Pair of Linear Equations in Two Variables Question Answers: NCERT Class 10 Mathematics

Current NCERT chapter
Academic session 2026-27
3 Solutions · 0 Topics · 0 Notes resources

This chapter is part of the verified current curriculum mapping for the academic session shown below.

Chapter Details

Chapter
3 — Pair of Linear Equations in Two Variables
NCERT Book
Mathematics
Language
English

Use this Class 10 Mathematics chapter hub to study Pair of Linear Equations in Two Variables from one place. SaralStudy currently provides 3 visible NCERT solution entries on this page.

NCERT Solutions

Review the SaralStudy solution entries currently available for this chapter.

Exercise 1
A:

Let the present age of Aftab and his father be x years and y years respectively.

                   According to question,

                   7 years ago, we have

                                                          x – 7 = 7 (y - 7)

          Or,                                          x – 7 = 7y – 49

          Or,                                          x – 7y = - 42       …………… (1)

          3 years from now, we have

                                                        (x + 3) = 3 (y + 3)

          Or,                                        x + 3 = 3y + 9

          Or,                                        x – 3y = 6            ……………. (2)

 

                 Graphical Representation

                 From equation (1),                   x – 7y = -42

                   Table value of x and y

x:

0

-42

-35

y:

6

0

1

                  From equation (2),     x – 3y = 6

                   Table value of x and y

x:

0

9

6

y:

-2

1

0

           Plotting the tables on the graph:

A:

Let the cost of one bat and one ball be x Rupees and y Rupees respectively.

                      According to first condition,

                                                          3x + 6y = ₨ 3900    …………….(1) 

                      According to second condition,              

                                                           x + 3y = ₨ 1300   …………….(2) 

                  Graphical Representation

                 From equation (1),                  3x + 6y = 3900

                   Table value of x and y

x:

100

300

700

y:

600

500

300

                  From equation (2),                   x + 3y = 1300

                   Table value of x and y

x:

100

700

400

y:

400

200

300

                   1 Unit = 100

A:

Let the cost of one kg apple be x ₨ and 1 kg grapes be y ₨.

                      According to first condition,

                                                                           2x + y = 160 ₨ …………..(1)

                      According to second condition,                 

                                                                           4x + 2y = 300

                                                                           2x + y = 150 ……………….(2)

                 Graphical Representation

                 Table for equation (1),                    2x + y = 160

                   Table value of x and y

x:

40

60

80

y:

60

40

0

                  Table for equation (2),                    2x + y = 150

                   Table value of x and y

x:

40

60

20

y:

70

30

110

      Exercise 3

A:

(i)                              x + y = 14                 …………….(1)

           x – y = 4                  …………….(2)

                                 From the equation (1), we get

                                   x = 14 - y                 …………….(3)

                                 Putting the value of x in equation (2), we get

                                    (14 - y) – y = 4                                                   

                                    14 – y – y = 4                                                       

                                     - 2y = - 10                                                    

                                   

                                       Putting the value of y in equation (3),

                                       x = 14 – 5 

                                       x = 9

                                        Hence,     x = 9 and y = 5

(iii)                             3x - y = 3                 …………….(1)

             9x – 3y = 9…………….(2)                                

                                 From the equation (1), we get

                                                  

                                  Putting the value of y in equation (2), we get

                                     9x – 3 (3x - 3) = 9

                                      9x – 9x + 9 = 9

                                      9 = 9, which is true.

                                  Therefore, pair of linear equation has infinite many solutions.

A:


A:

(i)                  Let the numbers be x and y, such that x > y

Therefore, according to question        

                      x - y = 26                 …………….(1)

x = 3y…………….(2)

                                   Putting the value of x from equation (2) to equation (1), we get

                                                                                    3y – y = 26

                                                                                    2y = 26

                                                                                     y = 13                  

                                    Putting the value in equation (2), we get

                                                                                     x = 3 x 13

                                                                                     x = 39   

                                    Hence, the numbers are 39 and 13.

(ii) Let one be xâ—¦ and other be yâ—¦ such that (xâ—¦ > yâ—¦)

Therefore, according to question          

                          xâ—¦ + yâ—¦ = 180â—¦…………….(1) (Supplementary angles)

xâ—¦ = 18 + yâ—¦…………….(2)

Putting the value of x from equation (2) to equation (1), we get                                             

18 + yâ—¦ + yâ—¦ = 180â—¦

2yâ—¦ = 162â—¦

Putting the value of y in equation (2), we get

                      xâ—¦ = 18 + 81

x = 99

(iv)                 Let the fixed charge be = ₨ x

Let the charge for 1 km distance be = ₨ y

According to first condition,

                                                                                   x + 10y = ₨ 105

                                                                                      x = 105 – 10y                  …………….(1)

                           According to second condition,

                                                                                     x + 15y = 155                  ………………(2) 

                                    Putting the value of x in equation (2), we get

                                                                                    105 – 10y + 15y = 155

                                                                                    5y = 50

                                                                                     y = 10 

                                    Putting the value of y in equation (2), we get

                                                                                     x + 15 x 10 = 155                               

                                                                                     x = 5

                                    Hence, the fixed charge for taxi is ₨ 5 and, the charge for one km distance is ₨ 50.

                                    Charge for 25 km distance

                                                                                     = 25 x 10 + 5

                                                                                     = ₨ 255

(vi)                 Let the age of Jacob be = x years

Let the age of Jacob’s father be = y years

After 5 years,

                                                Jacob’s age               x + 5 years

                                                   Son’s age                  y + 5 years            

                           According to question,

                       x + 5 = 3 (y + 5)           

x + 5 = 3y + 15

x = 3y + 10………………(1)

Five years ago,

(x- 5) = 7 (y - 5)

x – 5 = 7y – 35

x – 7y = -30……………….(2)

                                    Putting the value of x in equation (2), we get

                                                                                    3y + 10 - 7y = -30

                                                                                    -4y = -40

                                                                                     y = 10 

                                    Putting the value of y in equation (1), we get

                                                                                     x = 3 (10) + 10                                 

                                                                                     x = 40

                                    Hence, the present age of Jacob is 40 years and the age of his son is 10 years.


Exercise 2


A:

Given equations,

                            x + y + 1 = 0 …………………. (1)

                           3x + 2y + 2 = 0 ………………….. (2)

                           From the equation (1), we get

x

0

1

2

y

1

2

3

         From the equation (2), we get                

x

4

3

0

y

0

3

6

 The coordinates of vertices of triangle formed by these lines and the x- axis are (-1, 0), (4, 0), (2, 3).


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