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

Pair of linear equations in two variables means two linear equations that use the same two unknowns, usually x and y.
NCERT Solutions For Class 10 Maths Chapter 3 connect this chapter with graphical method, substitution method, elimination method and consistency of equations.

Chapter 3 Pair of Linear Equations in Two Variables helps students solve real-life situations using two equations formed from the same conditions. The chapter begins with examples such as rides and games at a fair, then explains how two lines can intersect, remain parallel or coincide on a graph.

NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables

Pair of Linear Equations in Two Variables

Medium

Q.

Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.

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Pair of Linear Equations in Two Variables

Medium

Q.

Find c if the system of equations cx + 3y + (3 – c) = 0; 12x + cy – c = 0 has infinitely many solutions?

[CBSE - 2019]

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Pair of Linear Equations in Two Variables

Easy

Q.

For the pair of linear equations $kx = y + 2$ and $6x = 2y + 3$, determine if there exists any real value of $k$ for which the system has infinitely many solutions. Justify your answer.

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Pair of Linear Equations in Two Variables

Easy

Q.

If $x = 1$ and $y = 2$ is a solution for the pair of linear equations $2x – 3y + a = 0$ and $2x + 3y – b = 0$, determine the relationship between the constants $a$ and $b$.

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Pair of Linear Equations in Two Variables

Medium

Q.

A fraction becomes $$\frac{1}{3}$$​ when 2 is subtracted from the numerator and it becomes $$\frac{1}{2}$$​ when 1 is subtracted from the denominator. Find the fraction.

[CBSE - 2019]

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Pair of Linear Equations in Two Variables

Medium

Q.

A father's age is three times the sum of the ages of his two children. After 5 years his age will be two times the sum of their ages. Find the present age of the father.

[CBSE - 2019]

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Pair of Linear Equations in Two Variables

Medium

Q.

Half of the difference between two numbers is 2. The sum of the greater number and twice the smaller number is 13. Find the numbers.

[CBSE - 2023]

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Pair of Linear Equations in Two Variables

Medium

Q.

Three years ago, Rashmi was thrice as old as Nazma. Ten years later, Rashmi will be twice as old as Nazma. How old are Rashmi and Nazma now?

[CBSE - 2024]

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Pair of Linear Equations in Two Variables

Easy

Q.

In Figure, ABCD is a rectangle. Find the values of x and y.

[CBSE - 2018]

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Pair of Linear Equations in Two Variables

Medium

Q.

Solve the following system of linear equations 7x – 2y = 5 and 8x + 7y = 15 and verify your answer.

[CBSE - 2024]

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Pair of Linear Equations in Two Variables

Medium

Q.

The coach of a cricket team buys 3 bats and 6 balls for ₹ 3900. Later, she buys another bat and 2 more balls of the same kind for ₹ 1300. Represent this situation algebraically and geometrically.

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Pair of Linear Equations in Two Variables

Easy

Q.

The value of k for which the system of equations x + y – 4 = 0 and 2x + ky = 3, has no solution, is

[CBSE - 2020]

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Pair of Linear Equations in Two Variables

Easy

Q.

The value of k for which the pair of equations kx = y + 2 and 6x = 2y + 3 has infinitely many solutions,

[CBSE - 2023]

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Pair of Linear Equations in Two Variables

Medium

Q.

In the given figure, graph of two linear equations are shown. The pair of these linear equations is : 

[CBSE - 2024]

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Pair of Linear Equations in Two Variables

Easy

Q.

If x = 1 and y = 2 is a solution of the pair of linear equations 2x – 3y + a = 0 and 2x + 3y – b = 0, then:

[CBSE - 2025]

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Pair of Linear Equations in Two Variables

Medium

Q.

ABCD is a cyclic quadrilateral. Find its all angles.

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Pair of Linear Equations in Two Variables

Medium

Q.

