How would I...?

Question

solve this problem. The sum of 4 consecutive even intergers is 44, find all 4 intergers.

Answer 1

Let x be the smallest integer. Then the other integers would be given by (x+2), (x+4), and (x+6).

x + (x+2) + (x+4) + (x+6) = 44
4x + 12 = 44
4x = 32
x = 8

So the integers are 8, 10, 12, 14.

Answer 2

If the four consecutive are all even , you first set them as

X, X+2, X+4, and X+6.

Then set up the equation as...

x + ( x + 2 ) + ( x + 4 ) + ( x + 6) = 44

Simplify the equation to...

4x + 12 = 44

Subtract the 12 from both sides to give you...

4x = 32

Divide both sides by 4 to get...

x = 8

So 8 is your first integer...

Now plug this into the original equation...

X=8, X+2=10, X+4=12, and X+6=14

Therefore your four consecutive even integers are 8, 10, 12, and 14.

Answer 3

x + x + 2 + x + 4 + x + 6 = 44
4x + 12 = 44
4x = 32
x = 8
so 8, 10 , 12, 14

Answer 4

hmm let me give it a shot
x=first consec. even int.
x+2=second " "
x+4=third" "
x+6=fourth" "
so the equasion is
x+(x+2)+(x+4)+(x+6)=44
4x+12=44
4x=32
x=8
then plug the 8 as x into the otheres to find what the numbers are. then do a check.

its been a while since ive done this so, dont rely on me. should listen inn class more.

Answer 5

let the numbers be y, y+2, y+4, y+6

y + (y+2) + (y+4) +(y+6) = 44
4y + 12 = 44
y = (44-12) / 4
y = 8

The numbers are 8, 10, 12, and 14

Answer 6

n + (n + 2) + (n + 4) + (n + 6) = 44

4n + 12 = 44

4n = 32

n = 8

8, 10, 12, 14

Answer 7

n + (n + 2) + (n + 4) + (n + 6) = 44
4n + 12 = 44
4n = 32
n = 8
integers are 8, 10, 12, 14.

Answer 8

2n + (2n + 2) + (2n + 4) + (2n + 6) = 44
8n + 12 = 44
8n = 32
n = 4
Integers are 8 , 10 , 12 , 14

Answer 9

i dunno but i got 2 points lol no i really dont know sorry