Algebra question: solving equations?

Question

What are 4 consecutive even integers such that the sum of the second and the fourth is 76?

Answer 1

the answer is i do not know

Answer 2

the answer is 36, 37, 38, 39

Answer 3

34, 36, 38, 40...

follow mistwing's.. that is the right one.

Answer 4

The sequence is: x, x + 2, x + 4, x + 6

Your question is: x + 2 + x + 6 = 76

Do the math

Answer 5

Let n be the second, then n + 4 is the fourth

n + n + 4 = 76

2n = 72

n = 36

So the four consecutive even integers are

34, 36, 38, 40

and 36 + 40 = 76

Answer 6

x-2       x           x + 2         x + 4

x + x + 4 = 76

2x = 72

x = 36

the answer

34 36 38 40

Answer 7

34,36,38,40

Let the consecutive even integers be 2x, 2x+2, 2x+4, 2x+6

2x+2 + 2x + 6 = 76

4x + 8 = 76

4x = 68

x = 17

so 2x = 34 , etc.

Good luck to you !

Answer 8

36,37,38,39

Answer 9

Hey Lily !

Lets consider the 4 consecutive even integers to be

  x

(x+2)

(x+4)

(x+6)

Even no's are 1,2,4,6,8,10.......... Since as we need only 4 consecutive even no's, so we've chosen the above four no's.

Now, the sum of second & fourth is 76.

i.e. (x+2) + (x+6) = 76

      x + 2 + x + 6 =76

         2x + 6 = 76

         2x = 76 - 6

         2x = 68

         x = 34

Now , as we've got the value of x, we just need to substitute it in our assumed integers,

x=34

(x+2) = 34 + 2 =36

(x+4) = 34 + 4 = 38

(x+6) = 34 + 6 = 40

Hence, the 4 consecutive even integers are 34, 36, 38 and 40.

Have a NICE day  

Cheers ^_^