Find three consecutive positive integers such that the product of the first and the third is 11 more than the second

3, 4, 5

Step-by-step explanation:

If the integers are x, x+1, and x+2, then:

x (x+2) = x+1 + 11

x² + 2x = x + 12

x² + x - 12 = 0

(x + 4) (x - 3) = 0

x = - 4 or 3

Since x is positive, x = 3. So the three integers are 3, 4, and 5.