Find three consecutive even integers so that the twice the sum of the second and third is twelve less than 6 t

Question

Find three consecutive even integers so that the twice the sum of the second and third is twelve less than 6 times the first.

Answer 1

three consecutive even integers are
x, x+2 and x+4

Equation is
2((x+2)+(x+4)) = 6x - 12

Solving for x
x = 12
So, 12, 14 and 16

Answer 2

Well, since the 3 even integers are consecutive, let's use x, x+2, and x+4.
So:

6x-12=2(x+2+x+4)
6x-12=2(2x+6)
6x-12=4x+12
6x-4x=12+12
2x=24
x=12
x+2=14
x+4=16

So the three integers are 12, 14, and 16.

Answer 3

x = first integer
x + 2 = second integer
x + 4 = third integer

2[(x + 2) + (x + 4)] = 6x - 12
2(2x + 6) = 6x - 12
4x + 12 = 6x - 12
24 = 2x
12 = x

The integers are 12, 14, and 16

Answer 4

x, x + 2, x + 4

2((x + 2) + (x + 4)) = 6x - 12
2(x + 2 + x + 4) = 6x - 12
2(2x + 6) = 6x - 12
4x + 12 = 6x - 12
-2x = -24
x = 12

ANS : 12, 14, and 16

Answer 5

x, y, and z

y = x+2
z = y+2
z = x+4
2y + 2z + 12 = 6x

right?

If so,

2y + 2z + 12 = 6x can be translated into 2x + 4 + 2x + 8 + 12 = 6x.

4x + 24 = 6x
2x = 24
x = 12

So they ARE 12, 14, and 16...