The sum of three consecutive even integers is 14 more than twice the first number. Find the three numbers.

Answer 1

The first number we will call x, we know it is even. The second number is x+ 2 and the third number is x+4

1st x

2nd x+2

3rd x+4

the sum of them is adding them: x + x + 2 + x + 4 = 3x +6

twice the first number is 2x

14 more than twice the first number is 2x + 14

now we set them equal and solve.

3x + 6 = 2x + 14

subtract 2x from both sides

x + 6 = 14

subtract 6 from both sides, then x = 8

if x = 8, then 3x + 6 is equal to 24 + 6 = 30

and 2x + 14 = 2*8 + 14 = 16+14 = 30

therefore the first number is 8, the second number is 10 and the third number is 12. Their sum is 30.

Answer 2

1st integer is X

2nd integer is X+2

3rd integer is X+4

The sum of them is 3x +6

Sium is 14 more than twice the first

3x+6 = 14 + 2x

x+6 = 14

x = 8

8, 10, and 12

the sum is 30

2(8)+14 = 16+14 = 30

yes

the answers in BOLD above are correct.

the integers are 8,10, and 12