Introduction
Linear Equations in Two Variables is a fundamental chapter in Class 9 mathematics that bridges the concepts you learned about linear equations in one variable with the coordinate geometry you studied.
What are Linear Equations in Two Variables?
A linear equation in two variables is an equation that can be expressed in the form:
ax + by + c = 0
where:
- a, b, and c are real numbers
- a and b are not both zero
- x and y are variables
This means at least one of a or b must be non-zero for the equation to be truly linear in two variables.
Examples of Linear Equations in Two Variables:
- x + y = 5
- 2x – 3y = 7
- 0.5x + 2y = 10
- x = 3y + 2
- 2x + y = 0
Examples That Are NOT Linear Equations in Two Variables:
- x² + y = 5 (contains squared term)
- xy + 3 = 0 (product of variables)
- x + 2/y = 5 (y in denominator)
Converting Equations to Standard Form
The standard form of a linear equation in two variables is ax + by + c = 0. Many equations can be written in different ways, but they can always be converted to this standard form.
Converting Equations to Standard Form
EQUATION CONVERSION PROCESS
Example 1: 2x + 3y = 4.37
↓
Move all terms to left side
↓
2x + 3y - 4.37 = 0
↓
Standard Form: ax + by + c = 0
Where: a = 2, b = 3, c = -4.37Example Conversions:
Example 1: 2x + 3y = 4.37
Standard form: 2x + 3y – 4.37 = 0
Here: a = 2, b = 3, c = -4.37
Example 2: x + 4 = 3y
Rearranging: x – 3y + 4 = 0
Here: a = 1, b = -3, c = 4
Example 3: 4 = 5x + 3y
Rearranging: -5x – 3y + 4 = 0 or 5x + 3y – 4 = 0
Here: a = 5, b = 3, c = -4
Example 4: 2x – y
Standard form: 2x – y = 0
Here: a = 2, b = -1, c = 0
Important Note:
Equations like 2x = 3 (containing only one variable) can also be expressed as linear equations in two variables:
2x = 3 becomes 2x + 0y – 3 = 0
where a = 2, b = 0, c = -3
Solutions of Linear Equations in Two Variables
What is a Solution?
A solution of a linear equation in two variables is a pair of values (one for x, one for y) that satisfies the equation. These solutions are written as ordered pairs (x, y).
Number of Solutions
- Linear equations in one variable: one solution
Solution Representation on Number Line (One Variable)
LINEAR EQUATION IN ONE VARIABLE
Equation: 2x - 5 = 0
Solution: x = 5/2 = 2.5
Number Line Representation:
←─────●─────●─────●─────●─────→
0 1 2.5 3 4
●
(unique point)
- Linear equations in two variables: infinitely many solutions
- We can choose any value for one variable
- Always get a corresponding value for the other variable
Imp Fact: Infinitely Many Solutions
Unlike linear equations in one variable which have a unique solution, a linear equation in two variables has infinitely many solutions. This is because we can choose any value for one variable and solve for the other.
Finding Solutions – Method 1: By Substitution
Understanding Solutions in Two Variables
LINEAR EQUATION IN TWO VARIABLES
Equation: x + 2y = 6
Substitution Method:
When x = 0: When y = 0: When x = 2:
0 + 2y = 6 x + 0 = 6 2 + 2y = 6
y = 3 x = 6 2y = 4
Solution: (0, 3) Solution: (6, 0) Solution: (2, 2)
When y = 1: When x = 4:
x + 2 = 6 4 + 2y = 6
x = 4 2y = 2
Solution: (4, 1) Solution: (4, 1)
Multiple solutions exist for ANY value chosen!ARTESIAN PLANE WITH SOLUTIONS
Y-axis
↑
4 |
|
3 ●(0,3)
|
2 | ●(2,2)
|
1 |
|
0 |────-------------●(6,0)─────→ X-axis
| 1 2 3 4 5 6
-1|
Each • represents a solution (x, y)
All solutions lie on a straight lineExample: Find four solutions of x + 2y = 6
Solution 1: Put x = 2
- 2 + 2y = 6
- 2y = 4
- y = 2
- Solution: (2, 2)
Solution 2: Put x = 0
- 0 + 2y = 6
- 2y = 6
- y = 3
- Solution: (0, 3)
Solution 3: Put y = 0
- x + 0 = 6
- x = 6
- Solution: (6, 0)
Solution 4: Put y = 1
- x + 2(1) = 6
- x + 2 = 6
- x = 4
- Solution: (4, 1)
So four solutions are: (2, 2), (0, 3), (6, 0), (4, 1)
Finding Solutions – Method 2: Easy Approach
The quickest way to find solutions is to:
- Put x = 0 and find the corresponding y value
- Put y = 0 and find the corresponding x value
This automatically gives two solutions.
EXERCISE 4.1 – Complete Solutions
Question 1
Problem: The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement. Take the cost of a notebook to be x and that of a pen to be y.
Answer:
Let x = cost of a notebook Let y = cost of a pen
According to the problem: The cost of a notebook is twice the cost of a pen
Therefore: x = 2y
In standard form: x – 2y = 0
This is the required linear equation in two variables.
