Find three consecutive positive integers such that the product of the first and third, minus the second, is 1 more than 7 times the third.

Define x :

Let the smallest number be x.
First number = x
Second Number = x + 1
Third Number = x + 2


Construct equation :

x(x+2) - (x+1) = 7(x+2) + 1


Solve x :

x(x + 2) - (x + 1) = 7(x + 2) + 1
x² + 2x - x - 1 = 7x + 14 + 1
x²- 6x - 16 = 0
(x+2)(x-8) = 0
x = -2 or x = 8

Since x is a positive integer, it cannot be negative.
⇒x = 8


The three numbers are 8, 9 and 10.

Logical equivalent of the give question is its contrapositive