Pair of Linear Equations Class 10 Important Questions for Board Exams

Pair of Linear Equations in Two Variables is one of the most important chapters in Class 10 Maths because it combines algebra, graphs, interpretation of solutions, and word problems in one place. This chapter is highly scoring in board exams when students clearly understand the standard form, the three main solving methods, and the meaning of unique solution, no solution, and infinitely many solutions.

Many students feel comfortable with direct equations, but they get confused in graphical questions, coefficient-comparison questions, and word problems based on age, fractions, or investments. The chapter becomes much easier when students first understand what a pair of linear equations really means and then practise each question type separately.

At Deeksha Vedantu, we always encourage students to prepare this chapter in a structured order: understand the meaning of the chapter name, revise the standard form, learn the solution conditions, then practise elimination, substitution, and graphical questions one by one. That is the best way to build confidence for board exams.

Chapter Overview at a Glance

This quick table helps students revise the whole chapter faster.

Quick Concept Table

Meaning of the Chapter Name

Before solving questions, students should understand the full chapter name clearly.

Meaning of Pair

Pair means two equations are given together.

Meaning of Linear Equation

A linear equation is an equation in which the highest power of the variable is 1.

Meaning of Two Variables

The equation contains two variables, usually x and y.

So, a pair of linear equations in two variables means two equations in x and y, each having degree 1.

Standard Form of a Linear Equation in Two Variables

A linear equation in two variables is written in the form:

ax + by + c = 0

Where:

• a, b, and c are real numbers
• a and b are not both zero

Example of Standard Form

2x + 3y – 7 = 0

This is a linear equation in two variables.

What Is the Solution of a Pair of Linear Equations

The solution is the value of x and y that satisfies both equations at the same time.

Meaning in Simple Words

If one ordered pair works in both equations, then that ordered pair is the solution of the pair.

Graphical Meaning of the Solution

When the two equations are represented by straight lines on a graph, the solution depends on how the lines behave.

Case 1: Intersecting Lines

If the two lines intersect at one point, then the pair has a unique solution.

Meaning

The coordinates of the intersection point are the only solution.

Case 2: Parallel Lines

If the two lines are parallel, they never meet.

Meaning

There is no solution.

Case 3: Coincident Lines

If the two lines lie exactly on each other, they are called coincident lines.

Meaning

There are infinitely many solutions.

Graphical Meaning Summary Table

Conditions for the Nature of Solutions

This is one of the most important theory areas in the chapter because students can identify the nature of solutions without solving the equations fully.

If the pair of linear equations is:

a₁x + b₁y + c₁ = 0

a₂x + b₂y + c₂ = 0

then the following conditions are used.

Case 1: Unique Solution Condition

If:

a₁/a₂ ≠ b₁/b₂

then the pair has a unique solution.

Graphical Meaning

The two lines intersect.

Case 2: No Solution Condition

If:

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

then the pair has no solution.

Graphical Meaning

The two lines are parallel.

Case 3: Infinitely Many Solutions Condition

If:

a₁/a₂ = b₁/b₂ = c₁/c₂

then the pair has infinitely many solutions.

Graphical Meaning

The two lines are coincident.

Nature of Solutions Condition Table

Why These Conditions Are Important

These conditions are often asked directly in MCQs, short answers, and value-based questions where students must find the value of a constant such as k.

Methods of Solving a Pair of Linear Equations

There are three main methods students must know.

Method Summary Table

Graphical Method

In this method, the two equations are represented on a graph and the intersection point is identified.

Substitution Method

In this method, one variable is written in terms of the other from one equation and substituted into the second equation.

Elimination Method

In this method, one variable is eliminated by making its coefficients equal and then adding or subtracting the equations.

Which Method Is Most Important for Boards

The graphical method and elimination method are especially important for board exams. Substitution is also useful, especially in word problems and when one equation is already easy to simplify.

Important Question 1: Solve by Elimination Method

Solve:

2x + 5y = -4

4x – 3y = 5

Given

• 2x + 5y = -4
• 4x – 3y = 5

Step 1: Make One Variable’s Coefficients Equal

To eliminate y, multiply the first equation by 3 and the second equation by 5.

Then we get:

6x + 15y = -12

20x – 15y = 25

Step 2: Add Both Equations

26x = 13

x = 13/26 = 1/2

Step 3: Put x in Any Original Equation

Using:

2x + 5y = -4

2(1/2) + 5y = -4

1 + 5y = -4

5y = -5

y = -1

Answer

x = 1/2 and y = -1

Important Question 2: Solve and Then Find a Required Expression

If:

2x + y = 23

4x – y = 19

find:

(a) 5y – 2x

(b) y/x – 2

Given

• 2x + y = 23
• 4x – y = 19

Step 1: Add the Equations

2x + y = 23

4x – y = 19

Adding:

6x = 42

x = 7

Step 2: Find y

2(7) + y = 23

14 + y = 23

y = 9

Step 3: Find Part (a)

5y – 2x = 5(9) – 2(7)

= 45 – 14 = 31

Step 4: Find Part (b)

y/x – 2 = 9/7 – 2

= 9/7 – 14/7

= -5/7

Answer

• 5y – 2x = 31
• y/x – 2 = -5/7

Important Question 3: Solve Graphically and Find Area

Solve graphically:

3x – 4y + 3 = 0

3x + 4y – 21 = 0

Also find the coordinates of the vertices of the triangular region formed by these lines and the x-axis, and calculate its area.

