# How do you find three consecutive even integers such that three times the largest is 34 more than the sum of the two smaller integers?

Jul 4, 2017

See a solution process below:

#### Explanation:

Let's call the smallest of the three consecutive integers: $n$

Because these are consecutive even integers it means we can write the other two integers as:

$n + 2$

$n + 4$

Now, we can write "three times the largest" as:

$3 \left(n + 4\right)$

And if this is equal to "34 more than the sum of the two smaller integers" we can write this as the equation:

$3 \left(n + 4\right) = 34 + n + n + 2$

We can now solve for $n$:

$3 \left(n + 4\right) = 34 + 1 n + 1 n + 2$

$3 \left(n + 4\right) = 1 n + 1 n + 2 + 34$

$3 \left(n + 4\right) = \left(1 + 1\right) n + 36$

$\textcolor{red}{3} \left(n + 4\right) = 2 n + 36$

$\left(\textcolor{red}{3} \cdot n\right) + \left(\textcolor{red}{3} \cdot 4\right) = 2 n + 36$

$3 n + 12 = 2 n + 36$

$- \textcolor{b l u e}{2 n} + 3 n + 12 - \textcolor{red}{12} = - \textcolor{b l u e}{2 n} + 2 n + 36 - \textcolor{red}{12}$

$\left(- \textcolor{b l u e}{2} + 3\right) n + 0 = 0 + 24$

$1 n = 24$

$n = 24$

Therefore the three consecutive even integers are:

$n = 24$

$n + 2 = 26$

$n + 4 = 28$