SOLUTION: Find the LARGEST of three consecutive integers such that 3 times the sum of the first and the third integer is equal to 30 more than 4 times the second integer.
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Question 853059: Find the LARGEST of three consecutive integers such that 3 times the sum of the first and the third integer is equal to 30 more than 4 times the second integer. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Find the LARGEST of three consecutive integers such that 3 times the sum of the first and the third integer is equal to 30 more than 4 times the second integer.
:
Let n = the largest of the consecutive numbers
then
(n-1) = the middle number
and
(n-2) = the smallest
:
3[n+(n-2)] = 4(n-1) + 30
3(2n-2) = 4n - 4 + 30
6n - 6 = 4n + 26
6n -4n + 26 + 6
2n = 32
n = 32/2
n = 16 is the largest number
:
:
See if that checks out in the statement: (the three numbers 14, 15, 16)
" 3 times the sum of the first and the third integer is equal to 30 more than 4 times the second integer."
3(14+16) = 4(15) + 30
3(30) = 60 + 30
