# How do you find four consecutive integers such that 2 times the sum of the first and fourth is 24 less than 6 times the third?

Jun 29, 2016

9,10,11,12

#### Explanation:

let x the first numbers,
so x+1 is the second,
x+2 the third,
x+3 the fourth

the sum of the first and fourth is
$x + x + 3 = 2 x + 3$

2 times this sum is
$2 \left(2 x + 3\right)$

6 times the third is 6(x+2)

so the requested equation to solve will be:
$2 \left(2 x + 3\right) = 6 \left(x + 2\right) - 24$

Let's solve it:

$4 x + 6 = 6 x + 12 - 24$

$6 x - 4 x = 24 - 12 + 6$

$2 x = 18$

$x = 9$