What is a Linear Equation in Two Variables?
Before understanding equations in two variables, it helps to recall the simpler case: a linear equation in one variable takes the form ax + b = 0, where a and b are real numbers and a ≠ 0. Examples: 3x + 5 = 0, 7x − 2 = 0.
An equation of the form ax + by + c = 0, where a, b, c are real numbers and a ≠ 0, b ≠ 0, and x, y are variables, is called a linear equation in two variables.
Here:
- a = coefficient of x
- b = coefficient of y
- c = constant term
ax + by + c = 0 (a ≠ 0, b ≠ 0)
Why "Linear"?
The highest power of any variable is 1. This produces a straight line when graphed, hence "linear".
Why "Two Variables"?
Both x and y appear, so a solution requires a pair of values one for each variable.
Infinitely Many Solutions
Unlike one-variable equations, a linear equation in two variables has infinitely many solution pairs (x, y).
Geometric Meaning
Every solution pair is a point on the line represented by the equation on the Cartesian plane.
Linear Equation in Two Variables Class 9 Maths Revision Notes PDF Download
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What is a Solution of a Linear Equation?
A solution (or root) of a linear equation ax + by + c = 0 is any ordered pair (x₀, y₀) of real numbers that satisfies the equation i.e., when substituted, both sides become equal.
Important: A linear equation in two variables has infinitely many solutions. Each solution corresponds to exactly one point on the straight line graph of the equation.
Verifying a Solution Method
To check whether a given pair (x₀, y₀) is a solution:
- Substitute x = x₀ and y = y₀ into the Left-Hand Side (LHS) of the equation.
- Compute the value of LHS.
- If LHS = RHS (= 0 for standard form), then the pair is a solution. Otherwise, it is not.
Quick Verification Examples
Q: Prove that x = 3, y = 2 is a solution of 3x − 2y = 5.
→ LHS = 3(3) − 2(2) = 9 − 4 = 5
→ RHS = 5
→ LHS = RHS
∴ (3, 2) is a solution of 3x − 2y = 5
Q: Prove that (1, 1) and (2, 5) are both solutions of 4x − y − 3 = 0.
→ For (1, 1): LHS = 4(1) − 1 − 3 = 0 = RHS ✓
→ For (2, 5): LHS = 4(2) − 5 − 3 = 0 = RHS ✓
∴ Both (1, 1) and (2, 5) satisfy 4x − y − 3 = 0
Q: Is x = 2, y = −1 a solution of 3x + 5y − 2 = 0?
→ LHS = 3(2) + 5(−1) − 2 = 6 − 5 − 2 = 1 ≠ 0
→ LHS ≠ RHS ∴ (2, −1) is NOT a solution
Graph of a Linear Equation
Every linear equation in two variables, when plotted on the Cartesian plane, produces a straight line. This line is the complete set of all solution pairs of the equation.
Case A: One Variable — ax = b or ay = b
Important Rule: If the equation is of the form ax = b, the graph is a vertical line parallel to the Y-axis at x = b/a.
If the equation is of the form ay = b, the graph is a horizontal line parallel to the X-axis at y = b/a.
Steps to draw ax = b or ay = b:
- Simplify to find the constant value of x or y.
- Plot that point on the appropriate axis.
- Plot a second point by choosing any value for the other variable.
- Draw a straight line through both points.
Q: Draw the graph of 2x + 5 = 0.
→ Simplify: 2x = −5 → x = −5/2 = −2.5
→ Plot A₁(−2.5, 0) and A₂(−2.5, 2) → Join them
→ vertical line at x = −2.5, parallel to Y-axis
∴ Graph is a vertical line x = −2.5
Q: Draw the graph of 3y − 15 = 0.
→ Simplify: 3y = 15
→ y = 5 →Plot B₁(0, 5) and B₂(3, 5)
→ Join them
→ horizontal line at y = 5, parallel to X-axis
∴ Graph is a horizontal line y = 5
Case B: General Form — ax + by + c = 0
Algorithm for graphing ax + by + c = 0:
- Express y in terms of x: y = −(ax + c) / b
- Choose 2–3 convenient values of x (like 0, 1, 2 or values that simplify fractions).
- Compute the corresponding y values.
- List the solution pairs in a table.
- Plot these points on the Cartesian plane and join them with a straight line.
Q: Draw the graph of x − 2y = 3. Find coordinates when (i) x = −5 and (ii) y = 0.
→ Solve for y: 2y = x − 3 → y = (x − 3)/2
| x | 0 | 3 | −2 |
|---|---|---|---|
| y = (x−3)/2 | −3/2 | 0 | −5/2 |
→ (i) When x = −5: y = (−5 − 3)/2 = −8/2 = −4
→ (ii) When y = 0: 0 = (x − 3)/2 → x = 3
∴ At x = −5, coordinate is (−5, −4). At y = 0, coordinate is (3, 0).
Q: Draw graphs of x + y = 4 and 2x − y = 2 on the same axes. Find the point of intersection.
→ For x + y = 4: y = 4 − x
| x | 0 | 2 | 4 |
|---|---|---|---|
| y | 4 | 2 | 0 |
→ For 2x − y = 2: y = 2x − 2
| x | 1 | 0 | 3 |
|---|---|---|---|
| y | 0 | −2 | 4 |
∴ The two lines intersect at point P = (2, 2)
Different Forms of a Line
a) Slope of a Line
If a line makes an angle θ with the positive direction of the x-axis, then the tangent of this angle is called the slope (gradient) of the line, denoted by m.
m = tan θ
b) Slope-Intercept Form
Slope-Intercept Formy = mx + c, where m is the slope and c is the y-intercept (the point where the line crosses the Y-axis).
When c = 0, the line passes through the origin and simplifies to y = mx.
c) Intercept Form
Intercept Formx/a + y/b = 1, where a = x-intercept (where line crosses X-axis) and b = y-intercept (where line crosses Y-axis).
Horizontal Line (y = k)
Slope m = 0. Parallel to X-axis. Example: y = 5 from 3y = 15.
Vertical Line (x = k)
Slope undefined. Parallel to Y-axis. Example: x = −2.5 from 2x + 5 = 0.
Line Through Origin
When c = 0 in y = mx + c. Always passes through (0, 0).
How to Solve System of Linear Equations Problems
When we have two linear equations in two variables simultaneously, we need to find a unique pair (x, y) that satisfies both. There are two primary algebraic methods:
Method 1: Elimination (Making Equal Coefficients)
Make the coefficient of one variable the same in both equations, then add or subtract the equations to eliminate that variable.
- Write the two equations clearly.
- Multiply each equation by a suitable constant so that the coefficient of one variable becomes equal in both.
- Add or subtract the equations to eliminate one variable.
- Solve the resulting single-variable equation.
- Substitute back into either original equation to find the other variable.
Q: Solve: 2x − 3y = 5 and 3x + 2y = 1
→ Multiply eq.(i) by 3: 6x − 9y = 15
→ Multiply eq.(ii) by 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 = 2 → x = 1 ∴ x = 1, y = −1
Method 2: Substitution Method
Express one variable in terms of the other from one equation, then substitute that expression into the second equation.
- From one equation, express x in terms of y (or y in terms of x).
- Substitute this expression into the other equation.
- Solve the resulting one-variable equation.
- Back-substitute to find the other variable.
Q: Solve: x + 4y = 14 and 7x − 3y = 5
→ From eq.(i): x = 14 − 4y
→ Substitute in eq.(ii): 7(14 − 4y) − 3y = 5
→ 98 − 28y − 3y = 5
→ 93 = 31y → y = 3
→ x = 14 − 4(3) = 14 − 12 = 2
∴ x = 2, y = 3
Quick Formulas of Linear Equaitons in Two Variables
| Concept | Formula / Form | Property |
|---|---|---|
| Standard Form | ax + by + c = 0 | a ≠ 0, b ≠ 0 |
| Slope-Intercept | y = mx + c | m = slope, c = y-intercept |
| Intercept Form | x/a + y/b = 1 | a = x-int, b = y-int |
| Through Origin | y = mx | c = 0 |
| Vertical Line | x = constant | Parallel to Y-axis |
| Horizontal Line | y = constant | Parallel to X-axis |
| Solutions | Infinitely many (x, y) pairs | Each is a point on the line |
Solved Examples of Linear Equations in Two Variables
The following examples cover all major question types asked in Class 9 examinations, progressing from basic to advanced.
Q: In 2x + 3y = 12, find y when x = 3.
→ 2(3) + 3y = 12 → 6 + 3y = 12
→ 3y = 6 → y = 2 y = 2
Q: Find the value of y in 5x − 2y = 10 when x = 0.
→ 5(0) − 2y = 10
→ −2y = 10 → y = −5 y = −5
→ Solution (0, −5)
Q: Write 3x = 7 − 2y in standard form ax + by + c = 0. Identify a, b, c.
→ 3x + 2y − 7 = 0
→ a = 3, b = 2, c = −7 Standard form: 3x + 2y − 7 = 0
Q: Draw the graph of y + 4 = 0.
→ y = −4 (horizontal line)
→ Plot (0, −4) and (3, −4); draw horizontal line through them Horizontal line parallel to X-axis at y = −4
Q: Express x − 3y + 6 = 0 in slope-intercept form. Find slope and y-intercept.
