Careers360 Logo
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables

Edited By Komal Miglani | Updated on Jun 26, 2025 10:49 PM IST | #CBSE Class 10th
Upcoming Event
CBSE Class 10th  Exam Date : 10 Jul' 2025 - 15 Jul' 2025

Solving a pair of linear equations is like finding where two friends agree to meet—on the same path, at the same point. A pair of linear equations in two variables consists of two linear equations in each corresponding x and y expression, and when we say "solving" the pair means we are looking for the point where they intersect on the Cartesian plane. NCERT Solutions for Class 10 Maths Chapter 3 will provide both clear and in-depth solutions that accurately answer all the exercise questions in the NCERT textbook. In this chapter, students will work with multiple methods of solving linear equations, such as graphical methods, algebraic methods, substitution methods, and elimination methods.

This Story also Contains
  1. NCERT Solution for Class 10 Maths Chapter 3 Solutions: Download PDF
  2. NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables (Exercise)
  3. Pair of Linear Equations in Two Variables Class 10- Notes
  4. Pair of Linear Equations in Two Variables Class 10 Chapter 3: Topics
  5. NCERT Solutions for Class 10 Maths: Chapter Wise
  6. NCERT Books and NCERT Syllabus
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables

A pair of linear equations shows that every problem has a point where paths cross—if we solve it correctly. These equations play a key role in algebra and find widespread use in various mathematical and real-world applications. Our academic team here at Careers360 comprises experienced experts with years of teaching experience who have developed these NCERT Solutions for Class 10 content based on the modified CBSE curriculum. For a detailed syllabus, study materials, and downloadable PDFs, check out this link: NCERT.

NCERT Solution for Class 10 Maths Chapter 3 Solutions: Download PDF

Download Solution PDF

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables (Exercise)

Below are the NCERT class 10 math chapter 3 solutions for exercise questions.

Class 10 Maths Chapter 3 Solutions Pair of Linear Equations in Two Variables Exercise: 3.1
Total Questions: 7
Page number: 28-29

Q1: Form the pair of linear equations in the following problems and find their solutions graphically.

(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

Answer:

Let the number of boys be x and the number of girls be y.

Now, according to the question,

Total number of students in the class = 10, i.e.

x+y=10.....(1)

And, given that the number of girls is 4 more than the number of boys it means; x=y+4

xy=4..........(2)

Different points (x, y) satisfying equation (1)

X

5

6

4

Y

5

4

6

Different points (x,y) satisfying equation (2)

X

5

6

7

y

1

2

3

Graph,

1635919752095

As we can see from the graph, both lines intersect at the point (7,3). that is x= 7 and y = 3, which means the number of boys in the class is 7 and the number of girls in the class is 3.

Q1: Form the pair of linear equations in the following problems and find their solutions graphically.

(ii) 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.

Answer:

Let the price of 1 pencil be x, and y be the price of 1 pen.

Now, according to the question

5x+7y=50......(1)

And

7x+5y=46......(2)

Now, the points (x,y) that satisfy the equation (1) are

X

3

-4

10

Y

5

10

0

And, the points (x,y) that satisfy the equation (2) are

X

3

8

-2

Y

5

-2

12

The Graph,

Graph

From the graph, both lines intersect at point (3,5), that is, x = 3 and y = 5, which means the cost of 1 pencil is 3 and the cost of 1 pen is 5.

Q2 (i): On comparing the ratios a1a2, b1b2and c1c2, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident: (i) 5x4y+8=0;7x+6y9=0

Answer:

Given Equations,

5x4y+8=0and7x+6y9=0

Comparing these equations with a1x+b1y+c1=0anda2x+b2y+c2=0 , we get

a1a2=57,b1b2=46andc1c2=89

It is observed that;

a1a2b1b2

It means that both lines intersect at exactly one point and have a unique solution.

