Class ten maths Pair of Linear Equa in Two Variables

Chapter 3: Pair of Linear Equations in Two Variables

TEXTBOOK’S EXERCISE 3.1

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 the present age of Aftab be x years and that of daughter be y years.

Seven years ago Aftab’s age = (x – 7) years

Seven years ago daughter’s age = (y – 7) years

According to the question,

After 3 years Aftab’s age = (x + 3) years.

After 3 years daughter’s age = (y + 3) years.

According to the question,

From (i), x – 7y = – 42

Points are A(0, 6) and B(7, 7)

From (ii), x – 3y = 6

Points are C(6, 0) and D(12, 2)

We plot these points on a graph paper and draw the lines. The graph is shown below.

We see that the lines intersect each other at E(42, 12).

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 a bat = Rs x and cost of a ball = Rs y

As per condition, 3x + 6y = 3900 …(i)

x + 2y = 1300 …(ii)

From (i), 3x = 3900 – 6y

⇒ y = 1/2 [1300 – x] …(iii)

Plot the points (0, 650), (300, 500) and (1300, 0).

From (ii), x + 2y = 1300

⇒ x = 1300 – 2y …(iv)

Plot the points (1300, 0), (600, 350) and (900, 200).

From the graph, it is clear that the two lines coincide.

Hence, both the equation are equivalent.

3. The cost of 2 kg of apples and 1 kg 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.

Sol. Let cost of 1 kg apples = Rs x, and

Cost of 1 kg grapes = Rs y

As per condition,

2x + y = 160 …(i)

4x + 2y = 300 …(ii)

If 2x + y = 160 ⇒ y = 160 – 2x …(iii)