Class 9 · Mathematics · Chapter 4
Linear Equations in Two Variables
Theory · Solved Examples · Exercises · MCQs
📋 Contents
- Linear Equation in One Variable
- Linear Equation in Two Variables
- Graphical Solution of a Linear Equation
- Different Forms of a Line
- Solution: One Variable
- Solution: Two Variables
- Lines Parallel to Axes
- Key Summary & Special Cases
- Let's Try — Practice Problems
- Exercise 1 (MCQs)
- Exercise 2 (Descriptive)
- Exercise 3 (Advanced)
1. Linear Equation in One Variable
An equation of the form ax + b = 0 where a and b are real numbers and x is a variable, is called a linear equation in one variable.
Terminology: Here a is called the coefficient of x and b is called the constant term.
Examples: 3x + 5 = 0, 7x – 2 = 0, etc.
Examples: 3x + 5 = 0, 7x – 2 = 0, etc.
The solution is obtained by isolating x:
ax + b = 0 ⟹ x = –b/a
Example
Solve: (3x/2) – (x/3) = 8 [image contained step-by-step working]
Multiply both sides by the LCM of 2 and 3, which is 6:
6 × (3x/2) – 6 × (x/3) = 6 × 8
9x – 2x = 48
7x = 48 ⟹ x = 48/7
Verification: substituting x = 48/7 back gives LHS = RHS ✓
Example
Solve: (x/2) – (x/3) = 3
LCM of 2 and 3 is 6. Multiply both sides by 6:
3x – 2x = 18
x = 18 Ans.
2. Linear Equation in Two Variables
An equation of the form ax + by + c = 0 where a, b, c are real numbers, a ≠ 0, b ≠ 0, and x, y are variables, is called a linear equation in two variables.
ax + by + c = 0 (a ≠ 0, b ≠ 0)
Any pair of values of x and y which satisfies the equation ax + by + c = 0 is called a solution of the equation.
Important: If a = 0 or b = 0, the equation reduces to a linear equation in one variable. A linear equation in two variables has infinitely many solutions.
Example 1
Prove that x = 3, y = 2 is a solution of 3x – 2y = 5.
LHS = 3x – 2y = 3 × 3 – 2 × 2 = 9 – 4 = 5 = RHS
Since x = 3, y = 2 satisfies 3x – 2y = 5, it is a solution. ✓
Example 2
Prove that x = 1, y = 1 as well as x = 2, y = 5 is a solution of 4x – y – 3 = 0.
Given equation: 4x – y – 3 = 0 …(i)
Check (x = 1, y = 1): LHS = 4(1) – 1 – 3 = 4 – 4 = 0 = RHS ✓
Check (x = 2, y = 5): LHS = 4(2) – 5 – 3 = 8 – 8 = 0 = RHS ✓
Both pairs satisfy the equation — both are solutions. This also illustrates that a linear equation in two variables has infinitely many solutions.
3. Graphical Solution of a Linear Equation
(A) Graph of ax = b or ay = b (One Variable)
- IObtain the linear equation.
- IIIf the form is ax = b (a ≠ 0), plot the point (b/a, 0) and one more point (b/a, k) where k is any real number. If the form is ay = b, plot (0, b/a) and (k, b/a) for any real k.
- IIIJoin the plotted points to obtain the required line.
Note: If the equation is in the form ax = b, the graph is a vertical line parallel to the Y-axis. If it is ay = b, the graph is a horizontal line parallel to the X-axis.
Example
Draw the graph of: (i) 2x + 5 = 0 (ii) 3y – 15 = 0
(i) Graph of 2x + 5 = 0:
Simplify: 2x = –5 ⟹ x = –5/2 = –2.5
Plot point A₁(–2.5, 0) and any other point A₂(–2.5, 3). Join them to get a vertical line at x = –2.5 (parallel to Y-axis).
(ii) Graph of 3y – 15 = 0:
Simplify: 3y = 15 ⟹ y = 5
Plot point B₁(0, 5) and B₂(3, 5). Join them to get a horizontal line at y = 5 (parallel to X-axis).
(B) Graph of ax + by + c = 0 (Two Variables)
- IObtain the linear equation ax + by + c = 0.
- IIExpress y in terms of x: y = (–ax – c)/b or x in terms of y.
- IIIChoose two or three values of x (or y), and calculate the corresponding values of y (or x). Record as coordinate pairs.
