Question

A father says to his son. "7 years ago my age was seven times your age, while 3 years hence my age will be three times your age". Find their present ages?

Answer 1

Step1: Assuming the ages of father and son by variables $\mathit{x}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{y}$

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

Step :2

	Present age	7 years ago	3 years after
Father	$x$	$x-7$	$x+3$
Son	$y$	$y-7$	y+3

Step :3 Applying the condition

According to the question,

Case :1

$$\begin{gathered} x-7=7(y-7) \\ \Rightarrow x-7=7y-49 \\ \Rightarrow x-7y=-49+7 \\ \Rightarrow x-7y=-42.................................\left(1\right) \end{gathered}$$

Case :2

$$\begin{gathered} x+3=3(y+3) \\ \Rightarrow x+3=3y+9 \\ \Rightarrow x-3y=9-3 \\ \Rightarrow x-3y=6.........................\left(2\right) \end{gathered}$$

Step :4 Solving for $x\&y$ using elimination method

$$\begin{gathered} \left(1\right)-\left(2\right)\Rightarrow -4y=-48 \\ \Rightarrow y=\frac{48}{4}=12 \end{gathered}$$

To get $x$, substitute $y=12$ in equation (1)

$$\begin{gathered} \left(1\right)\Rightarrow x-7\left(12\right)=-42 \\ \Rightarrow x-84=-42 \\ \Rightarrow x=84-42=42 \end{gathered}$$

Therefore the present age of father is $42$years and that of the son is$12$ years

Checking,

$$\begin{gathered} x-7=42-7=35 \\ y-7=12-7=535=7\left(5\right) \\ x+3=42+3=45 \\ y+3=12+3=1545=3\times 15 \end{gathered}$$