SOLUTION: how do you solve abs(y-4) + 1 = 3? When I solved it, I got y=6 and y=0. My textbook says it should be y=6 and y=2. What did I do wrong?

Algebra ->  Absolute-value -> SOLUTION: how do you solve abs(y-4) + 1 = 3? When I solved it, I got y=6 and y=0. My textbook says it should be y=6 and y=2. What did I do wrong?      Log On


   



Question 605983: how do you solve abs(y-4) + 1 = 3? When I solved it, I got y=6 and y=0. My textbook says it should be y=6 and y=2. What did I do wrong?
Found 2 solutions by lwsshak3, scott8148:
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
how do you solve abs(y-4) + 1 = 3?
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Since you do not know whether (y-4) is positive or negative, you must find a solution for both possibilities.
..
y-4+1=3
y=6
..
-(y-4)+1=3
-y+4+1=3
-y=-2
y=2
solutions: y=6 and 2

Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
abs(y-4) = 2 ___ y - 4 = ±2

y - 4 = 2 ___ y = 6

y - 4 = -2 ___ y = 2