SOLUTION: Find three consecutive integers such that the sum of the largest and four times the second is 42 more than the first.

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Question 1090549: Find three consecutive integers such that the sum of the largest and four times the second is 42 more than the first.
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
a,a+1,a+2
a+2+(4*(a+1))=a+42
a+2+4a+4=a+42
5a+6=a+42
4a=36
a=9
a+1=10
a+2=11
check
11+4*10=42+9
51=51
ok