Six years hence a man 's age will be three times his son 's age and three years ago,he was nine times as old as his son .find their present ages

6 years from now:

son's age ---- x
man's age --- 3x

present time: (subtract 6)
son's age --- x-6
man's age -- 3x-6

three years from now: (subtract 3)
son's age = x-9
man's age = 3x - 9

3x-9 = 9(x-9)
3x - 9 = 9x - 81
72 = 6x
x = 12

son's present age = 12-6 = 6 years
father's age = 36-6 = 30 years

To find the present ages of the man and his son, we can solve the given problem using algebraic equations.

Let's assume the present age of the man is M, and the present age of his son is S.

According to the first condition, "Six years hence, the man's age will be three times his son's age":

(M + 6) = 3(S + 6) -- Equation 1

And according to the second condition, "Three years ago, he was nine times as old as his son":

(M - 3) = 9(S - 3) -- Equation 2

Now, we have a system of two equations. We can solve them simultaneously to find the values of M and S.

Let's solve Equation 1 for M:

M + 6 = 3S + 18
M = 3S + 18 - 6
M = 3S + 12 -- Equation 3

Substitute Equation 3 into Equation 2:

(3S + 12) - 3 = 9(S - 3)
3S + 9 = 9S - 27
27 + 9 = 9S - 3S
36 = 6S
S = 6 -- Son's present age

Now substitute the value of S back into Equation 3 to find M:

M = 3S + 12
M = 3(6) + 12
M = 18 + 12
M = 30 -- Man's present age

Therefore, the present age of the man is 30 years, and the present age of his son is 6 years.