Pair of Linear Equations in Two Variables is an important Class 10 Maths chapter because it teaches students how to solve two conditions together. The chapter is practical, scoring, and closely linked to real-life situations like age, numbers, marks, and arrangements.
A pair means two equations, linear means the highest power of the variable is 1, and two variables usually means x and y. So the chapter is about solving two linear equations together and finding a common solution.
Chapter Overview at a Glance
Quick Concept Table
| Topic | Main idea |
| Linear equation in two variables | Equation of the form ax + by + c = 0 |
| Pair of linear equations | Two linear equations taken together |
| Solution of the pair | Common values of x and y satisfying both equations |
| Graphical method | Solve by plotting both lines on a graph |
| Unique solution | Two lines intersect at one point |
| No solution | Two lines are parallel |
| Infinitely many solutions | Two lines coincide |
| Substitution method | Put one variable in terms of the other |
| Elimination method | Eliminate one variable by addition or subtraction |
Standard Form and Meaning of Solution
A linear equation in two variables is generally written as:
ax + by + c = 0
Here, a is the coefficient of x, b is the coefficient of y, and c is the constant term.
The solution of a pair of linear equations is the pair of values of x and y that satisfies both equations at the same time.
Graphical Method
In the graphical method, both equations are plotted as straight lines. The nature of the solution depends on how the lines behave.
Graphical Meaning Table
| Type of lines | Number of solutions | Nature |
| Intersecting lines | One | Unique solution |
| Parallel lines | None | No solution |
| Coincident lines | Infinitely many | Infinite solutions |
How to Solve by Graphical Method
To solve graphically, find two points for each equation, plot them on graph paper, draw both lines, and locate the point of intersection.
Solved Example 1: Graphical Method
Solve graphically:
x + 3y = 6
2x – 3y = 12
Given
Equation 1 is x + 3y = 6 and equation 2 is 2x – 3y = 12.
Step 1: Find Two Points for the First Line
For x + 3y = 6:
If x = 0, then y = 2.
If y = 0, then x = 6.
So the points are (0, 2) and (6, 0).
Step 2: Find Two Points for the Second Line
For 2x – 3y = 12:
If x = 0, then y = -4.
If y = 0, then x = 6.
So the points are (0, -4) and (6, 0).
Answer
The two lines intersect at (6, 0), so x = 6 and y = 0.
Number of Solutions of a Pair of Linear Equations
If the pair of equations is:
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
then the number of solutions can be found using coefficient ratios.
Number of Solutions Summary Table
| Algebraic condition | Graphical meaning |
| a₁/a₂ ≠ b₁/b₂ | Intersecting lines |
| a₁/a₂ = b₁/b₂ ≠ c₁/c₂ | Parallel lines |
| a₁/a₂ = b₁/b₂ = c₁/c₂ | Coincident lines |
If a₁/a₂ ≠ b₁/b₂, the pair has a unique solution. If a₁/a₂ = b₁/b₂ ≠ c₁/c₂, it has no solution. If all three ratios are equal, it has infinitely many solutions.
Solved Example 2: Find the Nature of Solution
Find out whether the following pair has one solution, no solution, or infinitely many solutions:
5x – 4y + 8 = 0
7x + 6y – 9 = 0
Given
a₁ = 5, b₁ = -4, c₁ = 8
a₂ = 7, b₂ = 6, c₂ = -9
Step 1: Compare Ratios
a₁/a₂ = 5/7
b₁/b₂ = -4/6 = -2/3
Answer
Since 5/7 ≠ -2/3, the lines intersect at one point. So the pair has a unique solution.
Solved Example 3: Find k for No Solution
Find the value of k for which the system has no solution:
kx + 3y = k
12x + ky = k – 3
Step 1: Use the no-solution condition
For no solution:
a₁/a₂ = b₁/b₂ ≠ c₁/c₂
So:
k/12 = 3/k
k² = 36
k = ±6
Step 2: Check the second ratio condition
On checking c₁/c₂, only k = -6 satisfies the no-solution condition.
Answer
k = -6
Algebraic Methods
There are two main algebraic methods used in this chapter.
Algebraic Methods at a Glance
The substitution method works by expressing one variable in terms of the other and substituting it into the second equation. The elimination method works by removing one variable through addition or subtraction.
Substitution Method
In this method, one variable is made the subject in one equation and then substituted into the other equation.
Solved Example 4: Substitution Method
Solve:
y = x + 6
2x – y = 4
Step 1
From y = x + 6, write:
x = y – 6
Step 2
Substitute into the second equation:
2(y – 6) – y = 4
y – 12 = 4
y = 16
Step 3
Find x:
x = 16 – 6 = 10
Answer
x = 10 and y = 16
Elimination Method
In this method, one variable is eliminated by making coefficients suitable and then adding or subtracting the equations.
Solved Example 5: Elimination Method
Find x + y if:
2x + y = 23
4x – y = 19
Step 1
Add both equations:
6x = 42
x = 7
Step 2
Substitute into 2x + y = 23:
14 + y = 23
y = 9
Answer
x + y = 16
Word Problems Based on Pair of Linear Equations
These questions become easy when variables are chosen clearly and equations are formed carefully.
Word Problem Solving Method
Read the question carefully, choose the variables, form equations from the conditions, solve using substitution or elimination, and write the final answer in words.
Solved Example 6: Two-Digit Number Problem
The sum of the digits of a two-digit number is 9. Nine times the number is twice the number obtained by reversing the digits. Find the number.
