Question

Two years ago, a man's age was three times the square of his son's age. In three years time, his age will be four times his son's age. Find their present ages.

Answer 1

Let present age of father = x

and present age of son = y

Two years ago,

$x-2=3(y-2{)}^2$​​​​​​​ ..............(1)

Three years hence,

$x+3=4(y+3)$​​​​​​​ ............(2)

Now, from equation 2,

$x=4(y+3)-3$​​​​​​​

From equation 1,

=>$4(y+3)-3-2=3(y-2{)}^2$​​​​​​​

=>$4y+12-5=3(y^2+4-4y)$​​​​​​​

=> $4y+7=3y^2+12-12y$​​​​​​​

=> $3y^2+12-12y-4y-7=0$​​​​​​​

=>$3y^2-16y+5=0$​​​​​​​

=> $(y-5)\times (3y-1)=0$​​​​​​​

=> $y=5,\frac{1}{3}$​​​​​​​

Since age can not be in fraction,

So $y=5$​​​​​​​

From equation 2,

$x+3=4(5+3)$​​​​​​​

=> $x+3=4\times 8$​​​​​​​

=> $x+3=32$​​​​​​​

=> $x=32-3$​​​​​​​

=> $x=29$​​​​​​​

So, present age of father = 29

and present age of son = 5