There are three consecutive even numbers such that twice the first is 20 more then the second. Find the numbers

Question

I need the answer to the question above please help don't understand 

Answer 1

let's say "n" is a natural number. {1,2,3,4,..} To ensure we have an even number we will multiply "n" by 2. Two times any number will make an even number.

 

consecutive even numbers are like; 2, 4, 6, 8, 10 .. etc.  Add +2 to the previous  number to get the next consecutive.

 

1st even number = 2n

2nd even number = 2n + 2

3rd even number = 2n + 4

 

twice the first number (2n) is 20 more then the second (2n + 2).

 

2(2n) = 2n + 2 + 20

 

4n = 2n + 22

 

4n - 2n = 22

 

2n = 22

 

n = 11

 

Now use n = 11 to find the 3 consecutive even numbers.

 

1st even number = 2(11) = 22
2nd even number = 2(11) + 2 = 24
3rd even number = 2(11) + 4 = 26

 

22, 24, 26