- IVPlot the coordinate pairs on graph paper.
- VJoin the points. The line obtained is the graph of ax + by + c = 0.
Remember: A point which lies on the line is a solution of the equation. A point not on the line is not a solution.
Example 3 — Finding coordinates from graph
Draw the graph of x – 2y = 3. From the graph, find the coordinates when (i) x = –5 and (ii) y = 0.
Solve for y: 2y = x – 3 ⟹ y = (x – 3)/2
Compute values:
| x | 0 | 3 | –2 |
|---|---|---|---|
| y | –3/2 | 0 | –5/2 |
Plot these points and join them. From the graph:
(i) When x = –5: y = (–5 – 3)/2 = –4 ⟹ Point (–5, –4)
(ii) When y = 0: x = 3 ⟹ Point (3, 0)
Example 4 — Intersection of two lines
Draw the graphs of x + y = 4 and 2x – y = 2. Find their point of intersection.
Line (i): x + y = 4 ⟹ y = 4 – x
| x | 0 | 2 | 4 |
|---|---|---|---|
| y | 4 | 2 | 0 |
Line (ii): 2x – y = 2 ⟹ y = 2x – 2
| x | 1 | 0 | 3 |
|---|---|---|---|
| y | 0 | –2 | 4 |
Plot both lines on the same graph. The point of intersection P has coordinates x = 2, y = 2.
4. Different Forms of a Line
If a line makes an angle θ with the positive direction of the X-axis, then tan θ is called the slope of the line, denoted by m: m = tan θ.
(i) Slope–Intercept Form:
where m is the slope and c is the Y-intercept (the value where the line crosses the Y-axis).
y = mx + cwhere m is the slope and c is the Y-intercept (the value where the line crosses the Y-axis).
(ii) Line Through Origin:
When c = 0, the line always passes through the origin (0, 0).
y = mxWhen c = 0, the line always passes through the origin (0, 0).
(iii) Intercept Form:
where a is the X-intercept and b is the Y-intercept (intercepts on the positive directions of the respective axes).
x/a + y/b = 1where a is the X-intercept and b is the Y-intercept (intercepts on the positive directions of the respective axes).
5. Solution of Linear Equation in One Variable
For ax + b = 0 (a ≠ 0):
ax = –b ⟹ x = –b/a
Example
Solve: x/2 – x/3 = 3
LCM of 2 and 3 is 6. Multiply both sides by 6:
3x – 2x = 18
x = 18 Ans.
6. Solution of Linear Equations in Two Variables
Method 1: Elimination (Making Equal Coefficients)
Multiply the equations by suitable non-zero constants to make the coefficients of one variable equal, then add or subtract the equations to eliminate that variable.
Example
Solve: 2x – 3y = 5 and 3x + 2y = 1
Eq. (i): 2x – 3y = 5 ×3 ⟹ 6x – 9y = 15
Eq. (ii): 3x + 2y = 1 ×2 ⟹ 6x + 4y = 2
Subtract: (6x – 9y) – (6x + 4y) = 15 – 2 ⟹ –13y = 13
y = –1
Substitute y = –1 in eq. (i): 2x – 3(–1) = 5 ⟹ 2x + 3 = 5 ⟹ 2x = 2
x = 1, y = –1 Ans.
Method 2: Substitution Method
Express one variable in terms of the other from one equation, then substitute into the second equation.
Example
Solve: x + 4y = 14 …(i) and 7x – 3y = 5 …(ii)
From (i): x = 14 – 4y
Substitute in (ii): 7(14 – 4y) – 3y = 5
98 – 28y – 3y = 5 ⟹ 98 – 31y = 5 ⟹ 31y = 93
y = 3
Substitute y = 3 in (ii): 7x – 9 = 5 ⟹ 7x = 14
x = 2, y = 3 Ans.
7. Equations of Lines Parallel to the Axes
X-axis: Every point on the X-axis has y-coordinate = 0. The equation of the X-axis is y = 0.
Y-axis: Every point on the Y-axis has x-coordinate = 0. The equation of the Y-axis is x = 0.
A line of the form x = k (constant) is parallel to the Y-axis.
A line of the form y = k (constant) is parallel to the X-axis.
A line of the form y = k (constant) is parallel to the X-axis.
