Question 161590
    Find three consecutive odd integers such that the product of the first and second integer is 7 less then 10 times the third integer.
.
Let x = first consecutive odd integer
then
x+2 = second consecutive odd integer
x+4 = third consecutive odd integer
.
From: "product of the first and second integer is 7 less then 10 times the third integer"
x(x+2) = 10(x+4) - 7
x^2 + 2x = 10x + 40 - 7
x^2 + 2x = 10x + 33
x^2 - 8x - 33 = 0
.
Factoring gives us:
(x-11)(x+3) = 0
therefore,
x = {-3, 11}
.
Two possible solutions:
11, 13, 15
-3, -1, 1