# Class ten maths Pair of Linear Equa in Two Variables

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**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)

Plot the points (80, 0), (50, 60) and (0, 160).

From equation (ii), 4x + 2y = 300

y = 150 – 2x …(iv)

Plot the points (75, 0), (50, 50) and (0, 150).

We observe that the lines do not intersect anywhere, i.e., they are parallel.