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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 Mar 17, 2025 11:40 AM IST | #CBSE Class 10th
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A linear equation is a polynomial with degree 1. A pair of linear equations in two variables consists of two equations, each representing a straight line. These equations play a fundamental role in algebra and are widely used in various mathematical and real-world applications. They can be solved using different methods like substitution, elimination, and graphical representation. Linear equations have many real-life applications in fields like physics, economics, engineering, and computer science.

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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

This article on NCERT Class 10 Maths Chapter 3 Solutions of Pair of Linear Equations in Two Variables provides clear and step-by-step solutions for exercise problems in NCERT Class 10 Maths Book. These solutions of Pair of Linear Equations in Two Variables Class 10 are designed by Subject Matter Experts according to the latest CBSE syllabus, ensuring that students grasp the concepts effectively. NCERT solutions for other subjects and classes can be downloaded in NCERT solutions.

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Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables PDF Free Download

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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

The graph of a pair of linear equations in two variables is represented by two lines.

(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 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.

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 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).

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables (Intext Questions and 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

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 is x and the number of girls is y.

Now, according to the question,

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

x+y=10..(1)


And the number of girls is 4 more than the number of boys,i.e.

x=y+4xy=4.

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

X
5
6
4
Y
5
4
6

Different points (x,y) satisfying (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 x be the price of 1 pencil and y be the price of 1 pen,

Now, According to the question

5x+7y=50


And

7x+5y=46

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

X
3
-4
10
Y
5
10
0

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

X
3
8
-2
Y
5
-2
12

The Graph,

Graph

As we can see from the Graph, both line intersects at point (3,5) that is, x = 3 and y = 5 which means cost of 1 pencil is 3 and the cost of 1 pen is 5.

Q2 On comparing the ratios a1a2,b1b2andc1c2 find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident: (i) 5x4y+8=07x+6y9=0

Answer:

Given Equations,

5x4y+8=07x+6y9=0


Comparing these equations with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get

a1a2=57,b1b2=46 and c1c2=89


As we can see

a1a2b1b2

It means that both lines intersect at exactly one point.

Q2 On comparing the ratios a1a2,b1b2andc1c2 find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident: (ii) 9x+3y+12=018x+6y+24=0

Answer:

Given Equations,

9x+3y+12=018x+6y+24=0


Comparing these equations with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get

a1a2=918=12b1b2=36=12 and c1c2=1224=12


As we can see

a1a2=b1b2=c1c2

It means that both lines are coincident.

Q2 On comparing the ratios a1a2,b1b2andc1c2 , find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident: (iii) 6x3y+10=02xy+9=0

Answer:

Given Equations,

6x3y+10=02xy+9=0


Comparing these equations with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get

a1a2=62=3,b1b2=31=3 and c1c2=109


As we can see

a1a2=b1b2c1c2

It means that both lines are parallel to each other.

Q3 On comparing the ratios a1a2,b1b2andc1c2 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=52x3y=7


Comparing these equations with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get

a1a2=32,b1b2=23 and c1c2=57


As we can see

a1a2b1b2

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

Q3 On comparing the ratios a1a2,b1b2andc1c2, 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=84x6y=9


Comparing these equations with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get

a1a2=24=12b1b2=36=12 and c1c2=89


As we can see

a1a2=b1b2c1c2

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

Q3 On comparing the ratios a1a2,b1b2andc1c2, 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=79x10y=14


Comparing these equations with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get

a1a2=3/29=318=16b1b2=5/310=530=16 and c1c2=714=12


As we can see

a1a2b1b2

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

Q3 On comparing the ratios a1a2,b1b2andc1c2,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=1110x+6y=22


Comparing these equations with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get

a1a2=510=12b1b2=36=12 and c1c2=1122=12


As we can see

a1a2=b1b2=c1c2

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

Q3 On comparing the ratios a1a2,b1b2andc1c2 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=82x+3y=12


Comparing these equations with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get

a1a2=4/32=46b1b2=23 and c1c2=812=23


As we can see

a1a2=b1b2=c1c2

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

Q4 Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: (i) x+y=52x+2y=10

Answer:

