# How do you find two consecutive odd integers whose product is 1 less than 6 times their sum?

Nov 26, 2016

There are two sets of consecutive odd integers which meet the requirements of this problem:

11 and 13

or

-1 and 1

#### Explanation:

First, let's define our numbers in mathematical terms.

Let $x$ be the first odd integer. Then, because we are looking for consecutive odd integers, we can let the second integer be $x + 2$.

So this problem can now be rewritten as:

$x \cdot \left(x + 2\right) = \left(6 \cdot \left(x + \left(x + 2\right)\right)\right) - 1$

${x}^{2} + 2 x = \left(6 \cdot \left(2 x + 2\right)\right) - 1$

${x}^{2} + 2 x = 12 x + 12 - 1$

${x}^{2} + 2 x = 12 x + 11$

${x}^{2} + 2 x - 12 x - 11 = 12 x + 11 - 12 x - 11$

${x}^{2} - 10 x - 11 = 0$

Factoring this gives:

$\left(x - 11\right) \left(x + 1\right) = 0$

Solving first for $\left(x - 11\right)$ gives:

$\frac{\left(x - 11\right) \left(x + 1\right)}{x + 1} = \frac{0}{x + 1}$

$x - 11 = 0$

$x - 11 + 11 = 0 + 11$

$x = 11$

Solving for (x + 1) gives:

$\frac{\left(x - 11\right) \left(x + 1\right)}{x - 11} = \frac{0}{x - 11}$

$x + 1 = 0$

$x + 1 - 1 = 0 - 1$

$x = - 1$