SOLUTION: Find three consecutive positive integers such that that product of the first and third, minus the second, is 1 more than 6 times the third. The smallest integer is?

Question 847878: Find three consecutive positive integers such that that product of the first and third, minus the second, is 1 more than 6 times the third.
The smallest integer is?
Found 2 solutions by swincher4391, hamsanash1981@gmail.com:
Answer by swincher4391(1107) (Show Source): You can put this solution on YOUR website!
Let the consecutive integers be known as: n, n+1, and n+2.
We know that n(n+2) - (n+1) = 1 + 6*(n+2)
n^2 + 2n - n - 1 = 1 + 6n + 12
n^2 + n -1 = 6n + 13
n^2 -5n -14 = 0
(n-7)(n+2) = 0
n = 7
Then the numbers are 7, 8 ,9... 7 being the smallest.
Answer by hamsanash1981@gmail.com(151) (Show Source): You can put this solution on YOUR website!
Let the consecutive positive integers be x, x+1, x+2
then, x*(x+2) - (x-1)-1=6(x+2)

x = -2 is eliminated as it is given that integers are positive
When x = 7 the numbers are 7,8,9 and smallest number 7.
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