Q2 (ii): On comparing the ratios a1a2, b1b2and c1c2, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident: (ii) 9x+3y+12=0;18x+6y+24=0

Answer:

Given Equations,

9x+3y+12=0 and 18x+6y+24=0

Comparing these equations with a1x+b1y+c1=0anda2x+b2y+c2=0 , we get

a1a2=918=12,b1b2=36=12andc1c2=1224=12

It is observed that;

a1a2=b1b2=c1c2

It means that both lines are coincident and have infinitely many solutions.

Q2 (iii): On comparing the ratios a1a2, b1b2and c1c2 , find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident: (iii) 6x3y+10=0;2xy+9=0

Answer:

Given Equations,

6x3y+10=0 and2xy+9=0

Comparing these equations with a1x+b1y+c1=0anda2x+b2y+c2=0 , we get

a1a2=62=3,b1b2=31=3andc1c2=109

It is observed that;

a1a2=b1b2c1c2

It means that both lines are parallel and thus have no solution.

Q3 (i): On comparing the ratios a1a2, b1b2and c1c2, find out whether the lines representing the following pairs of linear equations are consistent, or inconsistent: (i) 3x+2y=5;2x3y=7

Answer:

Given Equations,

3x+2y=5 and 2x3y=7

Or, 3x+2y5=0 and 2x3y7=0

Comparing these equations with a1x+b1y+c1=0anda2x+b2y+c2=0 , we get

a1a2=32,b1b2=23andc1c2=57

It is observed that;

a1a2b1b2

It means that the given equations have a unique solution and thus the pair of linear equations is consistent.

Q3 (ii): On comparing the ratios a1a2, b1b2and c1c2, find out whether the lines representing the following pairs of linear equations are consistent, or inconsistent: (ii) 2x3y=8;4x6y=9

Answer:

Given Equations,

2x3y=8 and 4x6y=9

Or, 2x3y8=0 and 4x6y9=0

Comparing these equations with a1x+b1y+c1=0anda2x+b2y+c2=0 , we get

a1a2=24=12,b1b2=36=12andc1c2=89

It is observed that;

a1a2=b1b2c1c2

It means the given equations have no solution, and thus the pair of linear equations is inconsistent.

Q3 (iii): On comparing the ratios a1a2, b1b2and c1c2, find out whether the lines representing the following pairs of linear equations are consistent, or inconsistent: (iii) 32x+53y=7;9x10y=14

Answer:

Given Equations,

32x+53y=7 and 9x10y=14

Or, 32x+53y7=0 and 9x10y14=0

Comparing these equations with a1x+b1y+c1=0anda2x+b2y+c2=0 , we get

a1a2=3/29=318=16,b1b2=5/310=530=16andc1c2=714=12

It is observed that;

a1a2b1b2

It means the given equations have exactly one solution, and thus the pair of linear equations is consistent.

Q3 (iv): On comparing the ratios a1a2, b1b2and c1c2, find out whether the lines representing the following pairs of linear equations are consistent, or inconsistent: (iv) 5x3y=11;10x+6y=22

Answer:

Given Equations,

5x3y=11 and 10x+6y=22

Or, 5x3y11=0 and 10x+6y+22=0

Comparing these equations with a1x+b1y+c1=0anda2x+b2y+c2=0 , we get

a1a2=510=12,b1b2=36=12andc1c2=1122=12

It is observed that;

a1a2=b1b2=c1c2

It means the given equations have an infinite number of solutions, and thus a pair of linear equations is consistent.

Q3 (v): On comparing the ratios a1a2, b1b2and c1c2, find out whether the following pair of linear equations are consistent, or inconsistent (v) 43x+2y=8;2x+3y=12

Answer:

Given Equations,

43x+2y=8 and 2x+3y=12

Or, 43x+2y8=0 and 2x+3y12=0

Comparing these equations with a1x+b1y+c1=0anda2x+b2y+c2=0 , we get

a1a2=4/32=46=23,b1b2=23andc1c2=812=23

It is observed that;

a1a2=b1b2=c1c2

It means the given equations have an infinite number of solutions, and thus a pair of linear equations is consistent.

