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By Rohit RP
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Updated on 5 Aug 2026, 14:32 IST
NCERT Solutions for Class 10 Maths Chapter 3 provide complete, step-by-step answers to all questions in Exercises 3.1, 3.2 and 3.3. Students can use the exercise links below to find NCERT solutions based on the graphical method, substitution method, elimination method and real-life word problems.
Each answer explains how to choose the variables, form the two equations, select a suitable solving method and verify the final values. The chapter guide also covers consistent and inconsistent equations, intersecting, parallel and coincident lines, coefficient-ratio conditions and common mistakes students make while solving word problems.
The current NCERT Chapter 3 contains three exercises covering graphical and algebraic methods of solving a pair of linear equations.
| Exercise | Main topic | What the exercise covers |
| Exercise 3.1 | Graphical method and nature of solutions | Forming equations, drawing graphs, consistency and intersecting, parallel or coincident lines |
| Exercise 3.2 | Substitution method | Solving pairs algebraically and applying equations to ages, prices, fractions and fares |
| Exercise 3.3 | Elimination and mixed methods | Solving equations by elimination and substitution, followed by application-based problems |
These solutions follow the NCERT Class 10 Mathematics Syllabus marked Reprint 2026–27. In this edition, Chapter 3 contains Exercises 3.1, 3.2 and 3.3, followed by the chapter summary.
Some older textbooks and websites also show Exercises 3.4 to 3.7, the cross-multiplication method and a separate exercise on equations reducible to linear form. Those older-edition topics should be kept in a separate section so that they are not confused with the current exercise sequence.
A pair of linear equations consists of two first-degree equations containing the same two variables. The solution is the value of x and y that makes both equations true at the same time.
The general form is:
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
Here:
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
For example:
x + y = 5
x - y = 1
The values x = 3 and y = 2 satisfy both equations:
3 + 2 = 5
3 - 2 = 1
Therefore, (3, 2) is the solution of the pair.
Yes. The current chapter begins with the graphical method and uses graphs to explain unique, no and infinitely many solutions. Exercise 3.1 includes graph-based questions.
Cross-multiplication is not listed among the main algebraic methods in the current chapter summary. The current summary names substitution and elimination as the algebraic methods studied.
The summary does briefly mention that some situations can first appear non-linear and then be changed into a pair of linear equations. However, the current chapter does not provide a separate exercise for this material.
The current NCERT chapter has three exercises, each focusing on a different stage of learning.
| Exercise | Main focus | Skills practised |
| Exercise 3.1 | Graphical method and consistency | Forming equations, plotting lines and identifying the type of solution |
| Exercise 3.2 | Substitution method | Solving equations and modelling word problems |
| Exercise 3.3 | Elimination and mixed methods | Solving systems and application-based problems |
A pair of linear equations in two variables consists of two first-degree equations containing the same two variables.
The general form is:
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
Here:
Example:
2x + 3y = 12
x - y = 1
The values of x and y that satisfy both equations form the solution of the pair.
An ordered pair is written as (x, y). The first value is the value of x, and the second is the value of y.
For example, check whether (3, 2) satisfies:
x + y = 5
2x - y = 4
Substitute x = 3 and y = 2:
3 + 2 = 5
6 - 2 = 4
Both equations are true. Therefore, (3, 2) is a common solution.
Every ordered pair that satisfies a linear equation can be represented as a point on a coordinate plane. When these points are joined, they form a straight line.
For example:
x + y = 4
Some solutions are:
| x | y |
| 0 | 4 |
| 1 | 3 |
| 2 | 2 |
| 4 | 0 |
All these points lie on the same straight line.
A pair of linear equations can have one solution, no solution or infinitely many solutions.
A pair has a unique solution when the two lines intersect at exactly one point.
Example:
x + y = 5
x - y = 1
The lines intersect at (3, 2), so the system has one solution.
A pair has no solution when the two lines are parallel.
Example:
x + y = 4
2x + 2y = 10
The left side of the second equation is twice the left side of the first, but 10 is not twice 4. The lines have the same slope but different positions.
