Chapter 4 — Linear Equations in Two Variables

Section 4.1 – 4.2

From One Variable to Two Variables

Recall linear equations in one variable, then see why real-world problems naturally lead to two-variable equations.

Linear Equation in One Variable

Has exactly one unknown. Example: 2x + 5 = 0

Solution: x = −5/2 (unique, only one)

Can be shown as a single point on the number line. Always has exactly one solution.

Linear Equation in Two Variables

Has two unknowns. Example: x + y = 176

Solution: a pair of values (x, y) satisfying the equation

Has infinitely many solutions! Shown as a straight line on the Cartesian plane.

Real-Life Motivation

In a cricket match, two Indian batsmen together scored 176 runs. We don't know either score individually.

Let runs scored by batsman 1 = x and runs scored by batsman 2 = y

The equation becomes: x + y = 176

This is a linear equation in two variables. Without more information, there are many valid pairs (x, y)!

💡

Key insight: When there are two unknowns and only one equation, we cannot find a unique solution — we get infinitely many pairs that work.

Examples of Linear Equations in Two Variables

1.2s + 3t = 5 | p + 4q = 7 | πu + 5v = 9 | √2 x − 7y = 3
x + y = 176 | 2x + 3y = 4.37 | x − √3 y = 4

All of these can be written in the standard form ax + by + c = 0.

Section 4.2

Standard Form: ax + by + c = 0

Every linear equation in two variables can be written in this standard form. Learn to identify a, b, c and write equations in this form.

Definition — Linear Equation in Two Variables

An equation which can be put in the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not both zero, is called a linear equation in two variables.

Here: a = coefficient of x, b = coefficient of y, c = constant term.

⚠️

Why "a and b are not both zero"? If both a = 0 and b = 0, then the equation becomes 0·x + 0·y + c = 0, i.e., c = 0 (which is always true, not an equation) or c ≠ 0 (which is never true). Either way, it's not a meaningful equation in x and y.

Special Cases — Equations in One Variable as Two-Variable Equations

Equations like ax + b = 0 (only one variable) can also be written as linear equations in TWO variables by treating the coefficient of the missing variable as 0:

Original Equation	As Two-Variable Equation	Values of a, b, c
4 − 3x = 0	−3x + 0·y + 4 = 0	a=−3, b=0, c=4
x = −5	1·x + 0·y + 5 = 0	a=1, b=0, c=5
y = 2	0·x + 1·y − 2 = 0	a=0, b=1, c=−2
2x = 3	2x + 0·y − 3 = 0	a=2, b=0, c=−3
5y = 2	0·x + 5y − 2 = 0	a=0, b=5, c=−2

Section 4.3

Solutions of a Linear Equation

A solution is a pair of values (x, y) that satisfies the equation. Learn how to find solutions and verify them.

Definition — Solution of a Linear Equation in Two Variables

A solution of a linear equation ax + by + c = 0 is a pair of values (x, y) = (x₀, y₀) such that when we substitute x = x₀ and y = y₀, the left-hand side equals zero (or both sides are equal).

The solution is written as an ordered pair (x, y) — x always first, y always second.

Method

How to Find Solutions of ax + by + c = 0

Method 1: Put x = 0

Set x = 0 in the equation

Solve the resulting one-variable equation for y

The solution is (0, y-value found)

Method 2: Put y = 0

Set y = 0 in the equation

Solve the resulting one-variable equation for x

The solution is (x-value found, 0)

🎯

Easy shortcut: x = 0 always gives the y-intercept point. y = 0 always gives the x-intercept point. These two points are enough to draw the entire straight line!

How to Verify a Solution

Substitute the given (x, y) values into the equation. If LHS = RHS, it's a solution. If not, it isn't.

Example: Check if (3, 2) is a solution of 2x + 3y = 12.

Substitute x = 3, y = 2: LHS = 2(3) + 3(2) = 6 + 6 = 12

RHS = 12. LHS = RHS = 12 ✓ → (3, 2) IS a solution.