Q4 (i): Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: x+y=5;2x+2y=10

Answer:

Given Equations,

x+y=5 and 2x+2y=10

Comparing these equations with a1x+b1y+c1=0anda2x+b2y+c2=0 , we get

a1a2=12,b1b2=12andc1c2=510=12

It is observed that;

a1a2=b1b2=c1c2

It means the given equations have an infinite number of solutions, and thus a pair of linear equations is consistent.

The points (x,y) which satisfy both equations are

X

1

3

5

Y

4

2

0

Graph

Q4 (ii): Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: xy=8;3x3y=16

Answer:

Given Equations,

xy=8 and 3x3y=16

Comparing these equations with a1x+b1y+c1=0anda2x+b2y+c2=0 , we get

a1a2=13,b1b2=13=13andc1c2=816=12

It is observed that:

a1a2=b1b2c1c2

It means the given equations have no solution, and thus the pair of linear equations is inconsistent.

Q4 (iii): Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: 2x+y6=0;4x2y4=0

Answer:

Given Equations,

2x+y6=0 and 4x2y4=0

Comparing these equations with a1x+b1y+c1=0anda2x+b2y+c2=0 , we get

a1a2=24=12,b1b2=12=12andc1c2=64=32

It is observed that;

a1a2b1b2

It means the given equations have exactly one solution, and thus the pair of linear equations is consistent.

The points(x, y) satisfying the equation 2x+y6=0 are,

X

0

2

3

Y

6

2

0

And The points(x,y) satisfying the equation 4x2y4=0 are,

X

0

1

2

Y

-2

0

2

GRAPH:

1635919975632

As we can see, both lines intersect at point (2,2) and hence the solution of both equations is x = 2 and y = 2.

Q4 (iv): Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: 2x2y2=0;4x4y5=0

Answer:

Given Equations,

2x2y2=0,4x4y5=0

Comparing these equations with a1x+b1y+c1=0anda2x+b2y+c2=0 , we get

a1a2=24=12,b1b2=24=12andc1c2=25=25

It is observed that;

a1a2=b1b2c1c2

It means the given equations have no solution, and thus the pair of linear equations is inconsistent.

Q5: Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Answer:

Let l be the length of the rectangular garden and b be the width.

Now, according to the question, the length is 4 m more than its width, so we can write it as l=b+4

Or, lb=4....(1)

Also given Half Parameter of the rectangle = 36 it means l+b=36....(2)

Now, as we have two equations, add both equations, and we get,

l+b+lb=4+36

2l=40

l=20

We get the value of l, which is 20m

Now, putting this in equation (1), we get;

20b=4

b=204

b=16

Hence, the Length and width of the rectangle are 20m and 16m, respectively.

Q6 (i): Given the linear equation 2x+3y8=0, write another linear equation in two variables such that the geometrical representation of the pair so formed is: intersecting lines

Answer:

Given the equation,

2x+3y8=0

We know that the condition for the intersection of lines for the equations in the form a1x+b1y+c1=0 and a2x+b2y+c2=0 is,

a1a2b1b2

So any line with this condition can be 4x+3y16=0

Proof,

a1a2=24=12

b1b2=33=1

Hence, 121 it means a1a2b1b2

Therefore, the pair of lines has a unique solution, thus forming intersecting lines.

Q6 (ii): Given the linear equation 2x+3y8=0 , write another linear equation in two variables such that the geometrical representation of the pair so formed is parallel lines

Answer:

Given the equation,

2x+3y8=0

As we know that the condition for the parallel lines for the equations in the form a1x+b1y+c1=0 and a2x+b2y+c2=0 is,

a1a2=b1b2c1c2

So any line with this condition can be 4x+6y8=0

Proof,

a1a2=24=12

b1b2=36=12

c1c2=88=1

Hence, 12=121 it means a1a2=b1b2c1c2

Therefore, the pair of lines has no solutions; thus lines are parallel.

Q6 (iii): Given the linear equation 2x+3y8=0 , write another linear equation in two variables such that the geometrical representation of the pair so formed is: coincident lines

Answer:

Given the equation,

2x+3y8=0

As we know that the condition for the coincidence of the lines for the equations in the form a1x+b1y+c1=0 and a2x+b2y+c2=0 is,

a1a2=b1b2=c1c2

So any line with this condition can be 4x+6y16=0

Proof,

a1a2=24=12

b1b2=36=12

c1c2=816=12

Hence, 12=12=12 it means a1a2=b1b2=c1c2

Therefore, the pair of lines has infinitely many solutions; thus lines are coincident.

