Answer:
15,16,17
Step-by-step explanation:
The question states, The sum of three consecutive, positive, even integers is 18 more than twice the smallest.
Let a,b,c be positive integers
Let a be the smallest integer.
Because the answer says the sum of the integers has to be 18 more than 2 x the smallest integer. It means the situation can be represented as:
a + b + c = 2a + 18
Because these numbers are consecutive, meaning they are right besides each other. Every variable can be written in terms of a.
a = a
b = a + 1
c = a + 2
Now we can solve for a:
a + (a+1) + (a+2) = 2a + 18
a + a + 1 + a + 2 = 2a + 18
3a + 3 = 2a + 18
a + 3 = 18
a = 15
then we can fill in the rest.
b = (15) + 1 = 16
c = (15) + 2 = 17
Now we verify.
15 + 16 + 17 = 2(15) + 18 ?
15 + 16 + 17 = 48
2(15) + 18 = 48
We now have the solution. The consecutive numbers, are, and can only be
15,16,17