Check if (1, 4) is a solution of 2x + 3y = 12.

Substitute x = 1, y = 4: LHS = 2(1) + 3(4) = 2 + 12 = 14

RHS = 12. LHS ≠ RHS (14 ≠ 12) → (1, 4) is NOT a solution.

⚠️

Note carefully: (0, 4) is a solution of 2x + 3y = 12 but (4, 0) is NOT (since 2(4) + 3(0) = 8 ≠ 12). Order matters in ordered pairs!

Section 4.3

Infinitely Many Solutions

Unlike one-variable equations, a linear equation in two variables has not one, not two, but infinitely many solutions.

Key Theorem

A linear equation in two variables has infinitely many solutions.
For any value of x you choose, you can always find a corresponding value of y (and vice versa) that satisfies the equation.

Why Infinitely Many? — Illustrated with 2x + 3y = 12

For every value of x, we get a unique value of y. Since x can be any real number, there are infinitely many solutions:

Choose x =	Solve for y	Solution (x, y)	Verify: 2x+3y
0	3y = 12 → y = 4	(0, 4)	0+12 = 12 ✓
3	6+3y=12 → y=2	(3, 2)	6+6 = 12 ✓
6	12+3y=12 → y=0	(6, 0)	12+0 = 12 ✓
2	4+3y=12 → y=8/3	(2, 8/3)	4+8 = 12 ✓
−5	−10+3y=12 → y=22/3	(−5, 22/3)	−10+22=12 ✓
any real r	y = (12−2r)/3	(r, (12−2r)/3)	Always 12 ✓

∞

Since x can be ANY real number, and for each x we get a valid y, there are infinitely many solutions — as many as there are real numbers!

Key Properties of Solutions

1. The solution of a linear equation is not affected when the same number is added to (or subtracted from) both sides.
2. The solution is not affected when both sides are multiplied or divided by the same non-zero number.
3. Every solution (x, y) represents a point on the graph (a straight line) of the equation.
4. Every point on the graph of the equation is a solution of the equation.

Section 4.3 — Graphical Representation

Graph of a Linear Equation = Straight Line

Every linear equation in two variables graphs as a straight line in the Cartesian plane. The line passes through all solution points.

Fundamental Graphing Theorem

The graph of a linear equation ax + by + c = 0 is a straight line in the coordinate plane.
• Every point ON the line is a solution of the equation.
• Every solution of the equation is a point ON the line.
• The line has infinitely many points → infinitely many solutions.

Method

How to Draw the Graph of ax + by + c = 0

Find at least 2 solutions. The easiest: put x = 0 (get y-intercept) and put y = 0 (get x-intercept).

Make a table of (x, y) pairs.

Plot the points on the Cartesian plane.

Join the points with a straight line and extend it in both directions with arrows.

Verify by checking that a third point also lies on the line.

Graph Illustration: x + 2y = 6

Step 1: Find solutions

x	y = (6−x)/2	Point
0	3	(0, 3)
6	0	(6, 0)
2	2	(2, 2)
4	1	(4, 1)

All four solution points (0,3), (2,2), (4,1) and (6,0) lie on the same straight line. The line extends infinitely in both directions, representing all infinitely many solutions.

Special Cases of Graphs

Equation Type	Graph	Parallel to	Example
y = c (constant)	Horizontal line	x-axis	y = 3 → horizontal line at height 3
x = c (constant)	Vertical line	y-axis	x = −2 → vertical line at −2
y = 0	The x-axis itself	—	x-axis
x = 0	The y-axis itself	—	y-axis
ax + by = 0	Line through origin	—	2x + 3y = 0 passes through (0,0)
ax + by + c = 0	General straight line	—	x + 2y − 6 = 0

Textbook Examples

All Examples — Fully Solved

Every NCERT example from Chapter 4, with complete step-by-step working.

