SOLUTION: If 2|x+3|=4 and |y+1|/3=2, then |x+y| could equal each of the following EXCEPT 0,4,8,10,12. The answer given was 10. I am not sure i understand this. Can you please provide an e

Algebra ->  Absolute-value -> SOLUTION: If 2|x+3|=4 and |y+1|/3=2, then |x+y| could equal each of the following EXCEPT 0,4,8,10,12. The answer given was 10. I am not sure i understand this. Can you please provide an e      Log On


   

Question 937050: If 2|x+3|=4 and |y+1|/3=2, then |x+y| could equal each of the following EXCEPT
0,4,8,10,12. The answer given was 10. I am not sure i understand this. Can you please provide an explanation.
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Best is just to solve each equation, and then check the possibilities for |x+y|.

abs%28x%2B3%29=2
either -x-3=2
x%2B3=-2
x=-5
or x%2B3=2
x=-1

abs%28y%2B1%29=6
either -y-1=6
y%2B1=-6
y=-7
or y%2B1=6
y=5

Up to now you know x is -5 or -1
and
y is -7 or 5

There are FOUR possibilities for abs%28x%2By%29.
Which of them is in your list?

Try to follow my plan shown above.
(solution process result adjusted)