# The product of two consecutive odd integers is 29 less than 8 times their sum. Find the two integers. Answer in the form of paired points with the lowest of the two integers first?

May 11, 2018

$\left(13 , 15\right) \mathmr{and} \left(1 , 3\right)$

#### Explanation:

Let $x$ and $x + 2$ be the odd consecutive numbers, then

As per the question, we have

$\left(x\right) \left(x + 2\right) = 8 \left(x + x + 2\right) - 29$

$\therefore {x}^{2} + 2 x = 8 \left(2 x + 2\right) - 29$

$\therefore {x}^{2} + 2 x = 16 x + 16 - 29$

$\therefore {x}^{2} + 2 x - 16 x - 16 + 29 = 0$

$\therefore {x}^{2} - 14 x + 13 = 0$

$\therefore {x}^{2} - x - 13 x + 13 = 0$

$\therefore x \left(x - 1\right) - 13 \left(x - 1\right) = 0$

$\therefore \left(x - 13\right) \left(x - 1\right) = 0$

$\therefore x = 13 \mathmr{and} 1$

Now,

CASE I : $x = 13$

$\therefore x + 2 = 13 + 2 = 15$

$\therefore$ The numbers are (13, 15).

CASE II : $x = 1$

$\therefore x + 2 = 1 + 2 = 3$

$\therefore$ The numbers are (1, 3).

Hence, as there are two cases being formed here; the pair of numbers can be both (13, 15) or (1, 3).