$\begin{array}{l}\text{On\hspace{0.33em}}\mathrm{c}\text{omparing\hspace{0.33em}the\hspace{0.33em}ratios\hspace{0.33em}}\frac{{\text{a}}_{\text{1}}}{{\text{a}}_{\text{2}}}\text{,}\frac{{\text{b}}_{\text{1}}}{{\text{b}}_{\text{2}}}\text{,\hspace{0.33em}}\frac{{\text{c}}_{\text{1}}}{{\text{c}}_{\text{2}}}\text{,\hspace{0.33em}find\hspace{0.33em}out whether the lines representing the }\\ \text{following pairs of linear equations intersect at a point, are parallel or coincident:\hspace{0.33em}}\\ \text{(i) \hspace{0.33em}5x\hspace{0.33em}–\hspace{0.33em}4y\hspace{0.33em}+\hspace{0.33em}8 = 0 }\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}7x\hspace{0.33em}+\hspace{0.33em}6y\hspace{0.33em}–\hspace{0.33em}9 = 0}\\ \text{(ii) 9x\hspace{0.33em}+\hspace{0.33em}3y\hspace{0.33em}+\hspace{0.33em}12 = 0}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}18x\hspace{0.33em}+\hspace{0.33em}6y\hspace{0.33em}+\hspace{0.33em}24 = 0}\\ \text{(iii) 6x\hspace{0.33em}–\hspace{0.33em}3y\hspace{0.33em}+\hspace{0.33em}10 = 0}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}2x\hspace{0.33em}–\hspace{0.33em}y\hspace{0.33em}+\hspace{0.33em}9 = 0}\end{array}$

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Pair of Linear Equations in Two Variables

Medium

Q.

Form the pair of linear equations in the following problems, and find their solutions graphically.
(i)  10 students of Class X took part in 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.
(ii) 5 pencils and 7 pens together cost ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.

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Pair of Linear Equations in Two Variables

Medium

Q.

The cost of 2 kg of apples and 1 kg of grapes on a day was found to be ₹ 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ₹ 300. Represent the situation algebraically and geometrically.

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Pair of Linear Equations in Two Variables

Easy

Q.

Find the value of k for which the system of equations $x + y – 4 = 0$ and $2x + ky = 3$ has no solution.

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NCERT Solutions Class 10 Maths Chapter 3 cover Exercise 3.1, Exercise 3.2 and Exercise 3.3 in textbook order. Students first practise the graphical method Class 10 questions, then solve algebraic questions using the substitution method Class 10 and elimination method Class 10. These Class 10 Pair of Linear Equations in Two Variables Solutions also include consistency conditions and word problems from pair of linear equations in two variables Class 10. The chapter summary lists graphical and algebraic methods as the main ways to solve a pair of linear equations.

Key Takeaways

• Graphical Method: Two lines can intersect, coincide or remain parallel.
• Consistent Pair: A pair of equations has at least one solution.
• Substitution Method: One variable is written in terms of the other and substituted.
• Elimination Method: One variable is removed by adding or subtracting equations.

NCERT Solutions For Class 10 Maths Chapter 3 Structure 2026

NCERT Class 10 Maths Chapter 3 Exercise 3.1 Solutions

Exercise 3.1 focuses on forming pairs of linear equations, solving them graphically and checking whether the pair is consistent, inconsistent or dependent.

Q1. Form the pair of linear equations and find their solutions graphically.

Q1(i). 10 students of Class X took part in a Mathematics quiz. The number of girls is 4 more than the number of boys.

Let:

Number of boys = x

Number of girls = y

Total students:

x + y = 10

Girls are 4 more than boys:

y = x + 4

So:

y - x = 4

Pair of equations:

x + y = 10

y - x = 4

Solve:

From y = x + 4

x + x + 4 = 10

2x = 6

x = 3

Then:

y = 3 + 4

y = 7

Answer:

Number of boys = 3

Number of girls = 7

Q1(ii). 5 pencils and 7 pens together cost ₹50, while 7 pencils and 5 pens together cost ₹46.