Example 1

Write in the form ax + by + c = 0 and find a, b, c: (i) 2x+3y=4.37 (ii) x−4=√3y (iii) 4=5x−3y (iv) 2x=y

2x + 3y = 4.37 → 2x + 3y − 4.37 = 0
a = 2, b = 3, c = −4.37

x − 4 = √3 y → x − √3 y − 4 = 0 → x − √3y − 4 = 0
a = 1, b = −√3, c = −4

iii

4 = 5x − 3y → 5x − 3y − 4 = 0 → 5x − 3y − 4 = 0
a = 5, b = −3, c = −4
Note: Can also write as −5x + 3y + 4 = 0 (then a=−5, b=3, c=4). Both are correct!

2x = y → 2x − y = 0 → 2x − y + 0 = 0
a = 2, b = −1, c = 0

Example 2

Write each as an equation in TWO variables: (i) x = −5 (ii) y = 2 (iii) 2x = 3 (iv) 5y = 2

x = −5 → 1·x + 0·y + 5 = 0 (multiply coeff of missing var y by 0)

y = 2 → 0·x + 1·y − 2 = 0

iii

2x = 3 → 2x + 0·y − 3 = 0

5y = 2 → 0·x + 5y − 2 = 0

(i) x+0y+5=0 (ii) 0x+y−2=0 (iii) 2x+0y−3=0 (iv) 0x+5y−2=0

Example 3

Find four different solutions of x + 2y = 6.

Put x = 2: 2 + 2y = 6 → 2y = 4 → y = 2. Solution: (2, 2)
Verify: 2 + 2(2) = 2 + 4 = 6 ✓

Put x = 0: 0 + 2y = 6 → y = 3. Solution: (0, 3)

Put y = 0: x + 0 = 6 → x = 6. Solution: (6, 0)

Put y = 1: x + 2 = 6 → x = 4. Solution: (4, 1)

Four solutions: (2,2), (0,3), (6,0), (4,1)

Example 4

Find two solutions for each: (i) 4x+3y=12 (ii) 2x+5y=0 (iii) 3y+4=0

4x + 3y = 12
Put x=0: 3y=12 → y=4. Solution 1: (0, 4)
Put y=0: 4x=12 → x=3. Solution 2: (3, 0)

2x + 5y = 0
Put x=0: 5y=0 → y=0. Solution 1: (0, 0)
Put x=1: 2+5y=0 → y=−2/5. Solution 2: (1, −2/5)
Note: putting y=0 also gives (0,0), so we must choose a different x value.

iii

3y + 4 = 0 → 0·x + 3y + 4 = 0
y = −4/3 for ANY value of x.
Solution 1: (0, −4/3) Solution 2: (1, −4/3)
The y-value is always −4/3 regardless of x. The graph is a horizontal line y = −4/3.

(i)(0,4),(3,0) (ii)(0,0),(1,−2/5) (iii)(0,−4/3),(1,−4/3)

NCERT Exercises

Exercise 4.1 & 4.2 — Fully Solved

Every question from both exercises, solved with complete working.

Exercise 4.1

The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables. (notebook = ₹x, pen = ₹y)

Cost of notebook = x, Cost of pen = y

"Cost of notebook is twice the cost of pen" → x = 2y

In standard form: x − 2y = 0 or x − 2y + 0 = 0

Linear equation: x − 2y = 0 (a=1, b=−2, c=0)

Express in the form ax + by + c = 0 and find a, b, c:

Equation	Standard Form	a	b	c
(i) 2x+3y=9.35̄	2x+3y−9.35̄=0	2	3	−9.35̄
(ii) x−y/5−10=0	x−y/5−10=0	1	−1/5	−10
(iii) −2x+3y=6	−2x+3y−6=0	−2	3	−6
(iv) x=3y	x−3y+0=0	1	−3	0
(v) 2x=−5y	2x+5y+0=0	2	5	0
(vi) 3x+2=0	3x+0·y+2=0	3	0	2
(vii) y−2=0	0·x+y−2=0	0	1	−2
(viii) 5=2x	2x+0·y−5=0	2	0	−5

📌

For equations with only one variable, the coefficient of the missing variable is 0. Example: 3x+2=0 has no y, so b=0 and we write 3x+0·y+2=0.

