Two numbers a and b are graphed on a number line. it is always, sometimes, or never true that a- |b| < a+b and a > |b| explain

Consider the cases when A is not > |B|:

In these cases "(A- |B| < A+B) and (A > |B|)" is never true, because the second part of the proposition (sentence in "  ") is not true.

So consider the cases when A>|B|

A is clearly positive, since |B| is positive.

case 1: 

both A and B are positive, so |A|=A and |B|=B, and A>|B|, 

in this case 

A-|B|=A-B and clearly A-B<A+B as this is equivalent to -B<B, which is true.

case 2:

B is negative, so |B|=-B, 

thus A-|B|=A-(-B)=A+B

so A-|B|=A+B, thus A-|B| is not < A+B

case 3:

B=0,

A-|B|=A     and    A+B=A, 

so A-|B|=A+B, thus A-|B| is not < A+B.

sometimes: 

Precisely when A>B, and both A and B are positive.

69 i think because is the only one square