# How do you find three consecutive even integers whose sum is 48?

Mar 23, 2018

$\text{1st Integer} = 15$
$\text{2nd Integer} = 16$
$\text{3rd Integer} = 17$

#### Explanation:

Let's use $n$ to represent an integer (whole number). Since we need three integers, let's define them like this:

$\textcolor{b l u e}{n} =$1st integer
$\textcolor{red}{n + 1} =$2nd integer
$\textcolor{g r e e n}{n + 2} =$3rd integer

We know we can define the second and third integers as $n + 1$ and $n + 2$ due to the problem telling us that the integers are consecutive (in order)

Now we can make our equation since we know what it's going to equal:

$\textcolor{b l u e}{n} + \textcolor{red}{n + 1} + \textcolor{g r e e n}{n + 2} = 48$

Now that we've set up the equation, we can solve by combining like terms:

$3 n + 3 = 48$

$3 n = 45$ $\textcolor{b l u e}{\text{ ""Subtract " 3 " from both sides}}$

$n = 15$ $\textcolor{b l u e}{\text{ } \frac{45}{3} = 15}$

Now that we know what $n$ is, we can plug it back into our original definitions:

$\textcolor{b l u e}{n} = 15$ $\textcolor{b l u e}{\text{ 1st Integer}}$
$\textcolor{red}{15 + 1} = 16$ $\textcolor{red}{\text{ 2nd Integer}}$
$\textcolor{g r e e n}{15 + 2} = 17$ $\textcolor{g r e e n}{\text{ 3rd Integer}}$

$\textcolor{b l u e}{15} + \textcolor{red}{16} + \textcolor{g r e e n}{17} = 48$ $\text{ True}$