0, is |x-z| + |x| = |z|?
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??
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| + |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.
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
Your attitude determines your altitude
Smiling wins more friends than frowning