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.

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