Given Equations,

x+y=52x+2y=10


Comparing these equations with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get

a1a2=12b1b2=12 and c1c2=510=12


As we can see

a1a2=b1b2=c1c2

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

The points (x,y) which satisfies in both equations are

X
1
3
5
Y
4
2
0

Graph

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

Answer:

Given Equations,

xy=83x3y=16


Comparing these equations with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get

a1a2=13b1b2=13=13 and c1c2=816=12


As we can see

a1a2=b1b2c1c2

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

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

Answer:

Given Equations,

2x+y6=04x2y4=0


Comparing these equations with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get

a1a2=24=12b1b2=12=12 and c1c2=64=32


As we can see

a1a2b1b2

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

Now The points(x, y) satisfying the equation 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 intersects at point (2,2) and hence the solution of both equations is x = 2 and y = 2.

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

Answer:

Given Equations,

2x2y2=04x4y5=0


Comparing these equations with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get

a1a2=24=12


b1b2=24=12 and 


c1c2=25=25


As we can see

a1a2=b1b2c1c2

It means the given equations have no solution and thus 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.i.e.

l=b+4lb=4


Also given Half Parameter of the rectangle =36 i.e.

l+b=36.(2)


Now, as we have two equations, on adding both equations, we get,

l+b+lb=4+362l=40l=20


Putting this in equation (1),

20b=4b=204b=16

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

Q6 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: (i) intersecting lines

Answer:

Given the equation,

2x+3y8=0

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

a1a2b1b2


So Any line with this condition can be 4x+3y16=0
Here,

a1a2=24=12b1b2=33=1


As

121

the line satisfies the given condition.

Q6 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 (ii) parallel lines

Answer:

Given the equation,

2x+3y8=0


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

a1a2=b1b2c1c2


So Any line with this condition can be 4x+6y8=0
Here,

a1a2=24=12b1b2=36=12c1c2=88=1


As
12=121 the line satisfies the given condition.

Q6 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: (iii) coincident lines

Answer:

Given the equation,

2x+3y8=0


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

a1a2=b1b2=c1c2


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

a1a2=24=12b1b2=36=12c1c2=816=12


As
12=12=12 the line satisfies the given condition.

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


And

3x+2y12=0

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.

Pair of Linear Equations in Two Variables Class 10 Solutions Exercise: 3.2

Q1 Solve the following pair of linear equations by the substitution method. (i) x+y=14xy=4

Answer:

Given two equations,

x+y=14.(1)xy=4.(2)


Now, from (1), we have

y=14x


Substituting this in (2), we get

x(14x)=4x14+x=42x=4+14=18x=9


Substituting this value of x in (3)

y=14x=149=5


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

Q1 Solve the following pair of linear equations by the substitution method (ii) st=3s3+t2=6

Answer:

Given two equations,

st=3(1)s3+t2=6(2)


Now, from (1), we have

s=t+3.


Substituting this in (2), we get

t+33+t2=62t+6+3t6=65t+6=365t=30t=6


Substituting this value of t in (3)

s=t+3=6+3=9

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

Q1 Solve the following pair of linear equations by the substitution method. (iii) 3xy=39x3y=9

Answer:

Given two equations,

3xy=3(1)9x3y=9(2)


Now, from (1), we have

y=3x3..


Substituting this in (2), we get

9x3(3x3)=99x9x+9=99=9


This is always true, and hence this pair of the equation has infinite solutions.
As we have

y=3x3


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

Q1 Solve the following pair of linear equations by the substitution method. (iv) 0.2x+0.3y=1.30.4x+0.5y=2.3

Answer:

Given two equations,

0.2x+0.3y=1.3(1)0.4x+0.5y=2.3(2)


Now, from (1), we have

y=1.30.2x0.3


Substituting this in (2), we get

0.4x+0.5(1.30.2x0.3)=2.30.12x+0.650.1x=0.690.02x=0.690.65=0.04x=2


Substituting this value of x in (3)

y=(1.30.2x0.3)=(1.30.40.3)=0.90.3=3


Hence, Solution of the given equations is,

x=2 and y=3

Q1 Solve the following pair of linear equations by the substitution method. (v) 2x+3y=03x8y=0

Answer:

Given two equations,

2x+3y=0(1)3x8y=0(2)


Now, from (1), we have

y=2x3


Substituting this in (2), we get

3x8(2x3)=03x=8(2x3)3x=4x7x=0x=0


Substituting this value of x in (3)

y=2x3=0

Hence, Solution of the given equations is,

x = 0 , y = 0 .

