Answer:
2, 4 and 6 are consecutive even integers that fit the criterion.
Step-by-step explanation:
Model the problem as an equation
let a be first integer, b be second, c be third
8a = b+c+6
Since these are consecutive even integers:
b = a + 2
c = b + 2 = a + 4
Substitute these equivalents into the equation
8a = (a+2)+(a+4)+6
Combine like terms
8a = a+2+a+4+6
8a = 2a + 12
Isolate a
6a = 12 (divide both sides by 6)
a = 2
Sub a=2 into b=a+2 and c=a+4
b=a+2 = 2+2 = 4
b=4
c=a+4 = 2+4 = 6
c=6
