Question

There are 3 consecutive positive integers. the product of the first and third integer is 6 more than 6 times the second integer. what is the smallest integer? answer

Answer 1

let the three integers be x - 1 x and x + 1
(x - 1)(x + 1) = 6x + 6
x^2 - 1 = 6x + 6
x^2 - 6x - 7 = 0
(x - 7)(x + 1) = 0
x - 7 = 0
x = 7

x + 1 = 0
x = - 1

the solutions are (7,0) and (-1,0)