find three consecutive even integers so that the twice the sum of the second and third is twelve less than six times the first

n, n + 2, n + 4 - three consecutive even integers

the twice the sum of the second and third: 2[ (n + 2) + (n + 4) ]

twelve less than six times the first: 6n - 12

The equation:

2[ (n + 2) + (n + 4) ] = 6n - 12

2 (n + 2 + n + 4) = 6n - 12

2 (2n + 6) = 6n - 12 use distributive property

(2) (2n) + (2) (6) = 6n - 12

4n + 12 = 6n - 12 subtract 12 from both sides

4n = 6n - 24 subtract 6n from both sides

-2n = - 24 divide both sides by (-2)

n = 12

n + 2 = 12 + 2 = 14

n + 4 = 12 + 4 = 16

Answer: 12, 14, 16