A pair has infinitely many solutions when both equations represent the same line.
Example:
x + y = 4
2x + 2y = 8
The second equation is exactly twice the first. Every solution of the first equation also satisfies the second.
For:
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
use the following conditions:
| Coefficient condition | Number of solutions | Type of pair | Graph |
| a₁/a₂ ≠ b₁/b₂ | One | Consistent and independent | Intersecting lines |
| a₁/a₂ = b₁/b₂ ≠ c₁/c₂ | None | Inconsistent | Parallel lines |
| a₁/a₂ = b₁/b₂ = c₁/c₂ | Infinitely many | Dependent and consistent | Coincident lines |
The result 0 = 0 means that one equation is equivalent to the other. The system has infinitely many solutions.
The result 0 = k, where k is not zero, is a contradiction. The system has no solution.
Examples:
0 = 0 → infinitely many solutions
0 = 7 → no solution
Important: 0 = 0 does not mean x = 0 and y = 0.
The best method is the one that gives the solution clearly with the least difficult arithmetic.
| Method | Best used when | Advantage | Limitation |
| Graphical method | The question asks for a graph or line relationship | Shows the solution visually | The intersection may be approximate |
| Substitution method | One variable is already isolated or has coefficient 1 | Direct and easy to follow | Fractions may appear |
| Elimination method | Coefficients are equal, opposite or easy to match | Often the quickest algebraic method | Sign errors are common |
No. Correct graphical, substitution and elimination methods must describe the same common solution.
A graph may give a slightly approximate value when the intersection lies between grid points. Algebraic methods usually give the exact value.
Exercise 3.1 focuses on forming equations, drawing graphs and identifying whether two lines intersect, remain parallel or coincide.
The question descriptions below are paraphrased. Students should read the exact wording in their NCERT textbook before writing the final answer.
A quiz has 10 students, and the number of girls is four more than the number of boys.
Let:
Number of boys = x
Number of girls = y
First condition:
x + y = 10
Second condition:
y = x + 4
Therefore:
y - x = 4
The equations are:
x + y = 10
-y? No. Write the second equation correctly as:
-x + y = 4
Coordinate table for x + y = 10:
| x | y |
| 0 | 10 |
| 3 | 7 |
| 10 | 0 |
Coordinate table for -x + y = 4:
| x | y |
| 0 | 4 |
| 3 | 7 |
| -4 | 0 |
The two lines meet at (3, 7).
Answer:
Number of boys = 3
Number of girls = 7
Verification:
3 + 7 = 10
7 - 3 = 4
Common mistake: Writing x - y = 4 instead of y - x = 4 reverses the meaning.
Let:
Cost of one pencil = ₹x
Cost of one pen = ₹y
The two purchase conditions give:
5x + 7y = 50
7x + 5y = 46
Useful points for 5x + 7y = 50:
| x | y |
| 10 | 0 |
| 3 | 5 |
| -4 | 10 |
Useful points for 7x + 5y = 46:
| x | y |
| 3 | 5 |
| 8 | -2 |
| -2 | 12 |
The lines intersect at (3, 5).
Answer:
Cost of one pencil = ₹3
Cost of one pen = ₹5
Verification:
5(3) + 7(5) = 15 + 35 = 50
7(3) + 5(5) = 21 + 25 = 46
The lines can be classified by comparing a₁/a₂, b₁/b₂ and c₁/c₂.
Equations:
5x - 4y + 8 = 0
7x + 6y - 9 = 0
a₁/a₂ = 5/7
b₁/b₂ = -4/6 = -2/3
Since:
5/7 ≠ -2/3
The lines intersect at one point.
Answer: Intersecting lines with a unique solution.
Equations:
9x + 3y + 12 = 0
18x + 6y + 24 = 0
a₁/a₂ = 9/18 = 1/2
b₁/b₂ = 3/6 = 1/2
c₁/c₂ = 12/24 = 1/2
All three ratios are equal.
Answer: Coincident lines with infinitely many solutions.
