Find three consecutive even integers such that the sum of twice the first and three times the third is twenty more than four times the second. What is the SECOND number?

18

Step-by-step explanation:

let the consecutive integers be n , n + 2, n + 4 , then

2n + 3(n + 4) = 4(n + 2) + 20 ← distribute parenthesis and simplify both sides

2n + 3n + 12 = 4n + 8 + 20

5n + 12 = 4n + 28 ( subtract 4n from both sides )

n + 12 = 28 ( subtract 12 from both sides )

n = 16

1st number = 16

2nd number = n + 2 = 16 + 2 = 18

3rd number = n + 4 = 16 + 4 = 20