Ten years ago, a father was twelve times as old as his son and ten years hence, he will be twice as old as his son will be then. Find their present ages.
Let the present age of father be x years and the present age of his son be y years.
After 10 years, father's age will be $(x+10)$ years and son's age will be $(y+10)$ years. Thus using the given information, we have
$x+10=2(y+10)$
$\Rightarrow x+10=2 y+20$
$\Rightarrow x-2 y-10=0$
Before 10 years, the age of father was $(x-10)$ years and the age of son was $(y-10)$ years. Thus using the given information, we have
$x-10=12(y-10)$
$\Rightarrow x-10=12 y-120$
$\Rightarrow x-12 y+110=0$
So, we have two equations
$x-2 y-10=0$
$x-12 y+110=0$
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
$\frac{x}{(-2) \times 110-(-12) \times(-10)}=\frac{-y}{1 \times 110-1 \times(-10)}=\frac{1}{1 \times(-12)-1 \times(-2)}$
$\Rightarrow \frac{x}{-220-120}=\frac{-y}{110+10}=\frac{1}{-12+2}$
$\Rightarrow \frac{x}{-340}=\frac{-y}{120}=\frac{1}{-10}$
$\Rightarrow \frac{x}{340}=\frac{y}{120}=\frac{1}{10}$
$\Rightarrow x=\frac{340}{10}, y=\frac{120}{10}$
$\Rightarrow x=34, y=12$
Hence, the present age of father is 34 years and the present age of son is 12 years.
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