Q7: Draw the graphs of the equations xy+1=0and 3x+2y12=0 . Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Answer:

Given two equations,

xy+1=0.........(1)

And

3x+2y12=0.........(2)

The points (x,y) satisfying (1) are

X

0

3

6

Y

1

4

7

And The points(x,y) satisfying (2) are,

X

0

2

4

Y

6

3

0

GRAPH:

1635920037829

As we can see from the graph that both lines intersect at the point (2,3), And the vertices of the Triangle are ( -1,0), (2,3) and (4,0). The area of the triangle is shaded with a green color.

Class 10 Maths Chapter 3 Solutions Pair of Linear Equations in Two Variables Exercise: 3.2
Total Questions: 3
Page number: 33-34

Q1(i): Solve the following pair of linear equations by the substitution method.

x+y=14

xy=4

Answer:

Given two equations,

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

xy=4........(2)

Now, from (1), we have

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

Substituting this in (2), we get

x(14x)=4

x14+x=4

2x=4+14=18

x=9

Substituting this value of x in (3)

y=14x=149=5

Hence, the solution of the given equations is x = 9 and y = 5.

Q1(ii): Solve the following pair of linear equations by the substitution method

st=3

s3+t2=6

Answer:

Given two equations,

st=3..........(1)

s3+t2=6....... (2)

Now, from (1), we have

s=t+3........(3)

Substituting this in (2), we get

t+33+t2=6

2t+6+3t=36

5t+6=36

5t=30

t=6

Substituting this value of t in (3)

s=t+3=6+3=9

Hence, the solution of the given equations is s = 9 and t = 6.

Q1(iii): Solve the following pair of linear equations by the substitution method.

3xy=39

9x3y=9

Answer:

Given two equations,

3xy=3......(1)

9x3y=9.....(2)

Now, from (1), we have

y=3x3........(3)

Substituting this in (2), we get

9x3(3x3)=9

9x9x+9=9

9=9

This is always true, and hence this pair of equations has infinite solutions.

As we have

y=3x3,

One of many possible solutions is x = 1, and y = 0.

Q1(iv): Solve the following pair of linear equations by the substitution method.

0.2x+0.3y=1.3

0.4x+0.5y=2.3

Answer:

Given two equations,

0.2x+0.3y=1.3

0.4x+0.5y=2.3

Now, from (1), we have

y=1.30.2x0.3........(3)

Substituting this in (2), we get

0.4x+0.51.30.2x0.3=2.3

0.12x+0.650.1x=0.69

0.02x=0.690.65=0.04

x=2

Substituting this value of x in (3)

y=1.30.2x0.3=1.30.2×20.3=3

Hence, the solution of the given equations is

x = 2 and y = 3.

Q1(v): Solve the following pair of linear equations by the substitution method.

2x+3y=0

3x8y=0

Answer:

Given two equations,

2x+3y=0

3x8y=0

Now, from (1), we have

x=3y2........(3)

Substituting this in (2), we get

3(3y2)8y=0

3y222y=0

y(3222)=0

y=0

Substituting this value of y in (3)

x=0

Hence, the solution of the given equations is,

x = 0, and y = 0 .

Q1(vi): Solve the following pair of linear equations by the substitution method.

3x25y3=2

x3+y2=136

Answer:

Given,

3x25y3=2 ....... (1)

x3+y2=136 ........ (2)

From (1) we have,

x=(12+10y)9........(3)

Putting this in (2), we get,

(12+10y)93+y2=136

(12+10y)27+y2=136

(24+20y+27y)54=136

47y=141

y=3

Putting this value in (3), we get,

x=(12+10×3)9

x=2

Hence, x = 2 and y = 3.

Q2: Solve 2x + 3y = 11 and 2x − 4y = −24 and hence find the value of ‘m’ for which y = mx + 3.

