SOLUTION: john's sister is 3 years younger than john, john's father is 28 years older than john. five years ago the father's age was one year more than twice the sum of his children's ages t

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Question 206798: john's sister is 3 years younger than john, john's father is 28 years older than john. five years ago the father's age was one year more than twice the sum of his children's ages then. how old is each now?
i keep getting j-3 for sis j for j and j+28 for dad lost after that
Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
john's sister is 3 years younger than john, john's father is 28 years older than john. five years ago the father's age was one year more than twice the sum of his children's ages then. how old is each now?

John's age = j
Sister's age = j-3
Father's age = j+28

John's age 5 years ago = j-5
Sister's age 5 years ago = (j-3)-5 = j-3-5 = j-8
Father's age 5 years ago = (j+28)-5 = j+28-5 = j+23

Father's age 5 years age = 2(John's age 5 years ago + Sister's age 5 years ago) + 1

                    j+23 = 2[(j-5)+(j-8)]+1
                    j+23 = 2[j-5+j-8]+1
                    j+23 = 2[2j-13]+1
                    j+23 = 4j-26+1
                    j+23 = 4j-25                  
                      48 = 3j
                      16 = j

So,
John is 16
Sister is j-3 = 16-3 = 13
Father is j+28 = 16+28 = 44

Checking:

Five years ago John was 11, Sister was 8, and Father was 39.
Twice the sum of 11 and 8 is twice 19 or 38, and 1 more
than that is 39.  So it checks.
   
Edwin
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