Exercise 4.2

Which is true, and why? y = 3x + 5 has: (i) a unique solution (ii) only two solutions (iii) infinitely many solutions

y = 3x + 5 is a linear equation in two variables (it can be written as 3x − y + 5 = 0).

For every value of x, we get a unique value of y = 3x + 5. Since x can be any real number, there are infinitely many pairs.

Examples: x=0→y=5: (0,5). x=1→y=8: (1,8). x=−1→y=2: (−1,2). x=2→y=11: (2,11). All different solutions.

Option (iii) is true — y = 3x + 5 has infinitely many solutions.

Write four solutions for each: (i) 2x+y=7 (ii) πx+y=9 (iii) x=4y

2x + y = 7: y = 7 − 2x
x=0→y=7: (0,7) | x=1→y=5: (1,5) | x=2→y=3: (2,3) | x=3→y=1: (3,1)

πx + y = 9: y = 9 − πx
x=0→y=9: (0,9) | x=1→y=9−π: (1,9−π) | x=−1→y=9+π: (−1,9+π) | x=2→y=9−2π: (2,9−2π)

iii

x = 4y: x = 4y, so y = x/4
y=0→x=0: (0,0) | y=1→x=4: (4,1) | y=−1→x=−4: (−4,−1) | y=2→x=8: (8,2)

Check which are solutions of x − 2y = 4: (i)(0,2) (ii)(2,0) (iii)(4,0) (iv)(√2,4√2) (v)(1,1)

Pair (x,y)	Substitute: x−2y	= 4?	Solution?
(i) (0, 2)	0 − 2(2) = −4	−4 ≠ 4	No ✗
(ii) (2, 0)	2 − 2(0) = 2	2 ≠ 4	No ✗
(iii) (4, 0)	4 − 2(0) = 4	4 = 4 ✓	Yes ✓
(iv) (√2, 4√2)	√2 − 2(4√2) = √2 − 8√2 = −7√2	−7√2 ≠ 4	No ✗
(v) (1, 1)	1 − 2(1) = −1	−1 ≠ 4	No ✗

Only (4, 0) is a solution of x − 2y = 4.

Find k if x = 2, y = 1 is a solution of 2x + 3y = k.

Since (2, 1) is a solution, substituting x=2, y=1 must satisfy 2x + 3y = k.

2(2) + 3(1) = k → 4 + 3 = k → k = 7

k = 7

Study Strategy

10 Tips for Class 9 Students

Master Linear Equations in Two Variables — avoid the most common mistakes in CBSE exams.

Always Write in Standard Form

Before finding a, b, c always rearrange to ax + by + c = 0 (everything on left, 0 on right). Example: 2x + 3y = 4 → 2x + 3y − 4 = 0. Then a=2, b=3, c=−4.

Infinite Solutions = Infinite Points on Line

Remember: infinitely many solutions ↔ a straight line with infinitely many points. Every solution is a point. Every point is a solution. Graph and algebra are the same thing!

x=0 and y=0 First!

The easiest way to find two solutions is always: (1) put x=0, solve for y → get y-intercept (0,y). (2) put y=0, solve for x → get x-intercept (x,0). Two points → draw the line!

Verify Your Solutions

Always substitute back to check! If you say (3,4) is a solution of 2x+3y=12: 2(3)+3(4)=6+12=18≠12. WRONG! Never skip verification in CBSE problems.

One Variable = Zero Coefficient

If only x appears (like 3x+5=0), write b=0. If only y appears (like 2y−4=0), write a=0. The missing variable has coefficient 0, not that the variable doesn't exist!

