RBSE Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1

Question 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.
Solution:
Let the present age of Aftab be x years
and the present age of Aftab’s daughter be y years.
Algebraical Form : According to the first condition given in the problem
x – 7 = 7 (y – 7)
or x – 7 = 7y – 49
or x – 7y + 42 = 0
According to the second condition given in the problem
x + 3 = 3 (y + 3)
or x + 3 = 3y + 9
or x – 3y – 6 = 0
Graphical Form-
x – 7y + 42 = 0
x = 7y – 42
Substituting y = 5 in (1),
x = 7 × 5 – 42
= 35 – 42 = -7
Substituting y = 6 in (1),
x = 7 × 6 – 42
= 42 – 42 = 0
Substituting y = 7 in (1),
x = 7 × 7 – 42
= 49 – 42 = 7

Y	– 7	0	7
y	5	6	7

On plotting the points A (- 7, 5), B (0, 6), C (7, 7) and drawing a line joining then we get the graph of the equation x – 7y + 42 = 0

x – 3y – 6 = 0
x = 3y + 6 …(2)
Substituting y = 0 in (2)
x = 3 × 0 + 6
= 0 + 6 = 6
Substituting y = 3 in (2)
x = 3 × 3 + 6
= 9 + 6 = 15
Substituting y = – 2 in (2)
x = 3 × -2 + 6
= -6 + 6 = 0

x	6	15	0
y	0	3	– 2

Plotting the points D (6, 0), E (15, 3), F (0, – 2) and drawing a line joining them, we get the graph of the equation x – 3y – 6 = 0.

Question 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.
Solution:
Let the cost of a bat = ₹ x
and the cost of a ball = ₹ y
Algebraical Form—
According to the first condition of the problem,
3x + 6y = 3900
⇒ x + 2y = 1300
According to the second condition of the problem,
1x + 3y = 1300
Pair of linear equations in two variables is
x + 2y= 1300
and x + 3y = 1300
Graphical Form—
x + 2y = 1300
x + 3y = 1300
x = 1300 – 2y ….(1)
Putting y = 0 in (1)
x = 1300 – 2 (0)
⇒ x = 1300
Putting y = 500 in (1)
x = 1300 – 2 × 500
= 1300 – 1000 = 300
Putting y = 650 in (1)
x = 1300 – 2 × 650
= 1300 – 1300 = 0

x	1300	300	0
y	0	500	650

Plotting the points A (1300, 0), B (300, 500) and C (0, 650) and drawing a line joining them we get the graph of the equation x + 2y = 1300
x + 3y = 1300
x = 1300 – 3y ……..(2)
Putting y = 0 in (2)
x = 1300 – 3 × o = 1300
Putting y = 500 in (2)
x = 1300 – 3 × 500
= 1300 – 1500 = – 200
Putting y = 300 in (2)
x = 1300 – 3 × 300
= 1300 – 900 = 400

x	1300	– 200	400
y	0	500	300

Plotting the points A (1300, 0), E (-200, 500), F (400, 300) and drawing a line joining them we get the graph of the equation x + 3y = 1300

From the graph it is clear that both the lines intersect at A (1300, 0). Therefore, x = 1300 and y – 0 is the solution of the given pair of linear equations.

Question 3.
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.
Solution:
Let the cost of 1 kg of apples = ₹ x
and the cost of 1 kg of grapes = ₹ y
Algebraical Form :
According to the first condition of the problem
2x + 1y = 160
According to the second condition of the problem
4x + 2y = 300
∴ Pair of linear equations of two variables is
2x + y = 160
and 4x + 2y = 300
Graphical Form—
2 x + y= 160
2x= 160 – y
x = \(\frac{160-y}{2}\) ….(1)
Putting y = 0 in (1)
x = \(\frac{160-0}{2}\) = \(\frac{160}{2}\) = 80
Putting y = 60 in (1)
x = \(\frac{160-60}{2}\) = \(\frac{100}{2}\) = 50
Putting y = 160 in (1)
x = \(\frac{160-160}{2}\) = \(\frac{0}{2}\) = 0

x	80	50	0
y	0	60	160

Plotting the points A (80, 0), B (50, 60), C (0, 160) and drawing a line joining them we get the graph of the equation 2x + y = 160.

4x + 2y = 300
2x + y = 150
2x = 150 – y
x = \(\frac{150-y}{2}\) …..(2)
Putting y = 0 in (2) x = \(\frac{150-0}{2}\) = \(\frac{150}{2}\)
= 75
Putting y = 50 in (2) x = \(\frac{150-50}{2}\) = \(\frac{100}{2}\)
= 50
Putting y = 150 in (2) x = \(\frac{150-150}{2}\) = \(\frac{0}{2}\)
= 0

x	75	50	0
y	0	50	150

Plotting the points D (75, 0), E (50, 50), F (0. 150) and drawing a line joining them, we get the giaph of the equation 4x + 2y = 300.
From graph it is clear that both the lines intersect nowhere, i.e., they are parallel.