Equations:
6x - 3y + 10 = 0
2x - y + 9 = 0
a₁/a₂ = 6/2 = 3
b₁/b₂ = -3/-1 = 3
c₁/c₂ = 10/9
Since:
a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Answer: Parallel lines with no solution.
This question asks whether each pair is consistent or inconsistent.
3x + 2y = 5
2x - 3y = 7
a₁/a₂ = 3/2
b₁/b₂ = 2/-3
The ratios are unequal.
Answer: Consistent, with one unique solution.
2x - 3y = 8
4x - 6y = 9
Write in standard form:
2x - 3y - 8 = 0
4x - 6y - 9 = 0
a₁/a₂ = 1/2
b₁/b₂ = 1/2
c₁/c₂ = 8/9
Answer: Inconsistent, with no solution.
(3/2)x + (5/3)y = 7
9x - 10y = 14
a₁/a₂ = (3/2)/9 = 1/6
b₁/b₂ = (5/3)/(-10) = -1/6
The ratios are unequal.
Answer: Consistent, with one unique solution.
5x - 3y = 11
-10x + 6y = -22
The second equation is -2 times the first equation.
Answer: Dependent and consistent, with infinitely many solutions.
(4/3)x + 2y = 8
2x + 3y = 12
a₁/a₂ = (4/3)/2 = 2/3
b₁/b₂ = 2/3
c₁/c₂ = 8/12 = 2/3
Answer: Dependent and consistent, with infinitely many solutions.
x + y = 5
2x + 2y = 10
The second equation is twice the first.
Answer: Consistent and dependent. The lines coincide and have infinitely many solutions.
x - y = 8
3x - 3y = 16
For the coefficients:
a₁/a₂ = 1/3
b₁/b₂ = 1/3
For the constants in standard form:
c₁/c₂ = -8/-16 = 1/2
Answer: Inconsistent. The lines are parallel and have no solution.
2x + y - 6 = 0
4x - 2y - 4 = 0
Rewrite:
2x + y = 6
4x - 2y = 4
Multiply the first equation by 2:
4x + 2y = 12
Add it to the second equation:
4x + 2y = 12
4x - 2y = 4
8x = 16
x = 2
Substitute x = 2 into:
2x + y = 6
4 + y = 6
y = 2
Answer: Consistent, with the unique solution (2, 2).
2x - 2y - 2 = 0
4x - 4y - 5 = 0
a₁/a₂ = 1/2
b₁/b₂ = 1/2
c₁/c₂ = 2/5
Answer: Inconsistent. The lines are parallel and have no solution.
Let:
Width of the garden = x m
Length of the garden = y m
Half the perimeter is 36 m.
Therefore:
x + y = 36
The length is 4 m more than the width:
y = x + 4
Substitute:
x + x + 4 = 36
2x = 32
x = 16
Then:
y = 16 + 4 = 20
Answer:
Width = 16 m
Length = 20 m
Verification:
16 + 20 = 36
The given equation is:
2x + 3y - 8 = 0
Many correct answers are possible.
x + y - 1 = 0
Here:
2/1 ≠ 3/1
The lines intersect.
4x + 6y - 9 = 0
Here:
2/4 = 3/6
But:
-8/-9 ≠ 2/4
The lines are parallel.
4x + 6y - 16 = 0
This is twice the given equation.
The lines coincide.
The equations are:
x - y + 1 = 0
3x + 2y - 12 = 0
The first line meets the x-axis where y = 0:
x + 1 = 0
x = -1
First x-axis vertex = (-1, 0)
The second line meets the x-axis where y = 0:
3x - 12 = 0
x = 4
Second x-axis vertex = (4, 0)
Now solve the two lines together.
From:
x - y + 1 = 0
y = x + 1
Substitute into:
3x + 2y = 12
3x + 2(x + 1) = 12
5x + 2 = 12
5x = 10
x = 2
y = 3
Third vertex = (2, 3)
Answer:
The triangle has vertices:
(-1, 0), (4, 0) and (2, 3)
Optional check:
Base on the x-axis = 4 - (-1) = 5 units
Height = 3 units
Area = 1/2 × 5 × 3 = 7.5 square units
The graphical method finds the common solution by plotting both equations as straight lines and locating their point of intersection.