Q1 Solve the following pair of linear equations by the substitution method. (vi) 3x25y3=2x3+y2=136

Answer:

Given

3x25y3=2(1)x3+y2=136(2)


From (1) we have,

x=23(5y32)


Putting this in (2) we get,

13×23(5y32)+y2=13610y2749+y2=13620y5449+27y54=13647y54=136+4947y54=11754+245447y=117+2447y=141

y=14147y=3

putting this value in (3) we get,

x=23(5y32)x=23(5×332)x=23(52)x=23×3x=2


Hence x=2 and y=3.

Q2 Solve 2x+3y=11and 2x4y=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


Substituting this in (2), we get

2x4(112x3)=246x44+8x=7214x=447214x=28x=2


Substituting this value of x in (3)

y=(112x3)=112×(2)3=153=5


Hence, Solution of the given equations is,

x=2, and y=5

Now,

As it satisfies y=mx+3,

5=m(2)+32m=352m=2m=1

Hence Value of m is -1.

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

(i) 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 let the bigger number is y.

Now, According to the question,

yx=26(1)


And

y=3x(2)


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

3xx=262x=26x=13


Substituting this in (2)

y=3x=3(13)=39

Hence the two numbers are 13 and 39.

Q3 Form the pair of linear equations for the following problem and find their solution by substitution method (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Answer:

Let the larger angle be x and smaller angle be y

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

x+y=180(1)


Also given in the question,

xy=18(2)


Now, From (2) we have,

y=x18


Substituting this value in (1)

x+x18=1802x=180+182x=198x=99


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

y=x18=9918=81


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 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=38007x6


Substituting this value of y in (2)

3x+5(38007x6)=175018x+1900035x=1750×617x=105001900017x=8500x=850017x=500

Now, Substituting this value of x in (3)

y=38007x6=38007×5006=380035006=3006=50

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

Q3 Form the pair of linear equations for the following problems and find their solution by 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 denominator of the fraction is y

Now According to the question,

x+2y+2=91111(x+2)=9(y+2)11x+22=9y+1811x9y=4(1)


Also,

x+3y+3=566(x+3)=5(y+3)6x+18=5y+156x5y=3(2)

Now, From (1) we have

y=11x+49.


Substituting this value of y in (2)

6x5(11x+49)=354x55x20=27x=2027x=7


Substituting this value of x in (3)

y=11x+49=11(7)+49=819=9


Hence the required fraction is

xy=79

Q3 Form the pair of linear equations for the following problems and find their solution by 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+15x3y=10(1)


Also,

x5=7(y5)x5=7y35x7y=30(2)


Now,
From (1) we have,

x=10+3y


Substituting this value of x in (2)

10+3y7y=304y=3010

4y=40y=10


Substituting this value of y in (3),

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


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

Pair of Linear Equations in Two Variables Class 10 Solutions Exercise: 3.3

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

(i) 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


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

2x3y+3x+3y=4+155x=19x=195

Substituting this value in (1), we get

195+y=5y=5195y=65


Hence,

x=195 and y=65


Substitution method :
Given equations

x+y=52x3y=4.


Now, from (1) we have,

y=5x(3)

substituting this value in (2)

2x3(5x)=42x15+3x=45x=19x=195


Substituting this value of x in (3)

y=5x=5195=65


Hence,

x=195 and y=65

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

(ii) 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


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

3x+4y+4x4y=10+47x=14x=2

Putting this value in (2), we get

2(2)2y=22y=2y=1


Hence,

x=2 and y=1


Substitution method :
Given equations

3x+4y=10.(1) and 2x2y=2.(2)


Now, from (2) we have,

y=2x22=x1

substituting this value in (1)

3x+4(x1)=10

3x+4x4=107x=14x=2


Substituting this value of x in (3)

y=x1=21=1


Hence,

x=2 and y=1

Q1 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)9x=2y+79x2y7=0(2)


