Min Max, Critical Value with range GMAT, GRE

How many different values of y$\mathrm{ <span\; face="Tahoma,\; Geneva,\; sans-serif">satisfy</span><span\; face="Tahoma,\; Geneva,\; sans-serif"> </span>}$|y+3|=|8+y|+|4−y|$<span\; face="Tahoma,\; Geneva,\; sans-serif"\; style="font-style:\; inherit;">?</span>$
A. 0
B. 1
C. 2
D. 3
E. 4





The Critical Points Are: 
y + 3 = 0; 8 + y = 0 & 4 - y = 0

--> y = -8, - 3, 4

3 critical points --> 4 cases as shown


Case 1: y < -8
--> l8 + yl = -(8 + y); ly + 3l = -(y + 3) & l4 - yl = (4 - y)
--> |y + 3| = |8 + y| + |4 - y|
--> - (y + 3) = - (8 + y) + (4 - y)
--> - y - 3 = - 8 - y + 4 - y
--> y = -4 + 3 = - 1
--> y = -1; Not Possible

Case 2: -8 ≤ y < -3
--> l8 + yl = (8 + y); ly + 3l = -(y + 3) & l4 - yl = (4 - y)
--> |y + 3| = |8 + y| + |4 - y|
--> - (y + 3) = (8 + y) + (4 - y)
--> - y - 3 = 8 + y + 4 - y
--> - y = 12 + 3 = 15
--> y = -15; Not Possible

Case 3: -3 ≤ y ≤ 4
--> l8 + yl = (8 + y); ly + 3l = (y + 3) & l4 - yl = (4 - y)
--> |y + 3| = |8 + y| + |4 - y|
--> (y + 3) = (8 + y) + (4 - y)
--> y + 3 = 8 + y + 4 - y
--> y = 12 - 3 = 9
--> y = 9; Not Possible

Case 4: y > 4
--> l8 + yl = (8 + y); ly + 3l = (y + 3) & l4 - yl = -(4 - y)
--> |y + 3| = |8 + y| + |4 - y|
--> (y + 3) = (8 + y) - (4 - y)
--> y + 3 = 8 + y - 4 + y
--> - y = 4 - 3 = 1
--> y = -1; Not Possible


A