Question

) One year ago, a man was four times as old his son. After 6 years, his age exceeds twice his son’s age by 9 years. Find their present ages.

Answer 1

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Answer 2

Step-by-step explanation:

let present age of man =x

let present age of son =y

one year ago age of man= x-1

" " " " " son = y-1

According to question:

x-1 = 4(y-1)

x-4y = -3 (i.) eq.

After 6 years age of man= x+6

" " " " " son = y+6

According to question:

x+6 = 2(y+6)+9

x-2y = 15 (ii.) eq

Subtracting eq 2 from eq. 1

x-4y = -3

x-2y = 15

- + -

-2y = -18

y = 9

substituting the value of y in ii.) eq.

x-2×9 = 15

x-18 = 15

x = 15+18

x= 33

Present age of father = 33 years

Present age of son = 9years