Let:

Cost of one pencil = x

Cost of one pen = y

Equations:

5x + 7y = 50

7x + 5y = 46

Subtract the second equation from the first:

5x + 7y - 7x - 5y = 50 - 46

-2x + 2y = 4

y - x = 2

So:

y = x + 2

Substitute in:

5x + 7y = 50

5x + 7(x + 2) = 50

5x + 7x + 14 = 50

12x = 36

x = 3

Then:

y = 3 + 2

y = 5

Answer:

Cost of one pencil = ₹3

Cost of one pen = ₹5

NCERT Class 10 Maths Chapter 3 Exercise 3.1 Solutions for Consistency

For equations:

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

Use these conditions:

If a1/a2 ≠ b1/b2, the lines intersect and the pair has a unique solution.

If a1/a2 = b1/b2 = c1/c2, the lines coincide and the pair has infinitely many solutions.

If a1/a2 = b1/b2 ≠ c1/c2, the lines are parallel and the pair has no solution.

Q2. Compare the ratios and state whether the lines intersect, are parallel or coincident.

Q2(i). 5x - 4y + 8 = 0 and 7x + 6y - 9 = 0

Compare:

a1/a2 = 5/7

b1/b2 = -4/6

Since:

5/7 ≠ -4/6

Answer:

The lines intersect at a point.

Q2(ii). 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0

Compare:

a1/a2 = 9/18 = 1/2

b1/b2 = 3/6 = 1/2

c1/c2 = 12/24 = 1/2

Since:

a1/a2 = b1/b2 = c1/c2

Answer:

The lines are coincident.

Q2(iii). 6x - 3y + 10 = 0 and 2x - y + 9 = 0

Compare:

a1/a2 = 6/2 = 3

b1/b2 = -3/-1 = 3

c1/c2 = 10/9

Since:

a1/a2 = b1/b2 ≠ c1/c2

Answer:

The lines are parallel.

Q3. Find whether the following pairs of linear equations are consistent or inconsistent.

Q3(i). 3x + 2y = 5 and 2x - 3y = 7

Here:

a1/a2 = 3/2

b1/b2 = 2/-3

Since:

3/2 ≠ 2/-3

Answer:

The pair is consistent.

Q3(ii). 2x - 3y = 8 and 4x - 6y = 9

Write in standard form:

2x - 3y - 8 = 0

4x - 6y - 9 = 0

Compare:

a1/a2 = 2/4 = 1/2

b1/b2 = -3/-6 = 1/2

c1/c2 = -8/-9 = 8/9

Since:

a1/a2 = b1/b2 ≠ c1/c2

Answer:

The pair is inconsistent.

Q3(iii). 3x/2 + 5y/3 = 7 and 9x - 10y = 14

Multiply the first equation by 6:

9x + 10y = 42

Second equation:

9x - 10y = 14

Here, the coefficients are not proportional.

Answer:

The pair is consistent.

Q3(iv). 5x - 3y = 11 and -10x + 6y = -22

Write in standard form:

5x - 3y - 11 = 0

-10x + 6y + 22 = 0

Compare:

a1/a2 = 5/-10 = -1/2

b1/b2 = -3/6 = -1/2

c1/c2 = -11/22 = -1/2

Answer:

The pair is dependent and consistent.

Q3(v). 4x/3 + 2y = 8 and 2x + 3y = 12

Multiply the first equation by 3:

4x + 6y = 24

Multiply the second equation by 2:

4x + 6y = 24

Both equations are equivalent.

Answer:

The pair is dependent and consistent.

Class 10 Pair of Linear Equations in Two Variables Solutions for Graphical Method

A consistent pair may have one solution or infinitely many solutions. An inconsistent pair has no solution.

Q4. Which pairs are consistent or inconsistent? If consistent, obtain the solution graphically.

Q4(i). x + y = 5 and 2x + 2y = 10

The second equation is:

2(x + y) = 10

So:

x + y = 5

Both equations represent the same line.

Answer:

The pair is consistent and dependent. It has infinitely many solutions.

Q4(ii). x - y = 8 and 3x - 3y = 16

Multiply the first equation by 3:

3x - 3y = 24

But the second equation is:

3x - 3y = 16

The lines are parallel.

Answer:

The pair is inconsistent.

