# How do you find three consecutive odd integers such that he sum of the first and twice the second is 6 more than the third?

Jan 10, 2017

See process for finding the solution to this problem below:

#### Explanation:

First, let's define the three consecutive odd numbers.

The first, we can call $x$.

Because they are odd numbers we know they are every other number from $x$ so we need to add $2$

The second and third number will be $x + 2$ and $x + 4$

"twice the second" is the same as: $2 \left(x + 2\right)$

"the sum of the first and twice the second is then: $x + 2 \left(x + 2\right)$

This sum is "6 more than the third" or $\left(x + 4\right) + 6$

Putting this into an equation and solving for $x$:

$x + 2 \left(x + 2\right) = \left(x + 4\right) + 6$

$x + 2 x + 4 = x + 10$

$3 x + 4 = x + 10$

$3 x + 4 - \textcolor{red}{x} - \textcolor{b l u e}{4} = x + 10 - \textcolor{red}{x} - \textcolor{b l u e}{4}$

$3 x - \textcolor{red}{x} + 4 - \textcolor{b l u e}{4} = x - \textcolor{red}{x} + 10 - \textcolor{b l u e}{4}$

$2 x + 0 = 0 + 6$

$2 x = 6$

$\frac{2 x}{\textcolor{red}{2}} = \frac{6}{\textcolor{red}{2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} x}{\cancel{\textcolor{red}{2}}} = 3$

$x = 3$

The first number is 3 and therefore the next two consecutive odd numbers are 5 and 7