Choose values of x that make y easy to calculate.
For example:
2x + y = 6
If x = 0:
y = 6
If x = 2:
y = 2
Two points are (0, 6) and (2, 2).
Use the same clear scale on both axes unless a different scale is necessary and explicitly stated.
Example:
1 square = 1 unit on the x-axis
1 square = 1 unit on the y-axis
Two different points determine a straight line. A third point can be used as a check.
Label each line with its equation or a clear name such as L₁ and L₂.
A graph is drawn using a selected scale and pencil thickness. If the intersection has fractional or decimal coordinates, reading the exact value may be difficult.
For exact values, use substitution or elimination after estimating the answer from the graph.
Exercise 3.2 focuses on the substitution method and on forming equations from real-life situations.
Equations:
x + y = 14
x - y = 4
From the first equation:
x = 14 - y
Substitute into the second:
14 - y - y = 4
14 - 2y = 4
-2y = -10
y = 5
Then:
x = 14 - 5 = 9
Answer:
x = 9, y = 5
Verification:
9 + 5 = 14
9 - 5 = 4
Equations:
s - t = 3
s/3 + t/2 = 6
From the first equation:
s = t + 3
Substitute:
(t + 3)/3 + t/2 = 6
Multiply the complete equation by 6:
2(t + 3) + 3t = 36
2t + 6 + 3t = 36
5t = 30
t = 6
Then:
s = 6 + 3 = 9
Answer:
s = 9, t = 6
Equations:
3x - y = 3
9x - 3y = 9
The second equation is three times the first.
From the first equation:
y = 3x - 3
Substituting into the second gives a true identity:
9x - 3(3x - 3) = 9
9x - 9x + 9 = 9
9 = 9
Answer: Infinitely many solutions.
One possible way to write the solution set is:
x = k
y = 3k - 3
where k can be any real number.
Equations:
0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
Multiply both equations by 10:
2x + 3y = 13
4x + 5y = 23
From the first equation:
x = (13 - 3y)/2
Substitute into the second:
4[(13 - 3y)/2] + 5y = 23
2(13 - 3y) + 5y = 23
26 - 6y + 5y = 23
-y = -3
y = 3
Then:
x = (13 - 9)/2 = 2
Answer:
x = 2, y = 3
Equations:
√2x + √3y = 0
√3x - √8y = 0
Remember:
√8 = 2√2
From the first equation:
√2x = -√3y
x = -(√3/√2)y
Substitute into the second:
√3[-(√3/√2)y] - 2√2y = 0
-(3/√2)y - 2√2y = 0
Since:
2√2 = 4/√2
We get:
-(3/√2)y - (4/√2)y = 0
-(7/√2)y = 0
y = 0
Then:
x = 0
Answer:
x = 0, y = 0
Equations:
3x/2 - 5y/3 = -2
x/3 + y/2 = 13/6
Multiply the first equation by 6:
9x - 10y = -12
Multiply the second equation by 6:
2x + 3y = 13
From the second equation:
x = (13 - 3y)/2
Substitute into the first:
9[(13 - 3y)/2] - 10y = -12
Multiply by 2:
117 - 27y - 20y = -24
117 - 47y = -24
-47y = -141
y = 3
Then:
x = (13 - 9)/2 = 2
Answer:
x = 2, y = 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 into:
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 and m = -1
Let the larger number be x and the smaller number be y.
The difference is 26:
x - y = 26
One number is three times the other:
x = 3y
Substitute:
3y - y = 26
2y = 26
y = 13
x = 39
Answer: The numbers are 39 and 13.
Let:
Larger angle = x°
Smaller angle = y°
Supplementary angles add to 180°:
x + y = 180
The larger exceeds the smaller by 18°:
x - y = 18
From:
x = y + 18
Substitute:
y + 18 + y = 180
2y = 162
y = 81
x = 99
Answer: The angles are 99° and 81°.
Let:
Cost of one bat = ₹x
Cost of one ball = ₹y
The two purchases give:
7x + 6y = 3800
3x + 5y = 1750
From the second equation:
3x = 1750 - 5y
x = (1750 - 5y)/3
Substitute into the first:
7(1750 - 5y)/3 + 6y = 3800
Multiply by 3:
12250 - 35y + 18y = 11400
-17y = -850
y = 50
Then:
3x + 5(50) = 1750
3x + 250 = 1750
3x = 1500
x = 500
Answer:
Cost of one bat = ₹500
Cost of one ball = ₹50
Let:
Fixed taxi charge = ₹x
Charge per kilometre = ₹y
For 10 km:
x + 10y = 105
For 15 km:
x + 15y = 155
From the first equation:
x = 105 - 10y
Substitute into the second:
105 - 10y + 15y = 155
5y = 50
y = 10
Then:
x = 105 - 100 = 5
Fixed charge = ₹5
Charge per kilometre = ₹10
For 25 km:
Total charge = x + 25y
Total charge = 5 + 25(10)
Total charge = ₹255
Let:
Numerator = x
Denominator = y
The first fraction condition gives:
(x + 2)/(y + 2) = 9/11
Cross-multiply only to simplify this fractional equation:
11(x + 2) = 9(y + 2)
11x + 22 = 9y + 18
11x - 9y = -4
The second condition gives:
(x + 3)/(y + 3) = 5/6
6(x + 3) = 5(y + 3)
6x + 18 = 5y + 15
6x - 5y = -3
From:
6x - 5y = -3
x = (5y - 3)/6
Substitute into:
11x - 9y = -4
11(5y - 3)/6 - 9y = -4
Multiply by 6:
55y - 33 - 54y = -24
y - 33 = -24
y = 9
Then:
x = (45 - 3)/6 = 7
Answer: The fraction is 7/9.
Let:
Jacob’s present age = x years
His son’s present age = y years
Five years later:
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
From the first equation:
x = 3y + 10
Substitute into the second:
3y + 10 - 7y = -30
-4y = -40
y = 10
Then:
x = 3(10) + 10 = 40
Answer:
Jacob’s present age = 40 years
His son’s present age = 10 years
The substitution method solves a pair of equations by expressing one variable in terms of the other and placing that expression into the second equation.
If two expressions are equal, one can replace the other without changing the truth of an equation.
Suppose:
x = 7 - y
If another equation contains x, replacing x with 7 - y gives an equivalent condition.
Prefer the variable that:
Solve:
x + 2y = 11
3x - y = 13
From the first equation:
x = 11 - 2y
Substitute into the second:
3(11 - 2y) - y = 13
33 - 6y - y = 13
-7y = -20
y = 20/7
Then:
x = 11 - 40/7
x = 77/7 - 40/7
x = 37/7
Answer:
x = 37/7, y = 20/7
Exercise 3.3 uses elimination and substitution to solve pairs of equations and application-based problems.
Equations:
x + y = 5
2x - 3y = 4
Elimination method:
Multiply the first equation by 2:
2x + 2y = 10
Subtract the second equation:
(2x + 2y) - (2x - 3y) = 10 - 4
5y = 6
y = 6/5
Then:
x = 5 - 6/5
x = 25/5 - 6/5
x = 19/5
Substitution method:
From x + y = 5:
x = 5 - y
Substitute:
2(5 - y) - 3y = 4
10 - 5y = 4
y = 6/5
x = 19/5
Answer:
x = 19/5, y = 6/5
Equations:
3x + 4y = 10
2x - 2y = 2
Elimination method:
Multiply the first equation by 2:
6x + 8y = 20
Multiply the second equation by 3:
6x - 6y = 6
Subtract:
14y = 14
y = 1
Substitute:
2x - 2(1) = 2
2x = 4
x = 2
Substitution method:
From:
2x - 2y = 2
x - y = 1
x = y + 1
Substitute:
3(y + 1) + 4y = 10
7y + 3 = 10
y = 1
x = 2
Answer:
x = 2, y = 1
Equations:
3x - 5y = 4
9x - 2y = 7
Elimination method:
Multiply the first equation by 3:
9x - 15y = 12
Subtract the second equation:
(9x - 15y) - (9x - 2y) = 12 - 7
-13y = 5
y = -5/13
Substitute into:
3x - 5y = 4
3x - 5(-5/13) = 4
3x + 25/13 = 52/13
3x = 27/13
x = 9/13
Substitution method:
From:
3x - 5y = 4
x = (4 + 5y)/3
Substitute into:
9x - 2y = 7
9(4 + 5y)/3 - 2y = 7
3(4 + 5y) - 2y = 7
12 + 15y - 2y = 7
13y = -5
y = -5/13
x = 9/13
Answer:
x = 9/13, y = -5/13
Equations:
x/2 + 2y/3 = -1
x - y/3 = 3
Elimination method:
Multiply the first equation by 6:
3x + 4y = -6
Multiply the second equation by 3:
3x - y = 9
Subtract:
5y = -15
y = -3
Substitute:
x - (-3)/3 = 3
x + 1 = 3
x = 2
Substitution method:
From:
x - y/3 = 3
x = 3 + y/3
Substitute into:
x/2 + 2y/3 = -1
(3 + y/3)/2 + 2y/3 = -1
Multiply by 6:
9 + y + 4y = -6
5y = -15
y = -3
x = 2
Answer:
x = 2, y = -3
Let:
Numerator = x
Denominator = y
The first condition gives:
(x + 1)/(y - 1) = 1
x + 1 = y - 1
x - y = -2
The second condition gives:
x/(y + 1) = 1/2
2x = y + 1
2x - y = 1
Subtract the first equation from the second:
(2x - y) - (x - y) = 1 - (-2)
x = 3
Then:
3 - y = -2
y = 5
Answer: The fraction is 3/5.
Let:
Nuri’s present age = x years
Sonu’s present age = y years
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
Subtract the first equation from the second:
(x - 2y) - (x - 3y) = 10 - (-10)
y = 20
Then:
x - 2(20) = 10
x = 50
Answer:
Nuri is 50 years old.
Sonu is 20 years old.
Let:
Tens digit = x
Units digit = y
The number is:
10x + y
The reversed number is:
10y + x
The digit sum is 9:
x + y = 9
The second condition gives:
9(10x + y) = 2(10y + x)
90x + 9y = 20y + 2x
88x = 11y
8x = y
Substitute into:
x + y = 9
x + 8x = 9
9x = 9
x = 1
y = 8
Answer: The number is 18.
Verification:
9 × 18 = 162
Reversed number = 81
2 × 81 = 162
Let:
Number of ₹50 notes = x
Number of ₹100 notes = y
Total number of notes:
x + y = 25
Total value:
50x + 100y = 2000
Divide the second equation by 50:
x + 2y = 40
Subtract:
(x + 2y) - (x + y) = 40 - 25
y = 15
Then:
x = 25 - 15 = 10
Answer:
₹50 notes = 10
₹100 notes = 15
Let:
Fixed charge for the first three days = ₹x
Charge for each additional day = ₹y
For seven days, there are four additional days:
x + 4y = 27
For five days, there are two additional days:
x + 2y = 21
Subtract:
2y = 6
y = 3
Then:
x + 2(3) = 21
x = 15
Answer:
Fixed charge = ₹15
Charge for each additional day = ₹3
The elimination method removes one variable by adding or subtracting suitably adjusted equations.
Adding equal quantities to equal quantities preserves equality.
When a multiple of one equation is added to or subtracted from another equation, every common solution of the original system still satisfies the new equation.
Add the equations when one variable has opposite coefficients.
Example:
3x + 2y = 11
5x - 2y = 13
Adding eliminates y:
8x = 24
x = 3
Subtract when one variable has equal coefficients with the same sign.
Example:
4x + 3y = 18
4x - y = 10
Subtract the second from the first:
4y = 8
y = 2
Clear denominators before eliminating.