Q4(iii). 2x + y - 6 = 0 and 4x - 2y - 4 = 0

Equations:

2x + y = 6

4x - 2y = 4

From the first equation:

y = 6 - 2x

Substitute in the second equation:

4x - 2(6 - 2x) = 4

4x - 12 + 4x = 4

8x = 16

x = 2

Now:

y = 6 - 2(2)

y = 2

Answer:

The pair is consistent.

Solution:

x = 2, y = 2

Q4(iv). 2x - 2y - 2 = 0 and 4x - 4y - 5 = 0

Write:

2x - 2y - 2 = 0

4x - 4y - 5 = 0

Compare:

a1/a2 = 2/4 = 1/2

b1/b2 = -2/-4 = 1/2

c1/c2 = -2/-5 = 2/5

Since:

a1/a2 = b1/b2 ≠ c1/c2

Answer:

The pair is inconsistent.

Q5. Half the perimeter of a rectangular garden is 36 m. The length is 4 m more than the width. Find the dimensions.

Let:

Length = x

Width = y

Half perimeter:

x + y = 36

Length is 4 m more than width:

x = y + 4

So:

x - y = 4

Solve:

x + y = 36

x - y = 4

Add the equations:

2x = 40

x = 20

Then:

20 + y = 36

y = 16

Answer:

Length = 20 m

Width = 16 m

Q6. Given 2x + 3y - 8 = 0, write another linear equation so that the pair represents intersecting, parallel and coincident lines.

Given equation:

2x + 3y - 8 = 0

Q6(i). Intersecting lines

Choose a line whose coefficients are not proportional.

Example:

x + y - 1 = 0

Answer:

2x + 3y - 8 = 0 and x + y - 1 = 0 represent intersecting lines.

Q6(ii). Parallel lines

Choose proportional x and y coefficients but a different constant ratio.

Example:

4x + 6y + 5 = 0

Answer:

2x + 3y - 8 = 0 and 4x + 6y + 5 = 0 represent parallel lines.

Q6(iii). Coincident lines

Choose an equivalent equation.

Example:

4x + 6y - 16 = 0

Answer:

2x + 3y - 8 = 0 and 4x + 6y - 16 = 0 represent coincident lines.

Q7. Draw the graphs of x - y + 1 = 0 and 3x + 2y - 12 = 0. Find the vertices of the triangle formed by these lines and the x-axis.

First line:

x - y + 1 = 0

y = x + 1

When y = 0:

0 = x + 1

x = -1

So one x-axis point is:

(-1, 0)

Second line:

3x + 2y - 12 = 0

2y = 12 - 3x

y = (12 - 3x)/2

When y = 0:

12 - 3x = 0

x = 4

So second x-axis point is:

(4, 0)

Now find intersection of the two lines:

y = x + 1

Substitute in:

3x + 2y - 12 = 0

3x + 2(x + 1) - 12 = 0

3x + 2x + 2 - 12 = 0

5x - 10 = 0

x = 2

Then:

y = 2 + 1

y = 3

Intersection point:

(2, 3)

Answer:

The vertices of the triangle are:

(-1, 0), (4, 0), and (2, 3)

NCERT Class 10 Maths Chapter 3 Exercise 3.2 Solutions

Exercise 3.2 focuses on the substitution method Class 10 students use to solve a pair of linear equations. In this method, one variable is written in terms of the other and then substituted into the second equation.

Q1. Solve the following pairs of linear equations by the substitution method.

Q1(i). x + y = 14 and x - y = 4

From:

x + y = 14

x = 14 - y

Substitute in:

x - y = 4

14 - y - y = 4

14 - 2y = 4

-2y = -10

y = 5

Then:

x = 14 - 5

x = 9

Answer:

x = 9, y = 5

Q1(ii). s - t = 3 and s/3 + t/2 = 6

From:

s - t = 3

s = t + 3

Substitute in:

s/3 + t/2 = 6

(t + 3)/3 + t/2 = 6

Multiply by 6:

2(t + 3) + 3t = 36

2t + 6 + 3t = 36

5t = 30

t = 6

Then:

s = 6 + 3

s = 9

Answer:

s = 9, t = 6

Q1(iii). 3x - y = 3 and 9x - 3y = 9

Second equation:

9x - 3y = 9

Divide by 3:

3x - y = 3

Both equations are the same.

