Pair of Linear Equations in Two Variables : Exercise - 3.1 (Mathematics NCERT Class 10th)

Q.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.
Sol. Let present age of daughter= x years
and her father Aftab= y years
Then, algebraic representation is given by the following equations :
7(x – 7) = y – 7
⇒ 7x – 49 = y – 7
⇒ 7x – y = 42 .....(i)
⇒ y = 7x- 42
Put the different value of x and find corresponding value of y. Minimum two values required to draw graph for a linear equation.

x	6	7
y	0	7

and, 3(x + 3) = y + 3 

 ⇒ 3x + 9 = y + 3 

 ⇒ 3x – y = – 6 ......(ii) 

⇒ y = 3x + 6

x	0	1
y	6	9

To represent these equations graphically, we plot the points A(6, 0) and B(7, 7) to get the graph of equation (i) and the points C(0, 6) and D(1, 9) give the graph of equation (ii).

Q.2 The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 2 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically.

Sol. Let the cost of bat= Rs. x
and one ball= Rs. y. Then, the algebraic representation according to question :

3x + 6y = 3900
⇒ x + 2y = 1300 ...(1) (on deviding both side by 3)
⇒ x = 1300 - 2y

x	700	300
y	300	500

According to question
x + 3y = 1300 ...(2)
x = 1300 - 3y

x	400	100
y	300	400

To represent these equations graphically, we plot the points A(700, 300) and B(300, 500) to get the graph of equation (i) and the points C(400, 300) and D(100, 400) give the graph of equation (ii).

We observe that these lines intersect at point (1300, 0), which is the solution of both equation

Q.3 The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs. 160. After a months, the cost of 4 kg of apples and 2 kg of grapes is Rs. 300. Represent the situation algebraically and geometrically.

Sol. Let cost of 1 kg of apple is Rs. x and cost of 1 kg grapes is Rs.y. According to question: 2x + y = 160 ...(1) ⇒ y = 160 - 2x

x	50	60
y	60	40

According to question, 4x + 2y = 300 ⇒ 2x + y = 150 ...(2) ⇒ y = 150 - 2x

x	50	40
y	50	70

To represent these equations graphically, we plot the points A(50, 60) and B(60, 40) to get the graph of equation (1) and the points C(50, 50) and D(40,70) give the graph of equation (2).