Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” Is not this interesting? Represent this situation algebraically and graphically.

Given:

Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.”

To do:

We have to represent the above situation algebraically and graphically.

Solution:

Let $x$ be the present age of Aftab and $y$ be the present age of his daughter.

Age of Aftab 7 years ago$=x-7$.

Age of his daughter 7 years ago$=y-7$.

Age of Aftab 3 years later$=x+3$.

Age of his daughter 3 years later$=y+3$.

According to the question,

$x-7 = 7(y-7)$

$\Rightarrow x-7=7y-49$

$\Rightarrow x-7y+42= 0$.....(i)

$7y=x+42$

$y=\frac{x+42}{7}$

Also,

$x+3 = 3(y+3)$

$\Rightarrow x+3=3y+9$

$\Rightarrow x-3y-6=0$......(ii)

$3y=x-6$

$y=\frac{x-6}{3}$

To represent the above equations graphically we need at least two solutions for each of the equations.

For equation (i),

If $x=-7$ then $y=\frac{-7+42}{7}=\frac{35}{7}=5$

If $x=0$ then $y=\frac{0+42}{7}=6$

If $x=7$ then $y=\frac{7+42}{7}=\frac{49}{7}=7$

For equation (ii),

If $x=0$ then $y=\frac{0-6}{3}=\frac{-6}{3}=-2$

If $y=0$ then $0=\frac{x-6}{3}$

$\Rightarrow x=6$

If $x=3$ then $y=\frac{3-6}{3}=\frac{-3}{3}=-1$

The above situation can be plotted graphically as below:

The line AC represents the equation $x-7y+42=0$ and the line PR represents the equation $x-3y-6=0$.

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Updated on: 10-Oct-2022

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