Given
Let the tens digit be x and the ones digit be y.
Original number = 10x + y
Reversed number = 10y + x
Step 1: Form equations
x + y = 9
9(10x + y) = 2(10y + x)
Step 2: Simplify
90x + 9y = 20y + 2x
88x – 11y = 0
8x – y = 0
Step 3: Solve
From x + y = 9, we get y = 9 – x.
Substitute:
8x – (9 – x) = 0
9x = 9
x = 1, y = 8
Answer
The number is 18.
Solved Example 7: Age Problem
The age of a father is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the children’s ages. Find the father’s present age.
Given
Let father’s age = F and children’s total age = C.
Step 1: Form equations
F = 2C
F + 20 = C + 40
So F = C + 20
Step 2: Solve
2C = C + 20
C = 20
F = 40
Answer
The father’s present age is 40 years.
Solved Example 8: Competitive Examination Problem
In a competitive examination, 1 mark is awarded for each correct answer and 1/2 mark is deducted for every wrong answer. A student answered 120 questions and got 90 marks. Find the number of correct answers.
Given
Let correct answers = C and wrong answers = W.
Step 1: Form equations
C + W = 120
C – (1/2)W = 90
Step 2: Remove the fraction
2C – W = 180
Step 3: Solve
From C + W = 120, we get W = 120 – C.
Substitute:
2C – (120 – C) = 180
3C = 300
C = 100
Answer
The number of correct answers is 100.
Case Study Pattern
Case-study questions usually involve a real situation where students must first form equations and then solve them.
Solved Example 9: Examination Hall Case Study
There are students in two halls A and B. To make the numbers equal, 10 students are transferred from A to B. But if 20 students are transferred from B to A, the number in hall A becomes double that in hall B. Find the number of students in each hall.
Given
Let the students in hall A be x and in hall B be y.
Step 1: Form the first equation
x – 10 = y + 10
x – y = 20
Step 2: Form the second equation
x + 20 = 2(y – 20)
x – 2y = -60
Step 3: Solve
From x – y = 20 and x – 2y = -60:
y = 80
x = 100
Answer
Hall A has 100 students and hall B has 80 students.
Important Board Patterns from This Chapter
This chapter commonly gives graphical questions, nature-of-solution questions, substitution and elimination problems, and word problems based on age, numbers, marks, or arrangements.
Common Mistakes Students Make in This Chapter
Common Mistakes Table
| Mistake | Correct idea |
| Forgetting standard form | Convert properly before comparing ratios |
| Wrong sign while shifting terms | Check every step carefully |
| Mixing up no solution and infinite solutions | Compare all three ratios carefully |
| Choosing too many variables in word problems | Use the simplest possible variables |
| Doing graphs roughly | Plot properly on graph paper |
| Skipping final statement | Always write the answer in words |
Quick Revision Sheet
Quick Revision Table
| Topic | Formula or result |
| Standard form | ax + by + c = 0 |
| Unique solution | a₁/a₂ ≠ b₁/b₂ |
| No solution | a₁/a₂ = b₁/b₂ ≠ c₁/c₂ |
| Infinite solutions | a₁/a₂ = b₁/b₂ = c₁/c₂ |
| Graph of unique solution | Intersecting lines |
| Graph of no solution | Parallel lines |
| Graph of infinite solutions | Coincident lines |
| Methods to solve | Graphical, substitution, elimination |
Best Strategy to Score Well in This Chapter
Revise the three solution conditions regularly, practise both substitution and elimination, solve at least one word problem from each major type, practise one graph question on graph paper, and track sign errors separately.
Practice Questions
- Solve graphically:
x + 2y = 4
2x – y = 1
- Find whether the pair has a unique solution, no solution, or infinitely many solutions:
3x + 2y – 5 = 0
6x + 4y – 10 = 0
- Solve by substitution method:
y = 2x + 1
3x + y = 10
- Solve by elimination method:
2x + 5y = -4
4x – 3y = 5
- The sum of the digits of a two-digit number is 11. When the digits are reversed, the new number is 27 more than the original number. Find the number.
FAQs
Q1. What is a pair of linear equations in two variables?
It means two linear equations in x and y that are solved together.
Q2. What is the graphical meaning of a unique solution?
A unique solution means the two lines intersect at exactly one point.
Q3. What is the condition for no solution?
a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Q4. What is the condition for infinitely many solutions?
a₁/a₂ = b₁/b₂ = c₁/c₂
Q5. Which algebraic methods are used in this chapter?
The two main methods are substitution method and elimination method.
Q6. How do I decide whether to use substitution or elimination?
Use substitution when one variable is easy to isolate. Use elimination when coefficients can be cancelled quickly.
Q7. Why do students lose marks in word problems?
Usually because of wrong variable choice, incorrect equation formation, or missing the final statement.
Q8. How can I score full marks in this chapter?
Revise the conditions, practise both solving methods, solve word problems carefully, and present each step clearly.
Conclusion
Pair of Linear Equations in Two Variables is one of the most practical and scoring chapters in Class 10 Maths because it combines graph understanding, algebraic methods, and real-life applications. Once students understand the number of solutions and practise substitution, elimination, and word problems properly, the chapter becomes much easier.
The best preparation strategy is to move in order: learn the graphical meaning, revise the three conditions, practise algebraic methods, and then solve word problems with care.
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