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bazaltina [42]
1 year ago
12

2 years ago, father age was nine times the son age but 3 years later it will be 5 times only. find the present ages of the fathe

r and son​
Mathematics
1 answer:
Goshia [24]1 year ago
3 0

Answer:

<u><em>The Father is currently 47 and the Son is 7</em></u>

Step-by-step explanation:

Let F and S be the present ages of Father and Son, respectively.

We are told that <u>(F-2) = 9(S-2)</u> [2 years ago, father age was nine times the son age]

We also learn that <u>(F+3) = 5(S+3)</u>  [3 years later it will be 5 times only]

Take the first expression and isolate one of the variables (S or F).  I'll isolate F:

(F-2) = 9(S-2)

F = 9S - 16

Now use this in the second expression:

(F+3) = 5(S+3)

((9S-16)+3) = 5(S+3)

9S-13 = 5S+15

4S = 28

S = 7

Since F = 9S-16,

F = 9*(7)-16

F = 47

<u><em>Father is 47 and Son is 7</em></u>

CHECK:

Was the father 9 times the age of his son 2 years ago?

Father would have been 45 and son 5.  Yes, 9*5 = 45

In 3 years will he be 5 times older than his son?  Yes, Father would be 50 and son would be 10.  5*(10) = 50

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