Answer:

Given two equations,

2x+3y=11......(1)

2x4y=24.......(2)

Now, from (1), we have

y=112x3........(3)

Substituting this in (2), we get

2x4(112x3)=24

6x44+8x=72

14x=4472

14x=28

x=2

Substituting this value of x in (3)

y=112x3=112×(2)3=153=5

Hence, the solution of the given equations is,

x = −2, and y = 5.

Now,

As it satisfies y=mx+3,

5=m(2)+3

2m=35

2m=2

m=1

Hence, the value of m is -1.

Q3(i): Form the pair of linear equations for the following problem and find their solution by the substitution method.

The difference between the two numbers is 26, and one number is three times the other. Find them.

Answer:

Let two numbers be x and y, and the bigger number is y.

Now, according to the question,

yx=26......(1)

And

y=3x......(2)

Now, substituting the value of y from (2) in (1), we get,

3xx=26

2x=26

x=13

Substituting this in (2)

y=3x=3(13)=39

Hence, the two numbers are 13 and 39.

Q3(ii): Form the pair of linear equations for the following problem and find their solution by the substitution method

The larger of the two supplementary angles exceeds the smaller by 18 degrees. Find them.

Answer:

Let the larger angle be x and the smaller angle be y

Now, as we know, the sum of supplementary angles is 180. so,

x+y=1800.......(1)

Also given in the question,

xy=180.......(2)

Now, from (2) we have,

y=x180.......(3)

Substituting this value in (1)

x+x180=1800

2x=1800+180

2x=1980

x=990

Now, substituting this value of x in (3), we get

y=x180=990180=810

Hence, the two supplementary angles are

990 and 810.

Q3: Form the pair of linear equations for the following problems and find their solution by the substitution method.
(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.

Answer:

Let the cost of 1 bat is x and the cost of 1 ball is y.

Now, according to the question,

7x+6y=3800......(1)

3x+5y=1750......(2)

Now, from (1) we have

y=(38007x)6........(3)

Substituting this value of y in (2)

3x+5(38007x)6=1750

18x+1900035x=1750×6

17x=1050019000

17x=8500

x=850017

x=500

Now, Substituting this value of x in (3)

y=(38007x)6

y=38007×5006

y=380035006=50

Hence, the cost of one bat is 500 Rs and the cost of one ball is 50 Rs.

Q3: Form the pair of linear equations for the following problems and find their solution by the substitution method.
(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?

Answer:

Let the fixed charge is x and the per km charge is y.

Now, according to the question

x+10y=105.......(1)

And,

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

Now, from (1) we have,

x=10510y........(3)

Substituting this value of x in (2), we have

10510y+15y=155

5y=155105

5y=50

y=10

Now, substituting this value in (3)

x=10510y

=10510(10)

=105100=5

Hence, the fixed charge is 5 Rs and the per km charge is 10 Rs.

Now, Fair for 25 km :

x+25y=5+25(10)

=5+250=255

Hence, fair for 25km is 255 Rs.

Q3:.Form the pair of linear equations for the following problems and find their solution by the substitution method.
(v) A fraction becomes 911, if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it becomes 56. Find the fraction.

Answer:

Let the numerator of the fraction be x, and the denominator of the fraction be y

Now, according to the question,

x+2y+2=911

11(x+2)=9(y+2)

11x+22=9y+18

11x9y=4.........(1)

Also,

x+3y+3=56

6(x+3)=5(y+3)

6x+18=5y+15

6x5y=3...........(2)

Now, from (1) we have

y=11x+49.............(3)

Substituting this value of y in (2)

6x5(11x+49)=3

54x55x20=27

x=2027

x=7

Substituting this value of x in (3)

y=11x+49

=11(7)+49=819=9

Hence, the required fraction is

x=79

Q3: Form the pair of linear equations for the following problems and find their solution by the substitution method.
(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?

Answer:

Let x be the age of Jacob and y be the age of Jacob's son.

