Exercise - 3.1
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.” (Isn’t this interesting?) Represent this situation algebraically and graphically.
Let us denote age of Aftab x and age of his daughter y.
Then the algebraic representation is given by the following equations:
x – 7 = 7 (y – 7)
x - 7 = 7y - 49
x - 7y = -49 + 7
x - 7y = -42 .......(i)
x + 3 = 3 (y + 3)
x + 3 = 3y + 9
x - 3y = 9 - 3
x - 3y = 6 ......(ii)
Lets represent these equations in graphically:
x - 7y = -42 .......(i)
x - 3y = 6 ......(ii)
Plot the points A(0, 6), B(-42, 0) and P(0, -2), Q(6, 0), corresponding to the solutions in Table.
Now draw the lines AB and PQ, representing the equations x - 7y = -42 and x - 3y = 6, as shown in following Figure:
The coach of a cricket team buys 3 bats and 6 balls for Rs. 3900. Later, she buys another bat and 3 more balls of the same kind for Rs. 1300. Represent this situation algebraically and geometrically.
Let x and y be the bats and balls.
3x + 6y = 3900
x + 2y = 1300 .......(i)
x + 2y = 1300 ........(ii)
Lets represent these equations in graphically:
x + 2y = 1300 .......(i)
x + 2y = 1300 ........(ii)
Plot the points A(0, 650), B(1300, 0) and P(0, 650), Q(1300, 0), corresponding to the solutions in Table.
Now draw the lines AB and PQ, representing the equations x + 2y = 1300 and x + 2y = 1300, as shown in following Figure:

But, this is the same as Equation (i).
Hence the lines represented by Equations (i) and (ii) are coincident.
Therefore, Equations (i) and (ii) have infinitely many solutions.
The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs. 160.
After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs. 300. Represent the situation algebraically and geometrically.
Let us denote apples by x and grapes by y.
The algebraic representation is given by the following equations:
2x + y = 160 ...........(i)
4x + 2y = 300
2x + y = 150...........(ii)
Lets represent these equations in graphically:
2x + y = 160 ...........(i)
2x + y = 150...........(ii)
Plot the points A(0, 160), B(80, 0) and P(0, 150), Q(75, 0), corresponding to the solutions in Table.
Now draw the lines AB and PQ, representing the equations 2x + y = 160 and 2x + y = 150, as shown in following Figure: