Pair of Linear Equations in Two Variables Class 10 Notes PDF

The Pair of Linear Equations in Two Variables Class 10 Notes PDF explains how to form, graph and solve two linear equations using substitution and elimination. These notes also cover consistency conditions, unique and non-existent solutions, word problems, solved examples, MCQs, common mistakes and current CBSE syllabus guidance.

The official CBSE Class 10 Mathematics curriculum for 2026–27 includes graphical solutions, consistency and inconsistency, algebraic conditions for the number of solutions, substitution, elimination and simple situational problems.

Also Check: 

Complete Class 10 Maths Notes

CBSE Class 10 Maths Syllabus

Download Pair of Linear Equations Class 10 Notes PDF

The downloadable notes should combine the complete chapter explanation, formulas, graphs, examples and examination questions in one printable resource.

The current CBSE question-paper design gives Standard Mathematics a larger proportion of applying, analysing and evaluating questions than Basic Mathematics, so the two courses require different depths of practice.

Pair of Linear Equations in Two Variables Class 10 Notes PDF: Quick Revision

A pair of linear equations consists of two first-degree equations involving the same two variables, and its solution must satisfy both equations simultaneously.

General form

A linear equation in two variables can be written as:

ax + by + c = 0

Here, a and b must not both be zero.

A pair of linear equations can be written as:

a₁x + b₁y + c₁ = 0

a₂x + b₂y + c₂ = 0

Meaning of a solution

An ordered pair (x, y) is a solution when it satisfies both equations.

For example, consider:

x + y = 5

x − y = 1

Substitute x = 3 and y = 2:

3 + 2 = 5

3 − 2 = 1

Both statements are true, so:

(x, y) = (3, 2)

Three possible solution types

Coefficient conditions

For:

a₁x + b₁y + c₁ = 0

a₂x + b₂y + c₂ = 0

When any denominator is zero, do not divide by it. Compare cross-products instead.

For example:

a₁b₂ and a₂b₁

Main methods of solution

The official 2026–27 CBSE syllabus specifically names graphical, substitution and elimination methods.

What Is a Pair of Linear Equations in Two Variables?

A pair of linear equations represents two straight-line relationships involving the same two unknown quantities.

Consider:

2x + 3y = 12

x − y = 1

The first equation represents one straight line, and the second represents another straight line. The common point of the two lines, where it exists, gives the values of x and y that satisfy both equations.

Linear and non-linear equations

Important terms

Variable: An unknown quantity represented by a letter, usually x or y.

Coefficient: The number multiplying a variable.

Constant: A number without a variable.

Ordered pair: A solution written as (x, y).

Simultaneous equations: Another name for equations that must be satisfied at the same time.

Common solution: An ordered pair that satisfies both equations.

Example: identify the coefficients

Consider:

3x − 5y + 7 = 0

• Coefficient of x: 3
• Coefficient of y: −5
• Constant term: 7

How to Check the Solution of a Pair of Linear Equations

A proposed solution is verified by substituting its x- and y-values into both equations.

Check whether (2, 3) is a solution of:

2x + y = 7

x + 2y = 8

Equation 1

2x + y = 7

Substitute x = 2 and y = 3:

2(2) + 3 = 7

4 + 3 = 7

7 = 7

Equation 2

x + 2y = 8

Substitute x = 2 and y = 3:

2 + 2(3) = 8

2 + 6 = 8

8 = 8

Both equations are satisfied.

Therefore:

(x, y) = (2, 3)

Verification checklist

1. Substitute the x-value into both equations.
2. Substitute the y-value into both equations.
3. Simplify the left side.
4. Compare it with the right side.
5. Reject the ordered pair if even one equation is not satisfied.

CBSE’s competency-based item bank includes questions that ask students to determine whether an ordered pair satisfies a pair of equations, showing that verification is an assessed skill rather than an optional step.

Graphical Method of Solving Pair of Linear Equations

The graphical solution is the point where the graphs of the two linear equations intersect.

Steps of the graphical method

1. Select at least two convenient x-values for the first equation.
2. Calculate the corresponding y-values.
3. Plot the points and join them to form a straight line.
4. Repeat the process for the second equation.
5. Observe whether the two lines intersect, remain parallel or coincide.
6. Read the point of intersection as the solution.

Solved example

Solve graphically:

x + y = 5

x − y = 1

Step 1: Find points for x + y = 5

Rearrange:

y = 5 − x

Step 2: Find points for x − y = 1

Rearrange:

y = x − 1

Both lines pass through:

(3, 2)

Therefore:

x = 3

y = 2

Limitations of the graphical method

• A graph may give only an approximate answer.
• Fractional intersections can be difficult to read exactly.
• An inaccurate scale can change the apparent answer.
• Algebraic methods are usually more efficient when exact values are required.

The current syllabus nevertheless requires students to plot pairs of equations and interpret solutions graphically, including consistency and inconsistency.

Number of Solutions and Consistency Conditions

The number of solutions depends on whether the two lines intersect, remain parallel or represent the same line.

One unique solution

Condition:

a₁/a₂ ≠ b₁/b₂

The lines have different slopes and intersect at one point.

Example:

x + y = 5

x − y = 1

Here:

a₁/a₂ = 1/1 = 1

b₁/b₂ = 1/(−1) = −1

Since:

1 ≠ −1

The equations have one unique solution.

No solution

Condition:

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

The lines have the same direction but different positions, so they are parallel.

Example:

2x + 4y − 6 = 0

x + 2y − 5 = 0

Calculate:

a₁/a₂ = 2/1 = 2

b₁/b₂ = 4/2 = 2

c₁/c₂ = (−6)/(−5) = 6/5

Therefore:

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

The pair has no solution.

Infinitely many solutions

Condition:

a₁/a₂ = b₁/b₂ = c₁/c₂

The equations represent the same line.

Example:

2x + 4y − 6 = 0

x + 2y − 3 = 0

Calculate:

a₁/a₂ = 2/1 = 2

b₁/b₂ = 4/2 = 2

c₁/c₂ = (−6)/(−3) = 2

All three ratios are equal.

Therefore, the pair has infinitely many solutions.

Consistent and inconsistent pairs

Consistent pair: A pair with at least one solution.

This includes:

• One unique solution
• Infinitely many solutions

Inconsistent pair: A pair with no solution.

What elimination produces in each case

This graph–ratio–algebra connection helps students understand the conditions instead of memorising them as isolated rules.

Substitution Method for Pair of Linear Equations

The substitution method solves a pair by expressing one variable in terms of the other and replacing it in the second equation.

Steps

1. Select one equation.
2. Isolate x or y.
3. Substitute that expression into the other equation.
4. Solve the resulting equation in one variable.
5. Substitute the result back to find the second variable.
6. Verify the solution in both original equations.

Solved example

Solve:

x + y = 7

x − y = 1

From the first equation:

x + y = 7

x = 7 − y

Substitute x = 7 − y into the second equation:

x − y = 1

(7 − y) − y = 1

7 − 2y = 1

−2y = −6

y = 3

Substitute y = 3 into:

x = 7 − y

x = 7 − 3

x = 4

Therefore:

x = 4

y = 3

Verification

First equation:

4 + 3 = 7

Second equation:

4 − 3 = 1

Both equations are satisfied.

When substitution is efficient

Use substitution when:

• One variable already has coefficient 1 or −1.
• One equation is already written as x = … or y = ….
• Isolating a variable will not create difficult fractions.
• The question specifically asks for substitution.

Common substitution mistake

Incorrect:

x = 7 − y

Substituting only 7 and forgetting −y.

Correct:

Replace the entire x with:

(7 − y)

Elimination Method for Pair of Linear Equations

The elimination method solves a pair by making one variable’s coefficients equal and then adding or subtracting the equations.

Steps

1. Arrange both equations in the same variable order.
2. Select the variable to eliminate.
3. Multiply one or both equations if necessary.
4. Add or subtract the complete equations.
5. Solve for the remaining variable.
6. Substitute the value into either original equation.
7. Verify the final ordered pair.

Solved example

Solve:

2x + 3y = 12

3x − 2y = 5

To eliminate y, make its coefficients equal.

Multiply the first equation by 2:

2(2x + 3y = 12)

4x + 6y = 24

Multiply the second equation by 3:

3(3x − 2y = 5)

9x − 6y = 15

Add the equations:

4x + 6y = 24

9x − 6y = 15

13x = 39

x = 3

Substitute x = 3 into:

2x + 3y = 12

2(3) + 3y = 12

6 + 3y = 12

3y = 6

y = 2

Therefore:

x = 3

y = 2

Verification

First equation:

2(3) + 3(2) = 12

6 + 6 = 12

Second equation:

3(3) − 2(2) = 5

9 − 4 = 5

The solution is correct.

When elimination is efficient

Use elimination when:

• A variable already has equal coefficients.
• Coefficients can be matched using small multipliers.
• Both equations are in standard form.
• Substitution would introduce complicated fractions.

Common elimination mistake

When multiplying an equation, multiply every term, including the constant.

For example:

2x + 3y = 12

Multiplying by 2 gives:

4x + 6y = 24

It does not give:

4x + 3y = 12

Which Method Should You Use?

The best method is the one that produces the fewest and simplest calculations while preserving accuracy.

Quick method chooser

1. Is one equation already solved for x or y?
   Use substitution.
2. Are any coefficients equal or opposites?
   Use elimination.
3. Can coefficients be matched with small numbers?
   Use elimination.
4. Does the question ask for graphical interpretation?
   Use the graphical method.
5. Are both methods equally short?
   Choose either and verify the answer.

Substitution and elimination must produce the same solution when the calculations are correct because both methods solve the same pair of equations.

How to Form Equations from Word Problems

A word problem is solved by defining variables, translating each relationship into an equation and then solving the resulting pair.

CBSE’s competency-based materials include tasks that assess forming equations, interpreting costs, evaluating claims and reading graphs, so students should practise modelling situations rather than only solving ready-made equations.

Five-step word-problem process

1. Define the two unknown quantities.
2. Write their units.
3. Translate the first relationship into an equation.
4. Translate the second relationship into another equation.
5. Solve and check whether the answer makes sense in the original situation.

Phrase-to-equation translator

Example 1: prices of two items

Two pens and three notebooks cost ₹90. Three pens and two notebooks cost ₹85. Find the price of each item.

Let:

x = price of one pen

y = price of one notebook

First purchase:

2x + 3y = 90

Second purchase:

3x + 2y = 85

Multiply the first equation by 2:

4x + 6y = 180

Multiply the second equation by 3:

9x + 6y = 255

Subtract:

5x = 75

x = 15

Substitute x = 15 into:

2x + 3y = 90

2(15) + 3y = 90

30 + 3y = 90

3y = 60

y = 20

Therefore:

Price of one pen = ₹15

Price of one notebook = ₹20

Example 2: two-digit number

The sum of the digits of a two-digit number is 9. The number is 27 greater than the number formed by reversing its digits. Find the number.

Let:

x = tens digit

y = units digit

Sum of digits:

x + y = 9

Original number:

10x + y

Reversed number:

10y + x

The original is 27 greater:

10x + y = 10y + x + 27

Simplify:

9x − 9y = 27

x − y = 3

Now solve:

x + y = 9

x − y = 3

Add:

2x = 12

x = 6

Substitute into:

x + y = 9

6 + y = 9

y = 3

Therefore, the number is:

63

Check:

63 − 36 = 27

Example 3: rectangle dimensions

The perimeter of a rectangle is 50 cm. Its length is 5 cm more than its width. Find its dimensions.

Let:

x = length

y = width

Perimeter:

2x + 2y = 50

Simplify:

x + y = 25

Length relationship:

x = y + 5

Substitute:

y + 5 + y = 25

2y = 20

y = 10

Then:

x = 10 + 5

x = 15

Therefore:

Length = 15 cm

Width = 10 cm

Check:

2(15) + 2(10) = 50 cm

Parameter Questions Involving k

Parameter questions use the coefficient conditions to find values of k that produce one, no or infinitely many solutions.

Example: find k for no solution

Find k if the equations have no solution:

2x + 3y − 5 = 0

4x + ky − 12 = 0

For no solution:

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Here:

a₁/a₂ = 2/4 = 1/2

b₁/b₂ = 3/k

Set the first two ratios equal:

1/2 = 3/k

k = 6

Now compare constants:

c₁/c₂ = (−5)/(−12) = 5/12

Since:

1/2 ≠ 5/12

The equations have no solution when:

k = 6

Example: find k for a unique solution

For a unique solution:

a₁/a₂ ≠ b₁/b₂

Using the same pair:

2/4 ≠ 3/k

1/2 ≠ 3/k

Therefore:

k ≠ 6

Examination note

The 2025–26 Mathematics Standard sample paper included a parameter-based question asking when two lines were not parallel, showing how coefficient conditions may appear in objective questions.

Pair of Linear Equations Formula and Condition Sheet

The essential formulas identify the form of the equations, the number of solutions and the method used to solve them.

Cross-product alternative

When ratios contain a zero denominator, compare products.

A unique solution exists when:

a₁b₂ ≠ a₂b₁

This avoids dividing by zero and gives the same conclusion as comparing the first two ratios.

Important Questions with Answers

Important questions should cover solution conditions, algebraic methods, graphs, parameters and formation of equations from situations.

Question 1

Find the solution of:

x + y = 8

x − y = 2

Answer:

Add the equations:

2x = 10

x = 5

Substitute into x + y = 8:

5 + y = 8

y = 3

Solution:

(5, 3)

Question 2

Determine the number of solutions:

3x + 6y − 9 = 0

x + 2y − 5 = 0

Answer:

a₁/a₂ = 3/1 = 3

b₁/b₂ = 6/2 = 3

c₁/c₂ = (−9)/(−5) = 9/5

Therefore:

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

The pair has no solution.

Question 3

Solve by substitution:

2x + y = 9

x − y = 3

Answer:

From x − y = 3:

x = y + 3

Substitute:

2(y + 3) + y = 9

3y + 6 = 9

3y = 3

y = 1

x = 4

Solution:

(4, 1)

Question 4

Find k if the pair has infinitely many solutions:

2x + 4y − 6 = 0

kx + 8y − 12 = 0

Answer:

For infinitely many solutions:

2/k = 4/8 = (−6)/(−12)

4/8 = 1/2

Therefore:

2/k = 1/2

k = 4

Question 5

The sum of two numbers is 30, and their difference is 8. Find the numbers.

Answer:

Let the numbers be x and y.

x + y = 30

x − y = 8

Add:

2x = 38

x = 19

Then:

y = 11

The numbers are 19 and 11.

Pair of Linear Equations MCQs with Answers

MCQs test whether students can quickly identify forms, solution conditions and suitable solving methods.

Q1. Which is a linear equation in two variables?

A. x² + y = 5
B. xy = 8
C. 2x + 3y = 7
D. 1/x + y = 4

Answer: C

If two lines intersect at one point, the pair has:

A. No solution
B. One unique solution
C. Infinitely many solutions
D. Two solutions

Answer: B

If:

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

the pair has:

A. One solution
B. No solution
C. Infinitely many solutions
D. Exactly two solutions

Answer: B

Which method is usually easiest when one equation is x = 2y + 3?

A. Substitution
B. Graphical method only
C. Elimination only
D. Factorisation

Answer: A

The solution of x + y = 6 and x − y = 2 is:

A. (2, 4)
B. (3, 3)
C. (4, 2)
D. (6, 2)

Answer: C

Coincident lines represent:

A. No solution
B. One solution
C. Infinitely many solutions
D. An invalid equation

Answer: C

Assertion–Reason

Assertion: A consistent pair may have infinitely many solutions.

Reason: Coincident lines have every point in common.

Answer: Both statements are true, and the reason correctly explains the assertion.

Also Check: Pair of Linear Equations MCQs 

Common Mistakes in Pair of Linear Equations

The most frequent errors involve signs, incomplete multiplication, reversed word-problem relationships and incorrect ratio conditions.

Final checking routine

1. Arrange both equations in the same form.
2. Check all positive and negative signs.
3. Confirm that every term was multiplied.
4. Include units in word problems.
5. Substitute the final answer into both equations.
6. Check whether the result is realistic.

Current CBSE Syllabus and Exam Guidance

For 2026–27, students should prioritise graphs, consistency conditions, substitution, elimination and simple situational problems.

The official curriculum lists:

• Pair of linear equations in two variables
• Graphical method
• Consistency and inconsistency
• Algebraic conditions for the number of solutions
• Substitution
• Elimination
• Simple situational problems

Is cross-multiplication in the current syllabus?

The official 2026–27 CBSE syllabus does not specifically list cross-multiplication among the required methods; it names substitution and elimination. Older textbooks and online notes may still include cross-multiplication, so it should be labelled as a legacy or additional method rather than presented as compulsory syllabus content.

Is there a fixed chapter weightage?

CBSE assigns 20 marks to the complete Algebra unit, which includes Polynomials, Pair of Linear Equations, Quadratic Equations and Arithmetic Progressions. The curriculum does not guarantee a fixed number of marks for this chapter alone.

Basic and Standard Mathematics preparation

The 2026–27 design allocates approximately 24% of Standard Mathematics marks to application and 22% to higher-order skills, compared with 15% and 10% respectively in Basic Mathematics.

Question formats to practise

The latest official Class 10 sample-paper page available during this research is for 2025–26. Its Standard Mathematics paper includes MCQs, assertion–reason, very short answers, short answers, long answers and case-study questions. 

Also Check: Class 10 Maths sample papers