Now, according to the question

x+5=3(y+5)

x+5=3y+15

x3y=10........(1)

Also,

x5=7(y5)

x5=7y35

x7y=30.........(2)

Now,

From (1) we have,

x=10+3y...........(3)

Substituting this value of x in (2)

10+3y7y=30

4y=3010

4y=40

y=10

Substituting this value of y in (3),

x=10+3y=10+3(10)=10+30=40

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

Class 10 Maths Chapter 3 Solutions Pair of Linear Equations in Two Variables Exercise: 3.3
Total Questions: 2
Page number: 36-37

Q1(i): Solve the following pair of linear equations by the elimination method and the substitution method :

x+y=5 and 2x3y=4

Answer:

Elimination Method:

Given, equations

x+y=5............(1) and 2x3y=4........(2)

Now, multiplying (1) by 3 we get

3x+3y=15............(3)

Now, adding (2) and (3), we get

2x3y+3x+3y=4+15

5x=19

x=195

Substituting this value in (1), we get

195+y=5

y=5195

y=65

Hence,

x=195andy=65

Substitution method :

Given, equations

x+y=5............(1) and 2x3y=4........(2)

Now, from (1) we have,

y=5x.......(3)

Substituting this value in (2)

2x3(5x)=4

2x15+3x=4

5x=19

x=195

Substituting this value of x in (3)

y=5x=5195=65

Hence,

x=195andy=65

Q1(ii): Solve the following pair of linear equations by the elimination method and the substitution method :

3x+4y=10 and 2x2y=2

Answer:

Elimination Method:

Given, equations

3x+4y=10............(1) and 2x2y=2..............(2)

Now, multiplying (2) by 2 we get

4x4y=4............(3)

Now, adding (1) and (3), we get

3x+4y+4x4y=10+4

7x=14

x=2

Putting this value in (2) we get

2(2)2y=2

2y=2

y=1

Hence,

x=2andy=1

Substitution method :

Given, equations

3x+4y=10............(1) and 2x2y=2..............(2)

Now, from (2) we have,

y=2x22=x1.......(3)

Substituting this value in (1)

3x+4(x1)=10

3x+4x4=10

7x=14

x=2

Substituting this value of x in (3)

y=x1=21=1

Hence, x=2andy=1

Q1(iii): Solve the following pair of linear equations by the elimination method and the substitution method: (iii) 3x5y4=0 and 9x=2y+7

Answer:

Elimination Method:

Given, equations

3x5y4=0..........(1) and 9x=2y+7

9x2y7=0........(2)

Now, multiplying (1) by 3 we get

9x15y12=0............(3)

Now, subtracting (3) from (2), we get

9x2y79x+15y+12=0

13y+5=0

y=513

Putting this value in (1), we get

3x5(513)4=0

3x=42513

3x=2713

x=913

Hence,

x=913andy=513

Substitution method :

Given, equations

3x5y4=0..........(1) and 9x=2y+7

9x2y7=0........(2)

Now, from (2) we have,

y=9x72.......(3)

Substituting this value in (1)

3x5(9x72)4=0

6x45x+358=0

39x+27=0

x=2739=913

Substituting this value of x in (3)

y=9(9/13)72=81/1372=513

Hence, x=913andy=513

Q1(iv): Solve the following pair of linear equations by the elimination method and the substitution method :(iv) x2+2y3=1 and xy3=3

Answer:

Elimination Method:

Given, equations

x2+2y3=1........(1) and xy3=3............(2)

Now, multiplying (2) by 2, we get

2x2y3=6............(3)

Now, adding (1) and (3), we get

x2+2y3+2x2y3=1+6

5x2=5

x=2

Putting this value in (2), we get

2y3=3

y3=1

y=3

Hence,

x=2andy=3

Substitution method :

Given, equations

x2+2y3=1........(1) and xy3=3............(2)

Now, from (2) we have,

y=3(x3)......(3)

Substituting this value in (1)

x2+2(3(x3))3=1

x2+2x6=1

5x2=5

x=2

Substituting this value of x in (3)

y=3(x3)=3(21)=3

Hence, x=2andy=3

Q2(i): Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 12 if we only add 1 to the denominator. What is the fraction?

