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

where r the answers

step-by-step explanation:

y=3n + 1

step-by-step explanation:

this is wrong because it says the second number is one less and i added one instead