Now, multiplying (1) by 3, we get

9x15y12=0


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

9x2y79x+15y+12=013y+5=0y=513


Putting this value in (1) we get

3x5(513)4=0

3x=425133x=2713x=913


Hence,

x=913 and y=513


Substitution method :
Given equations

3x5y4=0.(1) and 9x=2y+79x2y7=0.(2)

Now, from (2) we have,

y=9x72

substituting this value in (1)

3x5(9x72)4=06x45x+358=039x+27=0x=2739=913


Substituting this value of x in (3)

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


Hence,

x=913 and y=513

Q1 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)xy3=3(2)

Now, multiplying (2) by 2, we get

2x2y3=6


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

x2+2y3+2x2y3=1+65x2=5x=2


Putting this value in (2), we get

2y3=3

y3=1y=3


Hence,

x=2 and y=3


Substitution method:
Given equations

x2+2y3=1(1) and xy3=3(2)


Now, from (2) we have,

y=3(x3)

substituting this value in (1)

x2+2(3(x3))3=1x2+2x6=15x2=5x=2


Substituting this value of x in (3)

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


Hence,

x=2 and y=3

Q2 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 1/2 if we only add 1 to the denominator. What is the fraction?

Answer:

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

Now, According to the question,

x+1y1=1x+1=y1(1)xy=2(2)


Also,

xy+1=122x=y+12xy=1


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

x=3


Putting this value in (1)

3y=2

y=5


Hence

x=3 and y=5


And the fraction is

35

Q2 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 age of Sonu be y.

Now, According to the question

x5=3(y5)x5=3y15(1)x3y=10(2)


Also,

x+10=2(y+10)x+10=2y+20(1)x2y=10(2)


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

y=20

putting this value in (2)

x2(20)=10x=50

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

Q2 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 10's digit be y.

Now, According to the question,

x+y=9(1)


Also

9(10y+x)=2(10x+y)90y+9x=20x+2y88y11x=08yx=0..(2)


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

9y=9y=1

now putting this value in (1)

x+1=9x=8

Hence the number is 18.

Q2 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=2000x+2y=40(2)


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

y=15


Putting this value in (1).

x+15=25x=10

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

Q2 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 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=27212y=6y=3


Putting this in (1)

x+4(3)=27x=2712=15

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

If interested, students can also check exercises here:

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NCERT Solutions for Class 10 Maths - Chapters Wise

Chapter No.

Chapter Name

Chapter 1

Real Numbers

Chapter 2

Polynomials

Chapter 3

Pair of Linear Equations in Two Variables

Chapter 4

Quadratic Equations

Chapter 5

Arithmetic Progressions

Chapter 6

Triangles

Chapter 7

Coordinate Geometry

Chapter 8

Introduction to Trigonometry

Chapter 9

Some Applications of Trigonometry

Chapter 10

Circles

Chapter 11

Constructions

Chapter 12

Areas Related to Circles

Chapter 13

Surface Areas and Volumes

Chapter 14

Statistics

Chapter 15

Probability

JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters


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

As NCERT Class 10 Maths Chapter 3 solutions are solved by the subject matter experts, the answers to all the questions are reliable. NCERT Class 10 Maths Chapter 3 Solutions give the step-by-step explanations to all the questions which makes it easy for the students to understand. Using the NCERT Class 10 Maths chapter 3 Solutions, students will be able to confirm the right answers once they are done solving the questions themselves.

How to Use NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables?

In this article, you have gone through the NCERT solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables and have a good knowledge of answering structurally. It's time to practice various kinds of problems based on a pair of linear equations in two variables.

After the completion of the NCERT syllabus, you can check the past 5-year papers of board exams. Class 10 Maths Chapter 3 Test Paper with Solution will increase your dealing ability with a variety of questions.

NCERT syllabus coverage and previous year papers are enough tools to get a good score in the board examination. After covering NCERT and the previous year papers of this chapter, you can jump to the next chapters.

NCERT Solutions of Class 10 - Subject Wise

NCERT Exemplar solutions - Subject wise


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.

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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.

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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

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