SOLUTION: find four consecutive odd integers such that twice the sum of the third and fourth integer exceeds three times the first by eighty-seven

Question 773519: find four consecutive odd integers such that twice the sum of the third and fourth integer exceeds three times the first by eighty-seven
Answer by pakhi(24) (Show Source): You can put this solution on YOUR website!
Let the four consecutive odd numbers be (2n-3),(2n-1),(2n+1),(2n+3)
So according to the condition of the problem
2{(2n+1)+(2n+3)} - 3(2n-3) = 87
or 2(4n+4) - 6n + 9 = 87
or 8n + 8 - 6n + 9 = 87
or 2n + 17 = 87
or 2n = 87 - 17 = 70
or n = 70/2 = 35
Therefore the numbers are
2n-3 = 2*35 - 3 = 70 - 3 = 67
2n-1 = 2*35 - 1 = 70 - 1 = 69
2n+1 = 2*35 + 1 = 70 + 1 = 71
2n+3 = 2*35 + 3 = 70 + 3 = 73
check: 2(71+73)-3*67 = 2*144-201 =288-201 = 87
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