Example
What type of graph will x + 4 = 0 and y – 4 = 0 produce?
x + 4 = 0 ⟹ x = –4 → a vertical line parallel to the Y-axis at x = –4.
y – 4 = 0 ⟹ y = 4 → a horizontal line parallel to the X-axis at y = 4.
8. Key Summary & Special Cases of ax + by + c = 0
🔑 Special Cases of ax + by + c = 0
- If b = 0 (a ≠ 0, c ≠ 0): Reduces to ax + c = 0, i.e., x = –c/a. Graph is a straight line parallel to the Y-axis passing through (–c/a, 0).
- If a = 0 (b ≠ 0, c ≠ 0): Reduces to by + c = 0, i.e., y = –c/b. Graph is a straight line parallel to the X-axis passing through (0, –c/b).
- If c = 0 and b = 0 (a ≠ 0): Reduces to ax = 0, i.e., x = 0. Graph is the Y-axis itself.
- If c = 0 and a = 0 (b ≠ 0): Reduces to by = 0, i.e., y = 0. Graph is the X-axis itself.
- If c = 0 (a ≠ 0, b ≠ 0): Reduces to ax + by = 0. Graph is a line passing through the origin.
🔑 Geometric Representation of ax + c = 0
- In one variable: A single point on the number line at x = –c/a.
- In two variables: A vertical line in the coordinate plane (written as a·x + 0·y = –c). The line is parallel to the Y-axis.
- A linear equation in two variables has infinitely many solutions.
- Every point on the graph of a linear equation in two variables is a solution of the equation.
Common Mistake: When a = 0 and b = 0, the equation becomes 0 = c. If c ≠ 0, there is no solution; if c = 0, every point is a solution. Always check before graphing.
9. Let's Try — Practice Problems with Solutions
Q1
Find two solutions for 2x + 3y = 9.
Rearrange: y = (9 – 2x)/3
Put y = 1: 1 = (9 – 2x)/3 ⟹ 3 = 9 – 2x ⟹ 2x = 6 ⟹ x = 3. Solution: (3, 1)
Put y = 2: 2 = (9 – 2x)/3 ⟹ 6 = 9 – 2x ⟹ 2x = 3 ⟹ x = 3/2. Solution: (3/2, 2) often written as (1, 2) variant when using integer substitution.
Hence (3, 1) and (1, 2) are two solutions from the infinitely many solutions.
Q2
Represent 2x + 3y = 9 in the form ax + by + c = 0 and determine a, b and c.
Rewrite: 2x + 3y – 9 = 0
Comparing with ax + by + c = 0: a = 2, b = 3, c = –9
Q3
Find the value of k if x = 2, y = 1 is a solution of 2x + 3y = k.
Substitute x = 2, y = 1: 2(2) + 3(1) = k
4 + 3 = k ⟹ k = 7
Q4
Express x in terms of y for 7x – 3y = 15. Check if the line intersects the y-axis at (0, –5).
7x = 15 + 3y ⟹ x = (15 + 3y)/7
Any point on the Y-axis has x = 0. Set x = 0: 0 = (15 + 3y)/7 ⟹ 15 + 3y = 0 ⟹ y = –5
Hence the line intersects the Y-axis at (0, –5). ✓
Q5
Draw the graph of 3x + y = 6. Find where it crosses the axes.
| x | 0 | 2 | 1 |
|---|---|---|---|
| y | 6 | 0 | 3 |
The line crosses the X-axis at C(2, 0) and the Y-axis at A(0, 6).
Q8
The taxi fare in a city: first km costs ₹8, subsequent distance costs ₹5/km. Taking distance as x km and total fare as ₹y, write a linear equation and draw its graph.
Equation: y = 8 + 5(x – 1) ⟹ y = 5x + 3
| x (km) | 1 | 2 | 3 |
|---|---|---|---|
| y (₹) | 8 | 13 | 18 |
Plot the points (1, 8), (2, 13), (3, 18) and join them to get the fare-distance graph (a straight line).
🔗 For more practice problems and video explanations on this topic, visit MyClass24.com.
Exercise 1 — Multiple Choice Questions
1. The graph of which line is parallel to the X-axis?
2. The graph of which line is parallel to the Y-axis?
3. The graph of which line lies in the 1st and 3rd quadrants?
4. The graph of which line lies in the 2nd and 4th quadrants?
5. Which line is equally inclined to both axes?
9. The graph of the line 2x + 3 = 0 lies in which quadrants?
11. Which of the following points lies on the line 3x – 2y = 2?
13. If (p, 4) lies on the line 3x – y = 5, then the value of p is?
16. A linear equation in two variables has how many solutions?
21. Which of the following is NOT a linear equation?
25. A man is thrice as old as his son. After 14 years, the man will be twice as old as his son. What is the son's present age?