Question 2
Problem: Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b, and c in each case.
i) 2x + 3y = 9.35
Standard form: 2x + 3y – 9.35 = 0
Values: a = 2, b = 3, c = -9.35
ii) x + y = 10
Standard form: x + y – 10 = 0
Values: a = 1, b = 1, c = -10
iii) -2x + 3y = 6
Standard form: -2x + 3y – 6 = 0
Values: a = -2, b = 3, c = -6
iv) x – 3y = 5
Standard form: x – 3y – 5 = 0
Values: a = 1, b = -3, c = -5
v) 2x – 5y = 0
Standard form: 2x – 5y + 0 = 0
Values: a = 2, b = -5, c = 0
vi) 3x + 2 = 0
Standard form: 3x + 0y + 2 = 0
Values: a = 3, b = 0, c = 2
vii) y – 2 = 0
Standard form: 0x + y – 2 = 0
Values: a = 0, b = 1, c = -2
viii) 5 = 2x – 4.3y
Rearranging: 2x – 4.3y – 5 = 0
Values: a = 2, b = -4.3, c = -5
EXERCISE 4.2 – Complete Solutions
Question 1
Problem: Which one of the following options is true, and why? y = 3x + 5 has:
i) a unique solution ii) only two solutions iii) infinitely many solutions
Answer: Option (iii) – infinitely many solutions is correct.
Explanation:
Because for every value of x we choose, we can find a corresponding value of y using the equation y = 3x + 5.
Examples:
- When x = 0, y = 5 → (0, 5)
- When x = 1, y = 8 → (1, 8)
- When x = 2, y = 11 → (2, 11)
- When x = -1, y = 2 → (-1, 2)
Since x can take any real value, we get infinitely many solutions.
Question 2
Problem: Write four solutions for each of the following equations.
i) 2x + y = 7
Solution 1: Put x = 0
- 2(0) + y = 7
- y = 7
- Solution: (0, 7)
Solution 2: Put x = 1
- 2(1) + y = 7
- 2 + y = 7
- y = 5
- Solution: (1, 5)
Solution 3: Put y = 1
- 2x + 1 = 7
- 2x = 6
- x = 3
- Solution: (3, 1)
Solution 4: Put y = -1
- 2x – 1 = 7
- 2x = 8
- x = 4
- Solution: (4, -1)
Four solutions: (0, 7), (1, 5), (3, 1), (4, -1)
ii) x + y = 9
Solution 1: Put x = 0
- 0 + y = 9
- y = 9
- Solution: (0, 9)
Solution 2: Put x = 3
- 3 + y = 9
- y = 6
- Solution: (3, 6)
Solution 3: Put y = 4
- x + 4 = 9
- x = 5
- Solution: (5, 4)
Solution 4: Put y = 0
- x + 0 = 9
- x = 9
- Solution: (9, 0)
Four solutions: (0, 9), (3, 6), (5, 4), (9, 0)
iii) x + 4y = 3
Wait, this seems incomplete in your question. Let me provide solutions assuming it should be x + 4y = 3:
Solution 1: Put x = 0
- 0 + 4y = 3
- y = 3/4 = 0.75
- Solution: (0, 0.75)
Solution 2: Put y = 0
- x + 0 = 3
- x = 3
- Solution: (3, 0)
Solution 3: Put x = -1
- -1 + 4y = 3
- 4y = 4
- y = 1
- Solution: (-1, 1)
Solution 4: Put x = 7
- 7 + 4y = 3
- 4y = -4
- y = -1
- Solution: (7, -1)
Four solutions: (0, 0.75), (3, 0), (-1, 1), (7, -1)
Question 3
Problem: Check which of the following are solutions of the equation x – 2y = 4 and which are not.
i) (0, 2)
Substituting x = 0, y = 2: 0 – 2(2) = 0 – 4 = -4 ≠ 4
NOT a solution
ii) (2, 0)
Substituting x = 2, y = 0: 2 – 2(0) = 2 – 0 = 2 ≠ 4
NOT a solution
iii) (4, 0)
Substituting x = 4, y = 0: 4 – 2(0) = 4 – 0 = 4 = 4
IS a solution
iv) (2, -1)
Substituting x = 2, y = -1: 2 – 2(-1) = 2 + 2 = 4 = 4
IS a solution
v) (1, -1.5)
Substituting x = 1, y = -1.5: 1 – 2(-1.5) = 1 + 3 = 4 = 4
IS a solution
Question 4
Problem: Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Answer:
Since (2, 1) is a solution of 2x + 3y = k, the point must satisfy the equation.
Substituting x = 2 and y = 1: 2(2) + 3(1) = k 4 + 3 = k k = 7
Therefore, k = 7
We can verify: 2x + 3y = 7 with x = 2, y = 1 gives 2(2) + 3(1) = 4 + 3 = 7 ✓
Imp Concepts to Remember
1. Standard Form
Any linear equation in two variables can be written as ax + by + c = 0
2. Two-Variable Requirement
For an equation to be linear in two variables, both a and b cannot be zero. At least one of them must be non-zero.
3. Solution Representation
Solutions are written as ordered pairs: (x, y)
4. Infinitely Many Solutions
A linear equation in two variables always has infinitely many solutions. We can choose any value for one variable and find the corresponding value of the other.
5. Finding Solutions Easily
The quickest method:
- Put x = 0 to find one solution
- Put y = 0 to find another solution
6. Verification
To check if a point is a solution, substitute the x and y values into the equation. If both sides are equal, it is a solution.
Why Linear Equations in Two Variables Matter
- Foundation for Advanced Mathematics: These equations form the basis for understanding systems of linear equations and linear algebra
- Real-World Applications: Many real-world situations involve two related quantities that can be expressed as linear equations
- Graphical Representation: Each solution can be plotted on the Cartesian plane, creating a line (which is why they’re called “linear”)
- Problem Solving: From cricket scores to mixture problems, linear equations in two variables help solve practical problems
Common Mistakes to Avoid
- Forgetting to rearrange: Not writing the equation in the form ax + by + c = 0 before identifying a, b, and c
- Confusing ordered pairs: Remember that (2, 3) is different from (3, 2)
- Missing solutions: Don’t assume there are only two solutions. There are infinitely many
- Calculation errors: Always verify your solution by substituting back into the original equation
- Forgetting conditions: Remember that a and b cannot both be zero
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