Answer:

Let the numerator of the fraction be x, and the denominator is y,

Now, according to the question,

x+1y1=1

x+1=y1

xy=2.........(1)

Also,

xy+1=12

2x=y+1

2xy=1..........(2)

Now, subtracting (1) from (2), we get

x=3

Putting this value in (1)

3y=2

y=5

Hence, x=3andy=5

And the fraction is: 35

Q2(ii): Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Answer:

Let the age of Nuri be x and the age of Sonu be y.

Now, according to the question

x5=3(y5)

x5=3y15

x3y=10.........(1)

Also,

x+10=2(y+10)

x+10=2y+20

x2y=10........(2)

Now, subtracting (1) from (2), we get

y=20

Putting this value in (2)

x2(20)=10

x=50

Hence, the age of Nuri is 50 and the age of Nuri is 20.

Q2(iii): Form the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method : (iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Answer:

Let the unit digit of the number be x and the 10's digit be y.

Now, according to the question,

x+y=9.......(1)

Also

9(10y+x)=2(10x+y)

90y+9x=20x+2y

88y11x=0

8yx=0.........(2)

Now adding (1) and (2), we get,

9y=9

y=1

Now putting this value in (1)

x+1=9

x=8

Hence, the number is 18.

Q2(iv): Form the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method : (iv) Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.

Answer:

Let the number of Rs 50 notes be x and the number of Rs 100 notes be y.

Now, according to the question,

x+y=25..........(1)

And

50x+100y=2000

x+2y=40.............(2)

Now, subtracting (1) from (2), we get

y=15

Putting this value in (1).

x+15=25

x=10

Hence, Meena received 10, 50 Rs notes and 15, 100 Rs notes.

Q2(v): Form the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method : (v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Answer:

Let the fixed charge be x, and per day charge is y.

Now, according to the question,

x+4y=27...........(1)

And

x+2y=21...........(2)

Now, Subtracting (2) from (1). We get,

4y2y=2721

2y=6

y=3

Putting this in (1)

x+4(3)=27

x=2712=15

Hence, the fixed charge is 15 Rs and the per-day charge is 3 Rs.

Also, read,

Aakash Repeater Courses

Take Aakash iACST and get instant scholarship on coaching programs.

Pair of Linear Equations in Two Variables Class 10- Notes

Linear Equations

Linear equations are polynomials with degree one. Eg: 3x=7,2x+6y=10

Types of Linear Equations

S. No.

Types of Linear Equation

General form

Description

Solutions

1.

Linear Equation in one Variable

ax + b = 0

Where a ≠ 0 and a & b are real numbers

One Solution

2.

Linear Equation in Two Variables

ax + by + c = 0

Where a ≠ 0 & b ≠ 0 and a, b & c are real numbers

Infinite Solutions possible

3.

Linear Equation in Three Variables

ax + by + cz + d = 0

Where a ≠ 0, b ≠ 0, c ≠ 0 and a, b, c, d are real numbers

Infinite Solutions possible

Simultaneous System of Linear Equations:

The simultaneous system of linear equations in two variables is in format,

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

The simultaneous system of linear equations can be solved using two methods,

1. Graphical Method
2. Algebraic Method

Graphical Method

Two lines represent the graph of a pair of linear equations in two variables.

(i) If the lines intersect at a point, then that point gives the unique solution of the two equations. In this case, the pair of equations is consistent.

(ii) If the lines coincide, then there are infinitely many solutions, each point on the line being a solution. In this case, the pair of equations is dependent (consistent).

(iii) If the lines are parallel, then the pair of equations has no solution. In this case, the pair of equations is inconsistent.

Algebraic Method:

The pair of linear equations can be solved using the algebraic method in two other methods, namely,
1. Substitution Method
2. Elimination Method

Substitution Method

In the substitution method, one of the linear equations is converted to an equation based on any one of the variables. Eg. The equation x−y=1 can be converted into x=y+1. Now, this value of x is substituted in the second linear equation, which makes the equation as a linear equation of one variable, which is much easier to solve.