Answer:

The pair has infinitely many solutions.

Q1(iv). 0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3

Multiply both equations by 10:

2x + 3y = 13

4x + 5y = 23

From:

2x + 3y = 13

2x = 13 - 3y

x = (13 - 3y)/2

Substitute in:

4x + 5y = 23

4[(13 - 3y)/2] + 5y = 23

2(13 - 3y) + 5y = 23

26 - 6y + 5y = 23

-y = -3

y = 3

Now:

2x + 3(3) = 13

2x + 9 = 13

2x = 4

x = 2

Answer:

x = 2, y = 3

Q1(v). √2x + √3y = 0 and √3x - √8y = 0

From:

√2x + √3y = 0

√2x = -√3y

x = -√3y/√2

Substitute in:

√3x - √8y = 0

√3(-√3y/√2) - √8y = 0

-3y/√2 - √8y = 0

Both terms are negative multiples of y, so:

y = 0

Then:

x = 0

Answer:

x = 0, y = 0

Q1(vi). 3x/2 - 5y/3 = -2 and x/3 + y/2 = 13/6

Clear denominators.

For the first equation, multiply by 6:

9x - 10y = -12

For the second equation, multiply by 6:

2x + 3y = 13

From:

2x + 3y = 13

x = (13 - 3y)/2

Substitute in:

9x - 10y = -12

9[(13 - 3y)/2] - 10y = -12

Multiply by 2:

117 - 27y - 20y = -24

117 - 47y = -24

-47y = -141

y = 3

Then:

2x + 3(3) = 13

2x + 9 = 13

2x = 4

x = 2

Answer:

x = 2, y = 3

Q2. Solve 2x + 3y = 11 and 2x - 4y = -24. Hence find m for which y = mx + 3.

Equations:

2x + 3y = 11

2x - 4y = -24

Subtract the second equation from the first:

2x + 3y - 2x + 4y = 11 - (-24)

7y = 35

y = 5

Substitute in:

2x + 3y = 11

2x + 15 = 11

2x = -4

x = -2

Now use:

y = mx + 3

5 = m(-2) + 3

5 = -2m + 3

2 = -2m

m = -1

Answer:

x = -2, y = 5

m = -1

NCERT Class 10 Maths Chapter 3 Exercise 3.2 Word Problems

These word problems are solved by first forming two linear equations and then applying the substitution method.

Q3(i). The difference between two numbers is 26 and one number is three times the other. Find them.

Let the larger number be x and the smaller number be y.

Equations:

x - y = 26

x = 3y

Substitute:

3y - y = 26

2y = 26

y = 13

Then:

x = 3(13)

x = 39

Answer:

The numbers are 39 and 13.

Q3(ii). The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Let:

Larger angle = x

Smaller angle = y

Supplementary angles:

x + y = 180

Difference:

x - y = 18

Add:

2x = 198

x = 99

Then:

y = 180 - 99

y = 81

Answer:

The angles are 99° and 81°.

Q3(iii). A cricket coach buys 7 bats and 6 balls for ₹3800. Later, she buys 3 bats and 5 balls for ₹1750. Find the cost of each bat and each ball.

Let:

Cost of one bat = x

Cost of one ball = y

Equations:

7x + 6y = 3800

3x + 5y = 1750

From the second equation:

3x = 1750 - 5y

x = (1750 - 5y)/3

Solving the pair gives:

x = 500

y = 50

Check:

7(500) + 6(50) = 3500 + 300 = 3800

3(500) + 5(50) = 1500 + 250 = 1750

Answer:

Cost of one bat = ₹500

Cost of one ball = ₹50

Q3(iv). Taxi charges include a fixed charge and a charge per km. For 10 km, charge is ₹105. For 15 km, charge is ₹155. Find fixed charge, charge per km and charge for 25 km.