27. A linear equation in two variables has at most how many solutions?
28. Which of the following are solutions of x – 2y = 2?
30. If x = 1, y = 1 is a solution of 9ax + 12ay = 63, then the value of a is?
✅ Answers to Exercise 1
Q1
(b)
Q2
(a)
Q3
(c)
Q4
(a)
Q5
(d)
Q6
(b)
Q7
(a)
Q8
(c)
Q9
(b)
Q10
(d)
Q11
(b)
Q12
(c)
Q13
(d)
Q14
(a)
Q15
(c)
Q16
(b)
Q17
(a)
Q18
(d)
Q19
(a)
Q20
(c)
Q21
(c)
Q22
(d)
Q23
(b)
Q24
(d)
Q25
(b)
Q26
(c)
Q27
(c)
Q28
(d)
Q29
(a)
Q30
(b)
Exercise 2 — Descriptive / Subjective Questions
- If x is the number of hours a labourer works and y is his wages in rupees, then y = 5x + 7. Draw the work–wages graph. From the graph, find the wages for 6 hours of work.
- If y = 100x, find the value of y when x = 6 (i.e., x = 6).
- Verify which of the following are solutions of x – 2y + 4 = 0 and which are not: (0, 2), (2, 0), (4, 0), (6, 1), (–2, 1).
- Find the value of b if (3, 4) is a solution of the equation 5x + by = 13.
- Express y in terms of x in 2x + 3y = 6. Find the point where the line cuts the Y-axis.
- Express x in terms of y in 3x – 2y = 12. Find the point where the line cuts the X-axis.
- Give the geometric representation of the equation 3x + 5 = 0 as an equation in (i) one variable and (ii) two variables.
- Give the geometric representation of 3y – 9 = 0 as an equation in (i) one variable and (ii) two variables.
- A man drives at a uniform speed of 90 km/h. Draw the time–distance graph. From the graph find the distance in (i) 1/2 hour and (ii) 2.5 hours.
- Nidhi and Nisha together contributed ₹300 towards the PM's Relief Fund. Express this as a linear equation in two variables and draw the graph.
- If x + 1/x = 3, find the value of x³ + x² + x + 1.
- Determine whether x = 5, y = 4 is a solution of x – 2y = –3.
- Solve: 8x – 5y = 34 and 3x – 2y = 13.
- Solve: 20x + 3y = 7 and 8y – 15x = 5.
- Solve: 2x – 3y – 3 = 0 and (x/3) + 4y + (1/2) = 0.
✅ Answers to Exercise 2
1. ₹37 2. 600 3. Solutions: (0, 2), (4, 0), (6, 1), (–2, 1); NOT a solution: (2, 0) 4. b = –1/4 5. y = (6 – 2x)/3; cuts Y-axis at (0, 2) 6. x = (2y + 12)/3; cuts X-axis at (4, 0) 9. (i) 45 km (ii) 225 km 10. x + y = 300 11. –104 12. Yes 13. x = 3, y = –2 14. x = 1/5, y = 1 15. x = 3, y = 1
📹 Watch step-by-step video solutions for Exercise 2 on MyClass24.com.
Exercise 3 — Advanced / Higher Order Thinking
- Draw the graph of 2x + 3y = 6 and use it to find the area of the triangle formed by the line and the coordinate axes.
- Draw the graph of 4x – y = 5 and 5y – 4x = 7 on the same graph paper and find the coordinates of their point of intersection.
- Find two numbers such that five times the greater exceeds four times the lesser by 22, and three times the greater together with seven times the lesser is 32.
- Draw the graph of x – y + 1 = 0 and 3x + 2y – 12 = 0 on the same graph. Calculate the area bounded by these lines and the X-axis.
- If p = 3x + 1 and q = 9x + 13 and p : q = 6 : 5, find the value of x.
✅ Answers to Exercise 3
1. Area = 3 sq. units 2. x = 2, y = 3 3. 6 and 2 4. 7.5 sq. units 5. x = –7
📝 Make Your Own Notes