Question

Six years ago ,the age of a father was thrice the age of his son. Six years hence ,his age will be double the age of his son. Find their present age?

Answer 1

Step :1 Assuming the ages of father and son by variables $\mathit{x}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{y}$

Let be $x$ be the age of father and $y$ be the present age of son.

Step :2 Prepare a table

	Present age	6 years ago	6 years after
Father	$x$	$x-6$	$x+6$
Son	$y$	$y-6$	$y+6$

Step: 3 Applying the condition,

According to the question,

Case :1

$$\begin{gathered} x-6=3(y-6) \\ \Rightarrow x-6=3y-18 \\ \Rightarrow x-3y=6-18 \\ \Rightarrow x-3y=-12..................................\left(1\right) \end{gathered}$$

Case :2

$$\begin{gathered} x+6=2(y+6) \\ \Rightarrow x+6=2y+12 \\ \Rightarrow x-2y=12-6 \\ \Rightarrow x-2y=6.....................\left(2\right) \end{gathered}$$

step: 4 Solving for $x$ and $y$ using elimination method

$\left(1\right)-\left(2\right)$$$\begin{gathered} \Rightarrow -y=-18 \\ \Rightarrow y=18 \end{gathered}$$

To find $x$ ,

Substitute$y=18in\left(1\right)$

$$\begin{gathered} \left(1\right)\Rightarrow x=3y-12 \\ =3\left(18\right)-12 \\ =54-12 \\ \Rightarrow x=42 \end{gathered}$$

Therefore present age of father is $42$ years and that of the son is $18$years