How many different values of y satisfy |y + 3| = |8 + y| + |4 - y| ?

Ugh, what an ugly question....
Anyway, back to it. Let's first find intervals for each modulus.
|y + 3|
If positive - |y + 3|, key point is y+3>=0 or y>=-3, so we get |y + 3|=y+3
If negative - |y + 3|, key point is y+3<0 or y<-3, so we get |y + 3|=-(y+3)=-y-3
|8 + y|
If positive - |8 + y|, key point is 8+y>=0 or y>=-8, so we get |8 + y|=8+y
If negative - |8 + y|, key point is 8+y<0 or y<-8, so we get |8 + y|=-(8+y)=-y-8
|4 - y|
If positive - |4 - y|, key point is 4-y>=0 or y<=4, so we get |4 - y|=4-y
If negative - |4 - y|, key point is 4-y<0 or y>4, so we get |4 - y|=-(4-y)=4+y
So, we get
\(1.\) \(y<-8\)
\(2.\) \(-8\leq{y}<-3\)
\(3.\) \(-3\leq{y}\leq4\)
\(4.\) \(y>4\)

As we see no number falls between the above intervals, hence A

How many different values of y satisfy |y+3|=|8+y|+|4−y| ?

A. 0
B. 1
C. 2
D. 3
E. 4

Answer: A

|y+3| = |8+y|+|4-y|
=> |y+8| -|y+3| + |y-4| = 0

range of y: |y+8| |y+3| |y-4|
y >= 4 : + + + => y+8 -y-3+y-4 = 0 => y = 1, but y can't be 1 because y has to be >=4 --> no solution
-3 <=y<4 : + + - => y+8 -y -3 -y +4 =0 => y = 9 , but y can't be 9 because y has to be < 4 --> no solution
-8 <=y<-3: + - - => y+8 +y +3 -y +4 =0 => y = -15 , but y can't be -15 because y has to be >= -8 --> no solution
y < -8 : - - - => y+8 -y-3+y-4 = 0 => y = 1, but y can't be 1 because y has to be < -8 --> no solution

How many different values of yy satisfy |y+3|=|8+y|+|4−y|

A. 0
B. 1
C. 2
D. 3
E. 4

if y is positive then |y+3| will always be less than |8+y| i.e. |y+3| can never be equal to |8+y|+|4−y|

If y is Negative then |y+3| will always be less than |4−y| i.e. |y+3| can never be equal to |8+y|+|4−y|

i.e. NO SOLUTION

Answer: Option A

How many different values of y satisfy |y+3|=|8+y|+|4−y|?
A. 0
B. 1
C. 2
D. 3
E. 4

Let us divide the number line into 4 different regions: -
Region 1 where y>4
Region 2 where -3<y<=4
Region 3 where -8<y<=-3
Region 4 where y<=-8

Region 1 where y>4
expression becomes
y+3 = y+8 + y-4
=> y = -1
Not possible in the region since y>4

Region 2 where -3<y<=4
y+3= y+8+4-y
=> y=9
Not possible in the region since -3<y<=4

Region 3 where -8<y<=-3
-y-3 = y+8 +4-y
=> y = -15
Not possible in the region since -8<y<=-3

Region 4 where y<=-8
-y-3 = -y-8 +4-y
=> y = -1
Not possible in the region since y<=-8

Therefore, it may be concluded that no value of y satisfies the above equation.
=> Zero solutions (0)

IMO A

|y+3|=|8+y|+|4−y|
Case 1---> y=< -8
-y-3=-8-y+4-y
y= -1 (Not possible)

Case 2 ---> -8<y=<-3
-y-3=8+y+4-y
y=9 (not possible)

Case 3-----> -3<y=<4
y+3=8+y+4-y
y=9(not possible)

Case 4-----> y>4
y+3=8+y+y-4
y=-1 (not possible)

y can take 0 value

IMO A

Okey
What i have learned is that when you see a question just as above where there is one variable (y in above case) and more than one modulas then follow the critical point approach.
Critical Point Approach:
Step 1: equate each term of modulas with 0.
In above case we will have three terms bcoz we have 3 modulas.
a) y+3=0-----> i.e. y=-3
b) 8+y=0-----> i.e. y=-8
c) 4-y=0-----> i.e. y=4

Step 2)
Put the equated value in number line and it becomes
_______-8______-3_____4______

Step 3)
As u see we have -8,-3,4 as determiners,
now i) try putting value beyond -8 (say 9) on our y+3----> -8+3=-5 (remember u can put any negative value beyond -8 and when you add that value to y+3 you get negative value.
Try putting the same picked value in 8+y----> 8+(-9)----> -1 and you will again get negative value. Now putting same value (-9) in 4-y----> u get 4-(-9)=13 which is positive in this case.

Now what you have to do is Put (-1) just before the terms on which u got negative while doing above step and (1) on terms where you got positive value.

it becomes
(-1)(y+3)= (-1)(8+y) +(1)(4-y)
solving this you will get y= -1
In the number line, you can see that we took -8 as limit and took -9 for the equation and while solving we got value as -1. The thing to consider here is that we had a limit of -infinitive to -8 but we got ans as -1 and -1 does not lies in that -infinitive to -8. So rule out this ans. Lets say if we had got value beyond -8 then we would have got the left side limiting value of our equation.

ii) in our i) we did put value beyond -8. Now in our second case we will put value which is between -8 and -3 bcoz we have case of beyond -8, between -8 and -3, between -3 and 4, and beyond 4.

Lets take (-5) and we will repeat same steps and we will get y+3---> -5+3=-2 (-ve)
8+y---> 8+(-5)=3 (+ve)
4-y---> 4-(-5)=9 (+ve)
then our equation becomes as
(-1)(y+3)= (1)(8+y)+(1)(4-y)
y=15 which do not fall in between -8 and -3.

when you repeat the same steps for -3 to 4 and beyond 4. You will come to see that neither of them will have values which lie withing the limiting number line range.

So, when none of the equation has satisfied the limiting range, you have eliminated all.
This way ans comes to 0.

(Note): If any two y= something have lied in the limiting number line lets say our beyond -8 term have given y=-12 and lets say our -3 to 4 would have given y= 2,then you would have got the range of the equation and that range would have been
-12<=y<=2. This means we would have (-12,-11,-10,-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2)= 15 values of Y satisfying the modulas equation of our question.
Hope this is perfectly understandable.
And any queries are welcome even on PM.

Posted from my mobile device

here the solution of the question

The three critical values are -8, -3 and 4

We can have 4 conditions:

1. y < -8:
-(y + 3) = - (8 + y) + (4 – y)
=> y = -1
This does not fall in the range of y < -3. Therefore, we can reject the solution.

2. -8 <= y < -3:
-(y + 3) = (8 + y) + (4 – y)
=> y = -15
This does not fall in the range of -8 <= y < -3. Therefore, we can reject the solution.

3. -3 <= y <4
(y + 3) = (8 + y) + (4 – y)
=> y = 9
This does not fall in the range of -3 <= y <4. Therefore, we can reject the solution.

4. y >= 4
(y + 3) = (8 + y) – (4 – y)
=> y = -1
This does not fall in the range of y >= 4. Therefore, we can reject the solution.

Therefore, no value of y satisfy. Answer A.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.