5= |b+2| + |3-b|

If b < -2,
The value of b+2 is less than 0. So, |b+2|=-(b+2) because |X| = -X if x is a negative number.
The value of 3-b is greater than 0. So, |3-b| = 3-b because |X| = X if X is a non negative number.
|b+2| + |3-b| = -(b+2) + 3-b = -2b + 1(which is not always equal to 5. It is equal to 5 when b = -2)

If b > 3,
The value of b+2 is greater than 0. So, |b+2|=(b+2) because |X| = X if X is a non negative number.
The value of 3-b is less than 0. So, |3-b|=-(3-b) because |X| = -X if x is a negative number.
|b+2| + |3-b| = (b+2) + -(3-b) = 2b - 1(which is not always equal to 5. It is equal to 5 when b = 3)

If -2<b<3
The value of b+2 is greater than 0. So, |b+2|=(b+2) because |X| = X if X is a non negative number.
The value of 3-b is greater than 0. So, |3-b| = 3-b because |X| = X if X is a non negative number.
|b+2| + |3-b| = (b+2) + 3-b = 5

Hence -2<b<3 is an always-true condition. I would have provided you with the EXACT answer, if you have posted the complete question but that is OK.

Let me know if you need any further help!

thank you for the explanation below is the exact question and in the question answers in the book they started with , a=5 when -2<= b<=3 before evaluate the statements, i want to know how they came up with these 2 numbers.

If a and b are integers and a= |b+2| + |3-b| , does a= 5 ?

1) b<3
2) b>-2

b = -2 and b = 3 are the only points where the sign of the terms b+2 and 3-b reverses.

If b < -2 then b + 2 < 0
If b > -2 then b + 2 > 0

Similarly

If b > 3 then 3 - b < 0
If b < 3 then 3 - b > 0

kassim wrote:thank you for the explanation below is the exact question and in the question answers in the book they started with , a=5 when -2<= b<=3 before evaluate the statements, i want to know how they came up with these 2 numbers.

If a and b are integers and a= |b+2| + |3-b| , does a= 5 ?

1) b<3
2) b>-2

Is |b+2| + |3-b| = 5?

|x-y| = the DISTANCE between x and y.
Thus:
|b+2| = |b-(-2)| = the distance between b and -2.
|3-b| = the distance between 3 and b.

Question rephrased:
Is the sum of the distance between b and -2 and the distance between b and 3 equal to 5?

Draw a number line:
...........-2.....................................3................

The distance between -2 and 3 is 5.
If b>3, then the distance between b and -2 will be greater than 5.
If b<-2, then the distance between b and 3 will be greater than 5.
Thus, if b>3 or b<-2, then the answer to the question stem is NO: the sum of the two distances will NOT be equal to 5.
For the sum of the two distances to be equal to 5, b must be BETWEEN -2 and 3, inclusive:

............-2<---|b+2|--->b<---|3-b|--->3................

As the number line shows, when b is between -2 and 3, inclusive, the sum of the two distances is equal to 5.

Question rephrased: Is b between -2 and 3, inclusive?

Statement 1: b<3
Statement 2: b>-2
When the two statements are combined, we know that b is between -2 and 3.
SUFFICIENT.

The correct answer is C.

Thank you That was very helpful