Find four consecutive odd integers such that the sum of the first, second, and fourth is 47

Question

I need to solve this problem for homework, but my teacher hasn't taught us the method.

Answer 1

Let the first odd integer be x.

Then, "consecutive odd integers" are x, (x+2), (x+4), (x+6), (x+8), etc.  [add 2 because they are odd]

 

"sum of the first, second, and fourth is 47"   means

                   x   +   (x+2)   +  (x+6)  =  47

 

3x + 8 = 47              [collect terms]

3x = 39                  [subtract 8 from both sides]

  x = 13                 [divide both sides by 3]

 

Now, find the other two numbers.

 

The three odd integers are:  13, 15, 19

 

Check:

  Is    13 + 15 + 19 = 47   ?

              47 = 47   ?yes

Answer 2

This is an edit based on Michael's comment.

 

Four odd numbers:

n, n + 2, n + 4, n + 6

n + n + 2 + n + 6 = 47

 

Can you solve for n?
What are the numbers?