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Answer:
Step-by-step explanation:
Given Six years before, the age of the mother was equal to the square of her son's age.
Let the age of the mother be x years and the age of the son be y years.
six years before means it will be x - 6 and square of the age of her son will be (y - 6)^2
So x - 6 = (y - 6)^2
we have the formula (a -b)^2 = a^2 - 2ab + b^2
x - 6 = y^2 - 12 y + 36
or x = y^2 - 12 y + 42
Given three years hence her age will be thrice the age of her son. So the equation will be
x + 3 = 3(y + 3)
x = 3y + 6
Substituting in x we have
3y + 6 = y^2 - 12 y + 42
y^2 - 15 y + 36 = 0
y^2 - 12y - 3y + 36 = 0
y(y - 12) - 3(y - 12) = 0
(y - 12)(y - 3) = 0
y = 12, 3
We take y = 12 since 6 yrs ago it will be 3 - 6 = -3 yrs.
taking y =12
Given x = 3y + 6 = 3(12) + 6 = 42
Now we get mother's age x = 42 yrs and son's age y = 12 yrs