Given

• 3x – 4y + 3 = 0
• 3x + 4y – 21 = 0

Step 1: Find Points for the First Line

For:

3x – 4y + 3 = 0

If y = 0:

3x + 3 = 0

x = -1

So one point is (-1, 0)

If y = 3:

3x – 12 + 3 = 0

3x = 9

x = 3

So another point is (3, 3)

Point Table for the First Line

Step 2: Find Points for the Second Line

For:

3x + 4y – 21 = 0

If y = 0:

3x – 21 = 0

x = 7

So one point is (7, 0)

If x = 3:

9 + 4y – 21 = 0

4y = 12

y = 3

So another point is (3, 3)

Point Table for the Second Line

Step 3: Identify the Intersection Point

The common point is:

(3, 3)

So the graphical solution is:

x = 3, y = 3

Step 4: Identify the Triangle Vertices

The triangle formed with the x-axis has vertices:

(-1, 0), (7, 0), (3, 3)

Step 5: Find the Area

Base = distance from -1 to 7 = 8 units

Height = 3 units

Area = 1/2 × 8 × 3 = 12 square units

Answer

• graphical solution = (3, 3)
• triangle vertices = (-1, 0), (7, 0), (3, 3)
• area = 12 square units

Important Question 4: Draw Graph and Find Where the Lines Meet the X-Axis

Draw the graphs of:

x + 2y = 5

2x – 3y = -4

Also find the points where the lines meet the x-axis.

Given

• x + 2y = 5
• 2x – 3y = -4

Step 1: First Equation Points

For x + 2y = 5

If y = 0, then x = 5

So one point is (5, 0)

If y = 1, then x = 3

So another point is (3, 1)

If y = 2, then x = 1

So another point is (1, 2)

Point Table for the First Line

Step 2: Second Equation Points

For 2x – 3y = -4

If y = 0, then 2x = -4, so x = -2

So one point is (-2, 0)

If y = 1, then 2x – 3 = -4, so 2x = -1, x = -1/2

So another point is (-1/2, 1)

If y = 2, then 2x – 6 = -4, so 2x = 2, x = 1

So another point is (1, 2)

Point Table for the Second Line

Step 3: Identify the Common Point

The common point is:

(1, 2)

Step 4: Find Where the Lines Meet the X-Axis

The first line meets the x-axis at:

(5, 0)

The second line meets the x-axis at:

(-2, 0)

Answer

• solution = (1, 2)
• x-axis intercept points = (5, 0) and (-2, 0)

Important Question 5: Age-Based Word Problem

Three years ago, Rashmi was three times as old as Nazma. Ten years later, Rashmi will be twice as old as Nazma. Find their present ages.

Given

• Three years ago, Rashmi was three times Nazma’s age
• Ten years later, Rashmi will be twice Nazma’s age

Step 1: Let Present Ages Be

Let Rashmi’s present age be x years.

Let Nazma’s present age be y years.

Step 2: Form First Equation

Three years ago:

x – 3 = 3(y – 3)

x – 3 = 3y – 9

x = 3y – 6

Step 3: Form Second Equation

Ten years later:

x + 10 = 2(y + 10)

x + 10 = 2y + 20

x = 2y + 10

Step 4: Equate Both Expressions of x

3y – 6 = 2y + 10

y = 16

Step 5: Find x

x = 2(16) + 10 = 42

Answer

• Rashmi’s present age = 42 years
• Nazma’s present age = 16 years

Important Question 6: Fraction Problem

If 1 is added to the numerator and 1 is subtracted from the denominator of a fraction, it becomes 1. If 1 is added only to the denominator, it becomes 1/2. Find the fraction.

Given

• (x + 1)/(y – 1) = 1
• x/(y + 1) = 1/2

Step 1: Let the Fraction Be x/y

Take the fraction as x/y.

Step 2: Use First Condition

(x + 1)/(y – 1) = 1

x + 1 = y – 1

y = x + 2

Step 3: Use Second Condition

x/(y + 1) = 1/2

2x = y + 1

Step 4: Substitute y = x + 2

2x = x + 2 + 1

2x = x + 3

x = 3

Then:

y = 3 + 2 = 5

Answer

The fraction is 3/5.

Important Question 7: Find k for No Solution or Infinite Solutions

Find the value of k for the pair:

2x + ky = 6

4x + 6y = 12

Given

• 2x + ky = 6
• 4x + 6y = 12

Step 1: Compare Ratios

Here:

a₁/a₂ = 2/4 = 1/2

b₁/b₂ = k/6

c₁/c₂ = 6/12 = 1/2

Step 2: Check Infinite Solution Condition

For infinitely many solutions:

a₁/a₂ = b₁/b₂ = c₁/c₂

So:

k/6 = 1/2

2k = 6

k = 3

Step 3: Interpret Correctly

When k = 3, all three ratios become equal.