Let:

Fixed charge = x

Charge per km = y

Equations:

x + 10y = 105

x + 15y = 155

Subtract:

5y = 50

y = 10

Then:

x + 10(10) = 105

x + 100 = 105

x = 5

For 25 km:

Charge = x + 25y

Charge = 5 + 25(10)

Charge = 255

Answer:

Fixed charge = ₹5

Charge per km = ₹10

Charge for 25 km = ₹255

Q3(v). A fraction becomes 9/11 if 2 is added to numerator and denominator. It becomes 5/6 if 3 is added to both. Find the fraction.

Let the fraction be:

x/y

First condition:

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

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

11x + 22 = 9y + 18

11x - 9y = -4

Second condition:

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

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

6x + 18 = 5y + 15

6x - 5y = -3

Solve:

11x - 9y = -4

6x - 5y = -3

The solution is:

x = 7

y = 9

Answer:

The fraction is 7/9.

Q3(vi). Five years hence, Jacob’s age will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. Find their present ages.

Let:

Jacob’s present age = x

Son’s present age = y

Five years hence:

x + 5 = 3(y + 5)

x + 5 = 3y + 15

x - 3y = 10

Five years ago:

x - 5 = 7(y - 5)

x - 5 = 7y - 35

x - 7y = -30

Subtract:

(x - 3y) - (x - 7y) = 10 - (-30)

4y = 40

y = 10

Then:

x - 3(10) = 10

x - 30 = 10

x = 40

Answer:

Jacob’s present age = 40 years

Son’s present age = 10 years

NCERT Class 10 Maths Chapter 3 Exercise 3.3 Solutions

Exercise 3.3 focuses on the elimination method Class 10 students use to remove one variable and solve the remaining equation.

Q1. Solve the following pairs of linear equations by elimination method and substitution method.

Q1(i). x + y = 5 and 2x - 3y = 4

From:

x + y = 5

Multiply by 2:

2x + 2y = 10

Second equation:

2x - 3y = 4

Subtract:

(2x + 2y) - (2x - 3y) = 10 - 4

5y = 6

y = 6/5

Then:

x + 6/5 = 5

x = 25/5 - 6/5

x = 19/5

Answer:

x = 19/5, y = 6/5

Q1(ii). 3x + 4y = 10 and 2x - 2y = 2

Simplify the second equation:

2x - 2y = 2

x - y = 1

So:

x = y + 1

Substitute in:

3x + 4y = 10

3(y + 1) + 4y = 10

3y + 3 + 4y = 10

7y = 7

y = 1

Then:

x = 1 + 1

x = 2

Answer:

x = 2, y = 1

Q1(iii). 3x - 5y - 4 = 0 and 9x = 2y + 7

Write:

3x - 5y = 4

9x - 2y = 7

Multiply the first equation by 3:

9x - 15y = 12

Second equation:

9x - 2y = 7

Subtract:

(9x - 15y) - (9x - 2y) = 12 - 7

-13y = 5

y = -5/13

Substitute in:

3x - 5y = 4

3x - 5(-5/13) = 4

3x + 25/13 = 4

3x = 52/13 - 25/13

3x = 27/13

x = 9/13

Answer:

x = 9/13, y = -5/13

Q1(iv). x/2 + 2y/3 = -1 and x - y/3 = 3

Clear denominators.

First equation:

x/2 + 2y/3 = -1

Multiply by 6:

3x + 4y = -6

Second equation:

x - y/3 = 3

Multiply by 3:

3x - y = 9

Subtract the second equation from the first:

(3x + 4y) - (3x - y) = -6 - 9

5y = -15

y = -3

Substitute in:

3x - y = 9

3x - (-3) = 9

3x + 3 = 9

3x = 6

x = 2

Answer:

x = 2, y = -3

Class 10 Maths Linear Equations Solutions for Elimination Word Problems

Exercise 3.3 word problems are formed using two conditions. After forming the pair of equations, elimination gives the answer quickly.

Q2(i). If 1 is added to the numerator and 1 is subtracted from the denominator, a fraction becomes 1. It becomes 1/2 if 1 is added only to the denominator. Find the fraction.

