• Skip to main content
  • Skip to secondary menu
  • Skip to primary sidebar
  • Skip to footer
  • RBSE Model Papers
    • RBSE Class 12th Board Model Papers 2022
    • RBSE Class 10th Board Model Papers 2022
    • RBSE Class 8th Board Model Papers 2022
    • RBSE Class 5th Board Model Papers 2022
  • RBSE Books
  • RBSE Solutions for Class 10
    • RBSE Solutions for Class 10 Maths
    • RBSE Solutions for Class 10 Science
    • RBSE Solutions for Class 10 Social Science
    • RBSE Solutions for Class 10 English First Flight & Footprints without Feet
    • RBSE Solutions for Class 10 Hindi
    • RBSE Solutions for Class 10 Sanskrit
    • RBSE Solutions for Class 10 Rajasthan Adhyayan
    • RBSE Solutions for Class 10 Physical Education
  • RBSE Solutions for Class 9
    • RBSE Solutions for Class 9 Maths
    • RBSE Solutions for Class 9 Science
    • RBSE Solutions for Class 9 Social Science
    • RBSE Solutions for Class 9 English
    • RBSE Solutions for Class 9 Hindi
    • RBSE Solutions for Class 9 Sanskrit
    • RBSE Solutions for Class 9 Rajasthan Adhyayan
    • RBSE Solutions for Class 9 Physical Education
    • RBSE Solutions for Class 9 Information Technology
  • RBSE Solutions for Class 8
    • RBSE Solutions for Class 8 Maths
    • RBSE Solutions for Class 8 Science
    • RBSE Solutions for Class 8 Social Science
    • RBSE Solutions for Class 8 English
    • RBSE Solutions for Class 8 Hindi
    • RBSE Solutions for Class 8 Sanskrit
    • RBSE Solutions

RBSE Solutions

Rajasthan Board Textbook Solutions for Class 5, 6, 7, 8, 9, 10, 11 and 12

  • RBSE Solutions for Class 7
    • RBSE Solutions for Class 7 Maths
    • RBSE Solutions for Class 7 Science
    • RBSE Solutions for Class 7 Social Science
    • RBSE Solutions for Class 7 English
    • RBSE Solutions for Class 7 Hindi
    • RBSE Solutions for Class 7 Sanskrit
  • RBSE Solutions for Class 6
    • RBSE Solutions for Class 6 Maths
    • RBSE Solutions for Class 6 Science
    • RBSE Solutions for Class 6 Social Science
    • RBSE Solutions for Class 6 English
    • RBSE Solutions for Class 6 Hindi
    • RBSE Solutions for Class 6 Sanskrit
  • RBSE Solutions for Class 5
    • RBSE Solutions for Class 5 Maths
    • RBSE Solutions for Class 5 Environmental Studies
    • RBSE Solutions for Class 5 English
    • RBSE Solutions for Class 5 Hindi
  • RBSE Solutions Class 12
    • RBSE Solutions for Class 12 Maths
    • RBSE Solutions for Class 12 Physics
    • RBSE Solutions for Class 12 Chemistry
    • RBSE Solutions for Class 12 Biology
    • RBSE Solutions for Class 12 English
    • RBSE Solutions for Class 12 Hindi
    • RBSE Solutions for Class 12 Sanskrit
  • RBSE Class 11

RBSE Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

April 8, 2022 by Safia Leave a Comment

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3

Question 1.
Solve the following pair of linear equations by the substitution method.
(i) x + y = 14
x – y = 4
(ii) s – t = 3
\(\frac{s}{3}\) + \(\frac{t}{2}\) = 6
(iii) 3x – y = 3
9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
(v) \(\sqrt{2}\)x + \(\sqrt{3}\)y = 0
\(\sqrt{3}\)x – \(\sqrt{8}\)y = 0
(vi) \(\frac{3 x}{2}-\frac{5 y}{3}\) = -2
\(\frac{x}{3}-\frac{y}{2}\) = \(\frac{13}{6}\)
Solution:
(i) The given pair of linear equations is
x + y = 14 …..(1)
and x – y = 4 (2)
From (2) x = 4 + y …..(3)
Substituting this value of x in equation (1)
4 + y + y = 14
or 2y = 14 – 4
or 2y = 10
or y = \(\frac{10}{2}\) = 5
Substituting this value of y in equation (3),
x = 4 + 5 = 9
Hence, x = 9 and y = 5

