An alternative approach: rephrase the question by thinking of absolute values as measures of distance.

✔ Absolute value of |a-b| is the distance between a and b on the number line. for instance |5-8| = |8-5| = distance between 5 and 8 on the number line.
✔ Absolute value of |a+b| = |a - (-b)| and is the distance between a and -b. For instance, |2+1| = |2-(-1)| = distance from 2 to -1 on the number line.

If x and y are integers and y = |x+3| + |4-x|, does y equal 7?

1) x < 4
2) x > -3

Rephrase the question: y is the sum of the distance between x & -3 and the distance between x & 4. Is the sum of these distances equal to 7? Think about where -3 and 4 are on the number line (they are 7 apart); the only way for the two distances to add up to 7 is for x to be between -3 and 4.

Rephrase: Is x between -3 and 4 inclusive?

Only together do the statements guarantee that x falls between these values. The answer is C.

Though i understand the above approaches,i don't get what was wrong with my approach.
I solved the above problem as below, but got [spoiler]ans E
[/spoiler]
for absolute value, we can take
case 1-> x is greater than or equal to 0 -> for x>0 ,y = (x+3)+(4-x)= 7
for x=0 ,y = (0+3)+(4-0)=7

case 2-> x is less than 0 -> y= -(x+3)+ [-(4-x)]= -7

so, i rephrased the qn as is y=7=? => is x greater than or equal to 0 ?

st1: Not sufficient
st2: Not sufficient

st1+ st2 : -3<x<4 => x can be -2,-1,0,1,2, and 3 => x can be +ve , 0 or -ve => not sufficient=> ans E

After i looked at the solution and solved again, i got that for each of the above values,i get y =7
but, i followed the rephrased question blindly, hence got the wrong answer.
What is wrong with my approach?

Please help.

ase 1-> x is greater than or equal to 0 -> for x>0 ,y = (x+3)+(4-x)= 7
for x=0 ,y = (0+3)+(4-0)=7

case 2-> x is less than 0 -> y= -(x+3)+ [-(4-x)]= -7

It's incorrect to claim that if x>0, then |4-x| is the same as (4-x). For instance, if x were 5, those two expressions would not have the same value.

Likewise, it's incorrect to claim that if x<0, then |4-x| is the same as -(4-x). For instance, if x were -1, then |4-x| would be 5, but -(4-x) would be -5.

The "switch" is not whether x itself is greater than or less than 0, but rather whether the entire contents of the absolute value brackets are greater than or less than 0

What we can say correctly is that if 4-x>0, then |4-x| = +(4-x), while if 4-x<0, then |4-x| = -(4-x)

buoyant wrote:If x and y are integers and y = |x+3| + |4-x|, does y equal 7?

1) x < 4
2) x > -3

[spoiler]OA: C[/spoiler]

An alternate approach is to determine the CRITICAL POINTS: the values where the EXPRESSIONS inside the absolute values are equal to 0.
Here, the critical points are x=-3 (where x+3 = 0) and x=4 (where 4-x=0).

First, test the critical points themselves.
If x=-3, then y = |x+3| + |4-x| = |-3+3| + |4-(-3)| = 0+7 = 7.
If x=4, then y = |x+3| + |4-x| = |4+3| + |4-4| = 7+0 = 7.
Thus, if x=-3 or x=4, the answer to the question stem is YES.

Next, test values to EACH SIDE of the critical points.
x<-3:
If x=-4, then y = |x+3| + |4-x| = |-4+3| + |4-(-4)| = 1+8 = 9.
In this range, y>7, so the answer to the question stem is NO.

-3<x<4:
If x=0, then y = |x+3| + |4-x| = |0+3| + |4-0| = 3+4 = 7.
In this range, y=7, so the answer to the question stem is YES.

x>4:
If x=5, then y = |x+3| + |4-x| = |5+3| + |4-5| = 8+1 = 9.
In this range, y>7, so the answer to the question stem is NO.

Question stem, rephrased:
Is -3≤x≤4?

Only when the statements are combined is it guaranteed that -3≤x≤4.

The correct answer is C.

Maybe an easier way:

If |a| > 0, |a| = a

If |a| < 0, |a| = -a

Now suppose that (x+3) and (4-x) are both positive:

If (x+3) is positive, then |x+3| = x + 3
If (4-x) is positive, then |4-x| = 4 - x

So if (x+3) and (4-x) are BOTH positive, |x+3| + |4-x| = (x+3)+(4-x) = 7

Hence the question is basically asking us "Are both (x + 3) and (4 - x) positive?" or, more simply, "Is x greater than or equal to -3 and less than or equal to 4?" Given both statements, we can answer affirmatively!