Let the fraction be:

x/y

First condition:

(x + 1)/(y - 1) = 1

x + 1 = y - 1

x - y = -2

Second condition:

x/(y + 1) = 1/2

2x = y + 1

2x - y = 1

Now solve:

x - y = -2

2x - y = 1

Subtract:

x = 3

Then:

3 - y = -2

y = 5

Answer:

The fraction is 3/5.

Q2(ii). Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. Find their present ages.

Let:

Nuri’s present age = x

Sonu’s present age = y

Five years ago:

x - 5 = 3(y - 5)

x - 5 = 3y - 15

x - 3y = -10

Ten years later:

x + 10 = 2(y + 10)

x + 10 = 2y + 20

x - 2y = 10

Now solve:

x - 3y = -10

x - 2y = 10

Subtract:

-y = -20

y = 20

Then:

x - 2(20) = 10

x - 40 = 10

x = 50

Answer:

Nuri’s present age = 50 years

Sonu’s present age = 20 years

Q2(iii). The sum of the digits of a two-digit number is 9. Nine times this number is twice the number obtained by reversing the digits. Find the number.

Let:

Ten’s digit = x

Unit’s digit = y

Number = 10x + y

Reversed number = 10y + x

Given:

x + y = 9

Also:

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

90x + 9y = 20y + 2x

88x = 11y

y = 8x

Substitute in:

x + y = 9

x + 8x = 9

9x = 9

x = 1

Then:

y = 8

Answer:

The number is 18.

Q2(iv). Meena withdraws ₹2000 in ₹50 and ₹100 notes. She gets 25 notes in all. Find the number of each type of note.

Let:

Number of ₹50 notes = x

Number of ₹100 notes = y

Total notes:

x + y = 25

Total amount:

50x + 100y = 2000

Divide by 50:

x + 2y = 40

Now solve:

x + y = 25

x + 2y = 40

Subtract:

y = 15

Then:

x + 15 = 25

x = 10

Answer:

₹50 notes = 10

₹100 notes = 15

Q2(v). A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹27 for 7 days, and Susy paid ₹21 for 5 days. Find the fixed charge and charge for each extra day.

Let:

Fixed charge = x

Charge for each extra day = y

For 7 days, extra days = 4:

x + 4y = 27

For 5 days, extra days = 2:

x + 2y = 21

Subtract:

2y = 6

y = 3

Then:

x + 2(3) = 21

x + 6 = 21

x = 15

Answer:

Fixed charge = ₹15

Charge for each extra day = ₹3

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

Chapter 3 uses three main methods: graphical method, substitution method and elimination method. The chapter also explains how coefficient ratios show whether the pair has a unique solution, no solution or infinitely many solutions.

Pair of Linear Equations in Two Variables Class 10

A pair of linear equations in two variables has two equations involving the same variables.

Copy-friendly form:

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

Graphical Method Class 10

The graphical method solves equations by drawing their lines on the coordinate plane.

Copy-friendly results:

Intersecting lines = one solution

Parallel lines = no solution

Coincident lines = infinitely many solutions

Substitution Method Class 10

The substitution method writes one variable in terms of the other.

Copy-friendly steps:

Find x in terms of y or y in terms of x.

Substitute in the other equation.

Solve the one-variable equation.

Substitute back to find the other variable.

Elimination Method Class 10

The elimination method removes one variable by adding or subtracting equations.

Copy-friendly steps:

Make one variable’s coefficients equal.

Add or subtract the equations.

Solve the remaining one-variable equation.

Substitute back to find the other variable.

Consistency Conditions

For:

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

Copy-friendly conditions:

If a1/a2 ≠ b1/b2, the pair has one solution.

If a1/a2 = b1/b2 = c1/c2, the pair has infinitely many solutions.

If a1/a2 = b1/b2 ≠ c1/c2, the pair has no solution.

Quick Formula Table for NCERT Solutions For Class 10 Maths Chapter 3

Useful Links for Class 10 Maths NCERT Solutions