Example:
x/2 + y/3 = 5
Multiply the complete equation by 6:
3x + 2y = 30
The safest way to solve a word problem is to define the unknowns first and translate one relationship at a time.
| Phrase | Algebraic meaning |
| A is 5 more than B | A = B + 5 |
| A is 5 less than B | A = B - 5 |
| A exceeds B by 5 | A - B = 5 |
| The difference is 5 | Larger - smaller = 5 |
| Twice A plus B | 2A + B |
| Five years hence | Present age + 5 |
| Five years ago | Present age - 5 |
| Fixed charge plus usage charge | Fixed charge + rate × usage |
| Two supplementary angles | A + B = 180° |
| Half the perimeter of a rectangle | Length + width |
Use present ages as the variables.
Example:
Present age of a parent = x
Present age of a child = y
“Five years ago, the parent was four times the child’s age” becomes:
x - 5 = 4(y - 5)
Always add or subtract the same number of years from both people’s present ages.
If:
Tens digit = x
Units digit = y
Then:
Original number = 10x + y
Reversed number = 10y + x
Digit restrictions:
If the digits “differ by 2,” the wording may require two cases:
x - y = 2
or:
y - x = 2
If x and y are prices per item, multiply each price by the number of items.
Example:
Three notebooks and two pens cost ₹110:
3x + 2y = 110
Many taxi, library, subscription and rental questions follow:
Total cost = fixed charge + rate × usage
Example:
A service charges ₹x as a fixed fee and ₹y per unit.
For 10 units costing ₹205:
x + 10y = 205
Let:
Numerator = x
Denominator = y
If 2 is added to both:
New fraction = (x + 2)/(y + 2)
Use brackets carefully.
Ask:
Consider:
x + y = 5
x - y = 1
The exact solution is x = 3 and y = 2.
For x + y = 5, use:
(0, 5) and (5, 0)
For x - y = 1, use:
(1, 0) and (3, 2)
The two lines intersect at:
(3, 2)
From:
x - y = 1
x = y + 1
Substitute into:
x + y = 5
y + 1 + y = 5
2y = 4
y = 2
x = 3
Add:
x + y = 5
x - y = 1
2x = 6
x = 3
Then:
3 + y = 5
y = 2
Students usually lose marks because of equation-formation or sign errors rather than because they do not know the method.
| Mistake | Why it is wrong | Correction |
| Checking only one equation | A common solution must satisfy both | Substitute into both equations |
| Treating 0 = 0 as x = 0, y = 0 | It is an identity | Interpret it as infinitely many solutions |
| Treating 0 = 5 as a solution | It is impossible | Conclude that there is no solution |
| Multiplying only one term | It changes the equation | Multiply every term |
| Reversing “5 less than” | The statement changes meaning | Translate it before solving |
| Writing a two-digit number as x + y | It ignores place value | Use 10x + y |
| Forgetting units | The result becomes incomplete | State rupees, years, metres or degrees |
| Rounding too early | It may change the final answer | Keep exact fractions until the end |
| Misreading a graph | Thick or inaccurate lines affect the point | Verify algebraically |
| Ignoring restrictions | The algebraic answer may be impossible | Check ages, digits, counts and denominators |
Competency-based questions test whether students can model, interpret and justify a situation rather than only repeat a procedure. CBSE publishes Grade 10 competency-focused Mathematics practice material that includes Pair of Linear Equations in Two Variables.
Plan A charges ₹100 per month plus ₹10 per GB.
Plan B charges ₹160 per month plus ₹4 per GB.
Answer:
Plan A:
y = 100 + 10x
Plan B:
y = 160 + 4x
Set the costs equal:
100 + 10x = 160 + 4x
6x = 60
x = 10
At 10 GB, both plans cost:
₹200
Below 10 GB, Plan A is cheaper.
Above 10 GB, Plan B is cheaper.
A school sells student tickets for ₹40 and adult tickets for ₹70. It sells 120 tickets and collects ₹6,300.
