Even consecutive integers

Question

Find the three consecutive even integers such that nine times of the first integer eight more than the sum of the second and the third integers

Answer 1

The even integers are consecutive, so let x equal the middle one. The lowest one becomes x - 2, and the highest one becomes x + 2. Nine times the first one is 9 times (x -2), or 9x - 18. The sum of the second and third integers is x + (x + 2), or 2x + 2. The problem states that this the first equation (9x - 18) is 8 more than the second one (2x + 2), so 9x - 18 = 2x + 2 + 8. Simplify and 9x - 18 = 2x + 10. Add 18 to both sides and 9x = 2x + 28. Now subtract 2x from both sides and 7x = 28. Divide both sides by 7 and x = 4. So 4 is the middle integer, and the series is 2, 4, and 6. Hope that helps.