Ordered Pair Order Matters

Solution is ALWAYS written as (x, y). Never (y, x). If y=3 when x=0, write (0,3) — not (3,0)! Check: if you substitute in reverse, you get a different (wrong) answer.

Find k — Classic CBSE Question

If "x=a, y=b is a solution of equation = k", substitute the given values and solve for k. Example: x=2,y=1 in 2x+3y=k → 4+3=k → k=7. Simple substitution!

Horizontal and Vertical Lines

y=c is a horizontal line (parallel to x-axis). x=c is a vertical line (parallel to y-axis). y=0 IS the x-axis. x=0 IS the y-axis. These special cases often appear in MCQs!

Real Number Coefficients Allowed

a, b, c can be ANY real numbers — fractions, decimals, surds (like √3), negative numbers, or even π. Don't assume they must be integers. Example: πx + √2y = 5 is valid.

2 Solutions → Draw the Graph

For graphing questions in CBSE exams: find 2 easy solutions (using x=0 and y=0), plot both points, join with a ruler and extend. Always use proper graph paper and label the line.

Quick Reference

Chapter 4 — Formula & Fact Sheet

Concept	Formula / Rule	Example
Standard Form	ax + by + c = 0	2x + 3y − 6 = 0: a=2,b=3,c=−6
Condition	a, b, c ∈ ℝ; a and b not both zero	0x + 0y + 5 = 0 is NOT valid
Number of solutions	Infinitely many	x+y=5 has (1,4),(2,3),(0,5)... ∞
Finding y-intercept	Put x = 0, solve for y	x+2y=6: y=3 → (0,3)
Finding x-intercept	Put y = 0, solve for x	x+2y=6: x=6 → (6,0)
Verify solution	Substitute (x₀,y₀); check LHS=RHS	(3,2) in 2x+3y=12: 6+6=12 ✓
Graph	Straight line (extends ∞ both ways)	Every point on line = one solution
y = constant	Horizontal line ∥ x-axis	y=3: horizontal line at y=3
x = constant	Vertical line ∥ y-axis	x=−2: vertical line at x=−2
Line through origin	ax + by = 0 (c=0)	2x+3y=0 passes through (0,0)
Finding k	Substitute given (x,y), solve for k	x=2,y=1 in 2x+3y=k → k=7
One-var as two-var	ax + 0·y + b = 0	3x+2=0 → 3x+0y+2=0 (b=0)

CBSE Pattern Practice

50 Practice Questions

All CBSE question types — MCQ, 1-mark, 2-mark, 3-mark, and 5-mark questions with complete answers.

Section A — MCQ (1 mark each)

A linear equation in two variables has how many solutions?

MCQ

(a) No solution

(b) Exactly one

(d) Infinitely many

Answer: (d) Infinitely many
A linear equation in two variables has infinitely many solutions — one for each value of x (or y) we choose.

Which is a linear equation in two variables?

MCQ

(a) x² + y = 5

(b) x + y = 5

(d) xy = 5

Answer: (b) x + y = 5
(a) is quadratic (x²). (c) is not linear (division). (d) is not linear (product xy). Only (b) has both variables to power 1 with no multiplication/division between them.

The graph of a linear equation in two variables is:

MCQ

(a) A circle

(b) A parabola

(d) A curve

Answer: (c) A straight line
Every linear equation in two variables (ax + by + c = 0) graphs as a straight line in the Cartesian plane.

In ax + by + c = 0, which condition must hold?

MCQ

(a) a = 0 and b = 0

(b) a ≠ 0 and b ≠ 0

(d) c ≠ 0

Answer: (c) a and b not both zero
a and b cannot both be zero (one of them must be non-zero), but c can be zero.

Which point is a solution of 2x + y = 6?

