Ten years later, A will be twice as old as B and five years ago,

Solution:

Let the present age of A be x years and the present age of B be y years.

After 10 years, $A$ 's age will be $(x+10)$ years and $B$ '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 5 years, the age of $A$ was $(x-5)$ years and the age of $B$ was $(y-5)$ years. Thus using the given information, we have

$x-5=3(y-5)$

$\Rightarrow x-5=3 y-15$

 

$\Rightarrow x-3 y+10=0$

So, we have two equations

$x-2 y-10=0$

$x-3 y+10=0$

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

$\frac{x}{(-2) \times 10-(-3) \times(-10)}=\frac{-y}{1 \times 10-1 \times(-10)}=\frac{1}{1 \times(-3)-1 \times(-2)}$

$\Rightarrow \frac{x}{-20-30}=\frac{-y}{10+10}=\frac{1}{-3+2}$

$\Rightarrow \frac{x}{-50}=\frac{-y}{20}=\frac{1}{-1}$

$\Rightarrow \frac{x}{50}=\frac{y}{20}=1$

$\Rightarrow x=50, y=20$

Hence, the present age of $A$ is 50 years and the present age of $B$ is 20 years.