(ii) The given pair of linear equations is
s – t = 3 ……(1)
and \(\frac{s}{3}+\frac{t}{2}\) = 6
or \(\frac{2 s+3 t}{6}\) = 6
or 2s + 3t = 36 …(2)
From (1), s = 3 + t …(3)
Substituting this value of 5 in equation (2),
2 (3 + t) + 3t = 36
or 6 + 2t + 3t = 36
or 6 + 5t = 36
or 5t = 36 – 6
or t = \(\frac{30}{5}\) = 6
Substituting this value of t in equation (3),
s = 3 + 6 = 9
Hence, s = 9 and t = 6

(iii) The given pair of linear equations is
3x – y = 3 …(1)
and 9x – 3y = 9 …(2)
From (1),
3x – 3 = y
or y = 3x – 3 …(3)
Substituting this value of y in equation (2),
9x – 3 (3x – 3) = 9
or 9x – 9x + 9 = 9
This statement is true for all values of x. So we cannot find any specific value of x in the form of solution. Therefore we cannot find out any value of y also. This situation has arisen because both the given equation are one and the same.
Hence there are infinitely many solutions of equations (1) and (2).

(iv) The given pair of linear equations is
0.2x + 0.3y = 1.3
RBSE Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 1
Substituting this value of x in equation (2)
4\(\left[\frac{13-3 y}{2}\right]\) + 5y = 23
26 – 6y + 5y = 23
-y = 23 – 26 = -3
y = 3
Substituting this value of y in equation (3)
RBSE Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 2
RBSE Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 3
RBSE Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 4
Hence, x = 2 and y = 3

RBSE Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

Question 2.
Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx +3.
Solution:
The given pair of linear
equations is 2x + 3y = 11 ….(1)
and 2x – Ay = – 24 ….(2)
From (2),
2x = 4y – 24
⇒ 2x = 2 (2y – 12)
⇒ x = 2y – 12 ….(3)
Substituting this value of x in equation (1)
2 (2y – 12) + 3y = 11
or 4y – 24 + 3y = 11
or 7y = 11 + 24
or 7y = 35
or y = \(\frac{35}{7}\) = 5
Substituting this value of y in equation (3)
x = 2 (5) – 12
= 10 – 12 = – 2
Now, take y = mx + 3
Substituting x = – 2, y = 5
5 = m (- 2) + 3
or 5 – 3 = -2 m
or 2 = -2m
or – 2m = 2
or m = – 1
Hence, x = – 2, y = 5 and m = – 1

RBSE Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

Question 3.
Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes \(\frac{9}{11}\), if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes \(\frac{5}{6}\). Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Solution:
(i) Let the two numbers be x and y.
According to the first condition,
x – y = 26 ….(1)
According to the second condition,
x = 3y …(2)
Substituting this value of x in equation (1)
3y – y= 26
or 2y = 26
or y = \(\frac{26}{2}\) = 13
Substituting this value of y in equation (2)
x = 3 × 13 = 39
Hence the two numbers are 39, 13.

(ii) Let the two supplementary angles be
x°, y° and x° > y°.
According to the first condition,
x° + y° = 180° ….(1)
According to the second condition,
x° = y° + 18° ……(2)
Substituting this value of x° in equation (1)
y° + 18° + y° = 180°
or 2y° = 180° – 18°
or 2y° = 162°
or y° = \(\frac{162^{\circ}}{2}\) = 81°
Substituting this value of y° in equation (2)
x° = 81° + 18° = 99°
Hence the required angles are 99°, 81°.

(iii) Let the cost of a bat = ₹ x
and the cost of a ball = ₹ y
According to the first condition,
7x + 6y= 3 800 ….(1)
According to the second condition,
3x + 5y = 1750 ….(2)
From (1), 7x = 3800 – 6y
or x = \(\frac{3800-6 y}{7}\) …(3)
Substituting this value of x in equation (2)
3\(\left[\frac{3800-6 y}{7}\right]\) + 5y = 1750
or \(\frac{11400-18 y+35 y}{7}\) = 1750
or 11400+ 17y = 1750 × 7
or 11400+ 17y = 12250
or 17y = 12250 – 11400
or 17y = 850
or y = \(\frac{850}{17}\) = 50
Substituting this value of y in equation (3)
RBSE Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 5
Hence,cost of one bat = ₹ 500
and cost of one ball = ₹ 50