So the pair has infinitely many solutions, not no solution.

Step 4: Understand the No-Solution Reminder

For no solution, the condition must be:

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

That is not possible here unless the constant term changes.

Answer

For k = 3, the pair has infinitely many solutions.

Important Question 8: Book Rental Case-Based Question

A shopkeeper charges a fixed rent for the first 2 days and an additional amount for each day after that. Amrita paid 22 rupees for keeping a book for 6 days. Radhika paid 16 rupees for keeping a book for 4 days. Find the fixed charge and the additional daily charge.

Given

• Fixed charge for first 2 days = x rupees
• Additional charge per extra day = y rupees

Step 1: Form the Equation for Amrita

Amrita kept the book for 6 days, which means 4 extra days after the first 2 days.

So:

x + 4y = 22

Step 2: Form the Equation for Radhika

Radhika kept the book for 4 days, which means 2 extra days after the first 2 days.

So:

x + 2y = 16

Step 3: Subtract the Equations

(x + 4y) – (x + 2y) = 22 – 16

2y = 6

y = 3

Step 4: Find x

x + 2(3) = 16

x + 6 = 16

x = 10

Answer

• fixed charge for first 2 days = 10 rupees
• additional charge per day = 3 rupees

Important Question 9: Reduction to a Linear Pair

Solve:

2/x + 3/y = 13

5/x – 4/y = -2

Given

• 2/x + 3/y = 13
• 5/x – 4/y = -2

Step 1: Use Substitution of Variables

Let:

1/x = m

1/y = n

Then the equations become:

2m + 3n = 13

5m – 4n = -2

Step 2: Eliminate n

Multiply the first equation by 4:

8m + 12n = 52

Multiply the second equation by 3:

15m – 12n = -6

Add:

23m = 46

m = 2

Step 3: Find n

2(2) + 3n = 13

4 + 3n = 13

3n = 9

n = 3

Step 4: Convert Back to x and y

1/x = 2

x = 1/2

1/y = 3

y = 1/3

Answer

x = 1/2 and y = 1/3

Important Question 10: Simple Value-Based Substitution

If (3, 2) is a solution of:

a²x + 2y – 20 = 0

find the value of a.

Given

• (x, y) = (3, 2)
• a²x + 2y – 20 = 0

Step 1: Substitute x = 3 and y = 2

a²(3) + 2(2) – 20 = 0

3a² + 4 – 20 = 0

3a² – 16 = 0

Step 2: Solve for a²

3a² = 16

a² = 16/3

Step 3: Find a

a = ±4/√3

Answer

a = ±4/√3

Common Board Question Patterns from This Chapter

This chapter usually repeats certain standard question styles.

Common Mistakes Students Make in Pair of Linear Equations

This chapter looks easy, but students often lose marks through avoidable mistakes.

Common Mistakes Table

Quick Revision Sheet for Pair of Linear Equations

This section is useful before board exams.

Quick Revision Table

Best Study Strategy for Board Exams

This chapter becomes much easier when revised in the right order.

Step-by-Step Revision Table

Practice Questions

This section helps students prepare through standard board-style patterns.

Important Practice Questions

1. Solve:

3x + 2y = 11

5x – y = 7

1. Solve graphically:

x + y = 4

x – y = 2

1. Find the value of k for which the equations have infinitely many solutions:

2x + ky = 6

4x + 6y = 12

1. If a fraction becomes 1 when 1 is added to the numerator and subtracted from the denominator, and becomes 1/2 when 1 is added to the denominator, find the fraction.
2. A person invested money in two schemes at different interest rates. Form equations and find the invested amounts.

FAQs

Q1. What is a pair of linear equations in two variables?

It means two linear equations in x and y, each having degree 1.

Q2. What is the standard form of a linear equation in two variables?

The standard form is ax + by + c = 0.

Q3. What does a unique solution mean graphically?

It means the two lines intersect at exactly one point.

Q4. What does no solution mean graphically?

It means the two lines are parallel and never meet.

Q5. What does infinitely many solutions mean graphically?

It means the two lines are coincident and lie on each other.

Q6. Which methods are used to solve pair of linear equations?

The main methods are graphical method, substitution method, and elimination method.

Q7. Which method is easiest for board exams?

Elimination method is often the fastest in many standard questions, while substitution is very useful in word problems.

Q8. How can I score well in this chapter?

You can score well by mastering the three solution conditions, practising elimination and graphical questions, and solving word problems carefully.

Conclusion

Pair of Linear Equations in Two Variables is one of the most useful and scoring chapters in Class 10 Maths because it connects algebraic solving, graphical interpretation, and practical word problems in one flow. Once students understand the standard form, the nature-of-solution conditions, and the main solving methods, the chapter becomes much easier.

The best way to prepare this chapter is to practise one method at a time, revise graph-based interpretation carefully, and solve board-style word problems with patience. At Deeksha Vedantu, we always remind students that this chapter becomes easy when the logic of the equations is understood clearly and the solution method is chosen wisely.