Let:
Student tickets = x
Adult tickets = y
Equations:
x + y = 120
40x + 70y = 6300
Divide the second by 10:
4x + 7y = 630
Multiply the first by 4:
4x + 4y = 480
Subtract:
3y = 150
y = 50
x = 70
Answer:
Student tickets = 70
Adult tickets = 50
A student solves:
2x + y = 8
2x - 3y = 4
The student subtracts and writes:
4y = 4
What is the error?
Correct subtraction:
(2x + y) - (2x - 3y) = 8 - 4
2x + y - 2x + 3y = 4
4y = 4
There is no error.
Therefore:
y = 1
Then:
2x + 1 = 8
x = 7/2
This type of question checks whether the student understands the sign change when subtracting brackets.
Without solving, classify:
3x + 6y = 12
x + 2y = 5
Compare:
a₁/a₂ = 3/1 = 3
b₁/b₂ = 6/2 = 3
c₁/c₂ = -12/-5 = 12/5
The first two ratios are equal, but the third is different.
Answer: No solution; the lines are parallel.
A student claims that (2, 3) solves:
x + y = 5
2x + y = 8
Check:
2 + 3 = 5
2(2) + 3 = 7, not 8
Answer: The ordered pair satisfies only the first equation, so it is not a solution of the pair.
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
| Condition | Result |
| a₁/a₂ ≠ b₁/b₂ | Unique solution |
| a₁/a₂ = b₁/b₂ ≠ c₁/c₂ | No solution |
| a₁/a₂ = b₁/b₂ = c₁/c₂ | Infinitely many solutions |
| Lines | Solution |
| Intersecting | One |
| Parallel | None |
| Coincident | Infinitely many |
Also Check: Class 10 Maths Formula Sheet | Printable Chapter 3 Pair of Linear Equations in Two Variables Notes
Older textbooks and websites may include additional exercises and methods. Keep these separate from the current Exercise 3.1–3.3 navigation.
Those pages may be based on an older textbook structure. Students should match online solutions with the exercise number and exact question in their own book.
For two equations:
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
an older formula is sometimes written as:
x/(b₁c₂ - b₂c₁) = y/(c₁a₂ - c₂a₁) = 1/(a₁b₂ - a₂b₁)
This formula should not replace conceptual understanding of substitution and elimination.
Use the method required by the current question or the method instructed by the teacher.
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It is a set of two first-degree equations containing the same two variables. A solution must satisfy both equations at the same time.
It can have one solution, no solution or infinitely many solutions. These cases correspond to intersecting, parallel and coincident lines.
A unique solution exists when a₁/a₂ ≠ b₁/b₂. The two lines intersect at one point.
There is no solution when a₁/a₂ = b₁/b₂ but c₁/c₂ is different. The lines are parallel.
There are infinitely many solutions when a₁/a₂ = b₁/b₂ = c₁/c₂. Both equations represent the same line.
A pair is consistent if it has at least one solution. It may have one solution or infinitely many solutions.
An inconsistent pair has no common solution. Its graph consists of two parallel lines.
It means the equations are dependent and represent the same condition. The pair has infinitely many solutions.
It is a false statement and shows that the equations contradict one another. The pair has no solution.
Substitution is usually easier when a variable is already isolated. Elimination is usually faster when coefficients are already equal, opposite or easy to match.
Choose the variable whose coefficients can be made equal with the smallest multipliers. This reduces arithmetic and sign errors.
Two different points are enough to draw a straight line. A third point can be used to check accuracy.
The intersection may fall between grid points, and the result depends on the scale and accuracy of the drawing. Algebraic methods provide exact values.
Substitute the values of x and y into both original equations. Both equations must become true.
Define the two unknowns, translate each condition separately and include units. Check that the equations describe the original situation before solving.
The tens digit x represents 10x, while the units digit y represents y. Therefore, the full number is 10x + y.
The current Reprint 2026–27 chapter contains Exercises 3.1, 3.2 and 3.3, followed by the summary.
Yes. Graphical representation and the nature of solutions are central parts of the current Exercise 3.1.
Yes. Systems of two linear equations are taught across many school boards, although exercise order, terminology and assessment style may differ.
Follow the method named in the question. When no method is specified, use a clear method that is part of the current course and show complete working.