→ −3y = −x − 6
→ y = x/3 + 2
→ Slope m = 1/3, y-intercept c = 2 y = (1/3)x + 2; m = 1/3, c = 2
Q: Does (4, 1) lie on the line 3x − 2y = 10?
→ LHS = 3(4) − 2(1) = 12 − 2 = 10 = RHS ✓
Yes, (4, 1) lies on 3x − 2y = 10
Q: Prepare a table of values for y = 2x + 1 for x = −1, 0, 1, 2.
→ x = −1: y = 2(−1)+1 = −1
→ x = 0: y = 1 | x = 1: y = 3 | x = 2: y = 5
| x | −1 | 0 | 1 | 2 |
|---|---|---|---|---|
| y | −1 | 1 | 3 | 5 |
Q: If (k, 3) is a solution of 2x − y + 1 = 0, find k.
→ 2k − 3 + 1 = 0
→ 2k = 2
→ k = 1 k = 1
Q: Solve: 4x + 3y = 24 and 3x − y = 5 by elimination.
→ Multiply eq.(ii) by 3: 9x − 3y = 15
→ Add: 4x + 3y + 9x − 3y = 24 + 15
→ 13x = 39
→ x = 3
→3(3) − y = 5
→ 9 − y = 5
→ y = 4 x = 3, y = 4
Q: The sum of two numbers is 30 and their difference is 10. Find them.
→ Let numbers be x and y: x + y = 30, x − y = 10
→ Adding: 2x = 40
→ x = 20
→ Substituting: 20 + y = 30
→ y = 10 The two numbers are 20 and 10
Q: Riya is 5 years older than Priya. In 10 years their combined age will be 45. Find current ages.
→ Let Priya's age = x.
Then Riya's = x + 5
→ (x + 10) + (x + 5 + 10) = 45
→ 2x + 25 = 45
→ x = 10 Priya = 10 years,
Riya = 15 years EXAMPLE 21Intercept form conversion
Q: Write 2x + 3y = 6 in intercept form and find x and y intercepts.
→ Divide by 6: x/3 + y/2 = 1
→ x-intercept = 3, y-intercept = 2 x/3 + y/2 = 1;
intercepts: a = 3, b = 2
Q: Are the lines x + 2y = 4 and 2x + 4y = 8 the same or different?
→ Divide second by 2: x + 2y = 4 (identical to first)
∴ The two equations represent the SAME line (coincident lines)
Q: The perimeter of a rectangle is 40 cm and its length is 4 more than its width. Find dimensions.
→ 2(l + w) = 40
→ l + w = 20; l = w + 4
→ (w + 4) + w = 20 → 2w = 16
→ w = 8, l = 12 Length = 12 cm, Width = 8 cm
Q: Solve x/2 = 3 + x/3.
→ x/2 − x/3 = 3
→ (3x − 2x)/6 = 3
→ x/6 = 3
→ x = 18 x = 18
Q: How many solutions does the equation 3x + y = 7 have? Give any three.
→ Infinitely many solutions exist. y = 7 − 3x
| x | 0 | 1 | 2 |
|---|---|---|---|
| y | 7 | 4 | 1 |
Three solutions: (0,7), (1,4), (2,1)
Q: 3 notebooks and 2 pens cost ₹85. 4 notebooks and 3 pens cost ₹115. Find the cost of each.
→ 3n + 2p = 85 ...(i) | 4n + 3p = 115 ...(ii)
→ Multiply (i)×3: 9n + 6p = 255; Multiply (ii)×2: 8n + 6p = 230
→ Subtract: n = 25. Then: 3(25) + 2p = 85 → 2p = 10 → p = 5 Notebook = ₹25, Pen = ₹5
Q: A boat covers 30 km upstream in 5 hours and 30 km downstream in 3 hours. Find speed of boat in still water and current.
→ Let boat speed = x, current = y. Upstream: x − y = 6; Downstream: x + y = 10
→ Adding: 2x = 16
→ x = 8; then y = 10 − 8 = 2 Boat speed = 8 km/h; Current speed = 2 km/h EXAMPLE 28Verify & find missing constant
Q: If x = a and y = 0 is a solution of 3x + 5y = 9, find a.
→ 3a + 5(0) = 9 → 3a = 9 → a = 3 a = 3
Q: Two lines 2x + ky = 5 and 3x − 2y = 7 intersect at a unique point. Show they're not parallel.
→ For parallel lines: a₁/a₂ = b₁/b₂
→ 2/3 = k/(−2)
→ k = −4/3
→If k ≠ −4/3,
lines intersect at a unique point (not parallel) Lines are NOT parallel for any k ≠ −4/3