# How do you find four consecutive even integers such that 4 times the sum of the first and second is 12 more than 6 times the fourth?

Jun 11, 2016

I found: $20 , 22 , 24 , 26$

#### Explanation:

Let us call our integers:
$2 n$
$2 n + 2$
$2 n + 4$
$2 n + 6$
and write:
$4 \left(2 n + 2 n + 2\right) = 12 + 6 \left(2 n + 6\right)$
rearranging:
$16 n + 8 = 12 + 12 n + 36$
$16 n - 12 n = 40$
$4 n = 40$
$n = \frac{40}{4} = 10$
so the integers should be:
$20 , 22 , 24 , 26$