Find two consecutive odd integers such that their product is 71 more than 3 times their sum.

Let's put this in terms of a system of equations. Let's call the numbers x and y, with x being the larger integer and y the smaller one. Since they are consecutive odd integers, we know that:
x = y + 2
The constraint is:
xy= 3(x+y) + 71         Simplify
xy = 3x + 3y + 71       Plug in (y+2) for x:
(y + 2) y = 3(y + 2) + 3y + 71         Simplify
y² + 2y = 3y + 6 + 3y + 71              Subtract 6y from  both sides, add 71+6
y² -4y = 77
y² - 4y - 77 = 0
Factor:
(y - 11) (y+7) = 0
y = 11 or y=-7

When y = 11, x = 13, since x = y+2
Plug in to make sure:
xy = 3(x+y) + 71
11*13 = 3 (11 + 13) + 71
143 = 3 (24) + 71
143 = 72 + 71
143 = 143               
Our values work! 
11, 13

250 is the answeralright