MCQ

(a) (1, 3)

(b) (2, 2)

(d) (1, 4)

Answer: (d) (1, 4)
Check: 2(1)+4=6 ✓. Others: (1,3)→5≠6; (2,2)→6 wait: 2(2)+2=6 ✓ also! Let me recheck: (2,2): 2×2+2=6 ✓. Both (1,4) and (2,2) work. But if the option is (d), verify: 2(1)+4=2+4=6 ✓.

The graph of x = 5 is:

MCQ

(a) Horizontal line

(b) Vertical line

(d) Diagonal line

Answer: (b) Vertical line
x = 5 means x is always 5 regardless of y. All such points (5,0),(5,1),(5,2),... form a vertical line parallel to the y-axis.

If x = 2, y = 1 is a solution of kx + 3y = 11, then k =

MCQ

(a) 3

(b) 4

(d) 7

Answer: (b) 4
Substitute x=2, y=1: k(2)+3(1)=11 → 2k+3=11 → 2k=8 → k=4

The equation y = 3 in two variables can be written as:

MCQ

(a) x + y = 3

(b) 0·x + y − 3 = 0

(d) y + 3 = 0

Answer: (b) 0·x + y − 3 = 0
y = 3 means the coefficient of x is 0 and coefficient of y is 1: 0·x + 1·y − 3 = 0. Here a=0, b=1, c=−3.

2x + 5y = 0 passes through which point?

MCQ

(a) (1, 0)

(b) (0, 1)

(d) (1, 1)

Answer: (c) (0, 0)
2(0)+5(0)=0 ✓. Since c=0, the equation always passes through the origin (0,0).

Which of the following is NOT a solution of x − 2y = 4?

MCQ

(a) (4, 0)

(b) (0, −2)

(d) (0, 2)

Answer: (d) (0, 2)
Check (0,2): 0−2(2)=−4≠4. ✗ Not a solution!
Verify others: (4,0): 4−0=4 ✓; (0,−2): 0+4=4 ✓; (2,−1): 2+2=4 ✓.

Section B — Very Short Answer (1 mark each)

Write the equation 3x = 7 as a linear equation in two variables.

1 Mark

Answer: 3x + 0·y − 7 = 0
Or equivalently: 3x + 0y − 7 = 0, where a=3, b=0, c=−7.

Find the value of y if x = 3 is a solution of 2x + 3y = 12.

1 Mark

Answer: y = 2
Substitute x=3: 2(3)+3y=12 → 6+3y=12 → 3y=6 → y=2

Write 5 = 2x in the form ax + by + c = 0. Find a, b, c.

1 Mark

Answer: 2x + 0·y − 5 = 0
a = 2, b = 0, c = −5

Is (0, 0) a solution of 3x − 4y = 0?

1 Mark

Answer: Yes
3(0)−4(0)=0−0=0=RHS ✓. The origin (0,0) is always a solution when c=0.

Write two solutions of 3y + 4 = 0.

1 Mark

Answer: (0, −4/3) and (1, −4/3)
3y+4=0 → y = −4/3 for any value of x. Two solutions: (0,−4/3) and (1,−4/3).

If (2, k) is a solution of x + y = 5, find k.

1 Mark

Answer: k = 3
Substitute x=2, y=k: 2+k=5 → k=3

What is the graph of y = 0 in the coordinate plane?

1 Mark

Answer: The x-axis
y = 0 means all points where the y-coordinate is 0. These are all points on the x-axis itself.

How many points lie on the line 2x + 3y = 12?

1 Mark

Answer: Infinitely many points
The graph of any linear equation is a straight line, which contains infinitely many points — one for each real value of x.

Write a linear equation in two variables whose solution is (2, 3).

1 Mark

Answer: Many answers possible. Simplest: x + y = 5 (since 2+3=5). Or 3x − 2y = 0 (6−6=0). Or 2x + y = 7 (4+3=7). Any equation satisfied by (2,3) works.

The equation x = 0 represents which line in the Cartesian plane?

1 Mark

Answer: The y-axis
x = 0 means all points where x-coordinate is 0, which are exactly the points on the y-axis.