(iv) Let the fixed charges of the taxi be ₹x and the charge of one km journey = ₹y
According to the first condition,
x + 10y = 105 ….(1)
According to the second condition,
x + 15y = 155 ….(2)
From (1),
x = 105 – 10y ….(3)
Substituting this value of x in equation (2)
105 – 10y + 15y = 155
or 5y = 155 – 105
or 5y = 50
or y = \(\frac{50}{5}\) = 10
Substituting this value of y in equation (3)
x = 105 – 10 × 10
= 105 – 100 = 5
Hence,the fixed charges of the taxi = ₹ 5
and the charge of 1 km journey = ₹ 10
Also, charges of 25 km joumery
= ₹ (10 × 25) + ₹ 5
= ₹ (250 + 5)
= ₹ 255

(v) Let the numerator of the fraction be x and the denominator be y.
∴ Required fraction = \(\frac{x}{y}\)
According to the first condition,
\(\frac{x+2}{y+2}\) = \(\frac{9}{11}\)
or 11 (x + 2) = 9 (y + 2)
or 11x + 22 = 9y + 18
or 11x = 9y + 18 – 22
or 11x = 9y – 4
or x = \(\frac{9 y-4}{11}\) …(1)
According to the second condition,
\(\frac{x+3}{y+3}\) = \(\frac{5}{6}\)
or 6 (x + 3) = 5(y + 3)
or 6x + 18 = 5y + 15
or 6x – 5y = 15 – 18
or 6x – 5y = – 3 ….(2)
Substituting this value of x in equation (1)
RBSE Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 6
or -y – 24 = -3 × 11
or -y = -33 + 24
or -y = -9
or y = 9
Substituting this value of y in equation (1)
x = \(\frac{9 \times 9-4}{11}\) = \(\frac{81-4}{11}\)
= \(\frac{77}{11}\) = 7
Hence, the required fraction is\(\frac{7}{9}\).

(vi) Let the present age of Jacob = x years
and the present age of Jacob’s sun = y years
Five years hence,
Age of Jacob = (x + 5) years
Age of his son= (y + 5) years
According to the first condition,
x + 5 = 3(y + 5)
or x + 5 = 3y + 15
or x = 3y + 15 – 5
or x = 3y + 10 ….(1)
Five years ago
Age of Jacob = (x – 5) years
Age of his son= (y – 5) years
According to the second condition,
x – 5 = 7(y – 5)
or x – 5 = 7y – 35
or x – 7y = – 35 + 5
or x – 7y = – 30 ….(2)
Substituting this value of x in equation (1)
3y + 10 – 7y= – 30
or – 4y = – 30 – 10
or -4y = – 40
or y = 10
Substituting this value of y in equation (1)
x = 3 (10) + 10
= 30 + 10 = 40
Hence,
present age of Jacob = 40 years
and present age of his son = 10 years

Share this:

  • Click to share on WhatsApp (Opens in new window)
  • Click to share on Twitter (Opens in new window)
  • Click to share on Facebook (Opens in new window)

Related

Filed Under: Class 10

Reader Interactions

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Primary Sidebar

Recent Posts

  • RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers Additional Questions
  • RBSE Solutions for Class 11 Psychology in Hindi Medium & English Medium
  • RBSE Solutions for Class 11 Geography in Hindi Medium & English Medium
  • RBSE Solutions for Class 3 Hindi
  • RBSE Solutions for Class 3 English Let’s Learn English
  • RBSE Solutions for Class 3 EVS पर्यावरण अध्ययन अपना परिवेश in Hindi Medium & English Medium
  • RBSE Solutions for Class 3 Maths in Hindi Medium & English Medium
  • RBSE Solutions for Class 3 in Hindi Medium & English Medium
  • RBSE Solutions for Class 4 Hindi
  • RBSE Solutions for Class 4 English Let’s Learn English
  • RBSE Solutions for Class 4 EVS पर्यावरण अध्ययन अपना परिवेश in Hindi Medium & English Medium

Footer

RBSE Solutions for Class 12
RBSE Solutions for Class 11
RBSE Solutions for Class 10
RBSE Solutions for Class 9
RBSE Solutions for Class 8
RBSE Solutions for Class 7
RBSE Solutions for Class 6
RBSE Solutions for Class 5
RBSE Solutions for Class 12 Maths
RBSE Solutions for Class 11 Maths
RBSE Solutions for Class 10 Maths
RBSE Solutions for Class 9 Maths
RBSE Solutions for Class 8 Maths
RBSE Solutions for Class 7 Maths
RBSE Solutions for Class 6 Maths
RBSE Solutions for Class 5 Maths
RBSE Class 11 Political Science Notes
RBSE Class 11 Geography Notes
RBSE Class 11 History Notes

Copyright © 2023 RBSE Solutions

 

Loading Comments...