Elimination Method

In the elimination method, the given system of equations is manipulated to eliminate one of the variables by adding or subtracting the equations.

For example,

Consider the system of equations: 2x+3y=8 4x−3y=10

To eliminate y, we add both equations:
(2x+3y)+(4x−3y)=8+10
6x=18
x=3

Now, substituting x=3 in the first equation:

2(3)+3y=86+3y=83y=2y=23
Thus, the solution is (3,23).

Pair of Linear Equations in Two Variables Class 10 Chapter 3: Topics

The topics discussed in the NCERT Solutions for class 10, chapter 3, Pair of linear equations in two variables are:

  • Introduction
  • Graphical Method of Solution of a Pair of Linear Equations
  • Algebraic Methods of Solving a Pair of Linear Equations
  • Substitution Method
  • Elimination Method

NCERT Solutions for Class 10 Maths: Chapter Wise


Access all NCERT Class 10 Maths solutions from one place using the links below.

Also, read,

NCERT Solutions of Class 10 Subject Wise

Students can use the following links to check the solutions to Maths and science-related questions.

NCERT Exemplar Solutions Subject-wise

After completing the NCERT textbooks, students should practice exemplar exercises for a better understanding of the chapters and clarity. The following links will help students find exemplar exercises.

NCERT Books and NCERT Syllabus

Students can use the following links to check the latest NCERT syllabus and read some reference books.

Frequently Asked Questions (FAQs)

1. How do you solve a pair of linear equations using the substitution method?

In the substitution method, one of the linear equations is converted to an equation based on any one of the variables. Eg. The equation xy=1 can be converted into x=y+1. Now, this value of x is substituted in the second linear equation, which makes the equation as a linear equation of one variable, which is much easier to solve.

2. What is the elimination method in Chapter 3 of Class 10 Maths?

The elimination method is one of the algebraic methods used to solve linear equations. 

In the elimination method, the given system of equations is manipulated to eliminate one of the variables by adding or subtracting the equations.

For example, 

Consider the system of equations: 2x+3y=8, 4x3y=10

To eliminate y, we add both equations:
(2x+3y)+(4x3y)=8+10
6x=18
x=3

Now, substituting x=3 in the first equation:
2(3)+3y=8
6+3y=8
3y=2
y=23

Thus, the solution is (3,23).

3. How to solve linear equations graphically as per NCERT Class 10 Maths?

The graph of a pair of linear equations in two variables is represented by two lines. Graph the linear equations and find the intersection points. The solution of the linear equations depends on the intersection points. 

(i) If the lines intersect at a point, then that point gives the unique solution of the two equations. 

(ii) If the lines coincide, then there are infinitely many solutions — each point on the line being a solution. 

(iii) If the lines are parallel, then the pair of equations has no solution.

4. Can two linear equations have infinitely many solutions?

Yes, two linear equations have infinitely many solutions when they represent the same line. For the linear equations a1x+b1y+c1 and a2x+b2y+c2 to have infinitely many solutions, a1a2=b1b2=c1c2

5. How to determine if a system of equations is consistent or inconsistent?

(i) If the lines intersect at a point, then that point gives the unique solution of the two equations. In this case, the pair of equations is consistent.

(ii) If the lines coincide, then there are infinitely many solutions — each point on the line being a solution. In this case, the pair of equations is dependent (consistent).

(iii) If the lines are parallel, then the pair of equations has no solution. In this case, the pair of equations is inconsistent.

Articles

Upcoming School Exams

View All School Exams

Explore Top Universities Across Globe

Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Hello

Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.

1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.

2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.

3. Counseling and Seat Allocation:

After the KCET exam, you will need to participate in online counseling.

You need to select your preferred colleges and courses.

Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.

4. Required Documents :

Domicile Certificate (proof that you are a resident of Karnataka).

Income Certificate (for minority category benefits).

Marksheets (11th and 12th from the Karnataka State Board).

KCET Admit Card and Scorecard.

This process will allow you to secure a seat based on your KCET performance and your category .

check link for more details

https://medicine.careers360.com/neet-college-predictor

Hope this helps you .

Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

View All

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

Back to top