SOLUTION: find three consecutive integers such that the sum of the first and the third increased by 8 is 40 more than the second.

SOLUTION: find three consecutive integers such that the sum of the first and the third increased by 8 is 40 more than the second.

Algebra.Com

Question 788756: find three consecutive integers such that the sum of the first and the third increased by 8 is 40 more than the second.
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
let the integers be (n-1) , n, (n+1)
(n-1)+(n+1) +8 =n+40
2n+8=n+40
n=40-8
n=32
The numbers are
31 32 , 33

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