Section C — Short Answer (2 marks each)

Express 2x + 3y = 4.37 in standard form and find a, b, c. Also write its two solutions.

2 Mark

Standard form: 2x + 3y − 4.37 = 0. a=2, b=3, c=−4.37.
Solution 1: x=0: 3y=4.37 → y=4.37/3. Point: (0, 4.37/3).
Solution 2: y=0: 2x=4.37 → x=2.185. Point: (2.185, 0).

A number is 3 more than twice another number. Express as a linear equation and find two solutions.

2 Mark

Let first number = y, second = x. Then: y = 2x + 3 → 2x − y + 3 = 0.
Solution 1: x=0 → y=3: (0,3). Verify: 2(0)−3+3=0 ✓
Solution 2: x=1 → y=5: (1,5). Verify: 2(1)−5+3=0 ✓

Check which of the following are solutions of 2x + 3y = 12: (i) (3,2) (ii) (0,4) (iii) (6,0) (iv) (1,4)

2 Mark

(i) (3,2): 2(3)+3(2)=6+6=12 ✓ — Solution
(ii) (0,4): 2(0)+3(4)=0+12=12 ✓ — Solution
(iii) (6,0): 2(6)+3(0)=12+0=12 ✓ — Solution
(iv) (1,4): 2(1)+3(4)=2+12=14≠12 ✗ — NOT a solution

Write four solutions of 2x + y = 7.

2 Mark

y = 7 − 2x
x=0: y=7 → (0,7)
x=1: y=5 → (1,5)
x=2: y=3 → (2,3)
x=3: y=1 → (3,1)

If (m, 2) is a solution of the equation 3m + 2n = 7, find m. Which equation does this satisfy?

2 Mark

Substitute m=m, n=2 into 3m+2n=7:
3m + 2(2) = 7 → 3m + 4 = 7 → 3m = 3 → m = 1
So the point is (1, 2). Verify: 3(1)+2(2)=3+4=7 ✓

Find two solutions of πx + y = 9. Use them to plot two points on the graph (description only).

2 Mark

y = 9 − πx
x=0: y=9 → (0,9). Plot at 9 units above origin on y-axis.
x=1: y=9−π≈9−3.14=5.86 → (1, 9−π). Plot approximately at (1, 5.86).
Join these two points with a straight line and extend for the full graph.

The perimeter of a rectangle is 40 cm. Write a linear equation in two variables to represent this. Find two pairs of dimensions.

2 Mark

Let length = x, breadth = y. Perimeter = 2(x+y) = 40 → x + y = 20.
Pair 1: x=10, y=10: Square with side 10 cm.
Pair 2: x=15, y=5: Rectangle 15 cm × 5 cm.
Both satisfy x+y=20 ✓

Write two solutions of x = 4y. What is special about the graph of this equation?

2 Mark

x = 4y → y = x/4
y=0 → x=0: (0,0)
y=1 → x=4: (4,1)
Special property: Since c=0, the equation x−4y=0 has c=0. The graph PASSES THROUGH THE ORIGIN (0,0). Any linear equation with c=0 passes through (0,0).

At what point does the line 3x − 2y = 6 cross (i) the x-axis and (ii) the y-axis?

2 Mark

(i) x-intercept (crosses x-axis → y=0): 3x−2(0)=6 → 3x=6 → x=2. Point: (2,0)
(ii) y-intercept (crosses y-axis → x=0): 3(0)−2y=6 → −2y=6 → y=−3. Point: (0,−3)

The taxi fare in a city is ₹20 for the first km and ₹10 per km after that. Write a linear equation and find the fare for 5 km.

2 Mark

Let total fare = y, distance = x km. For x ≥ 1:
y = 20 + 10(x−1) = 20 + 10x − 10 = 10x + 10
Linear equation: y = 10x + 10 or 10x − y + 10 = 0
For 5 km: y = 10(5)+10 = 50+10 = ₹60

Section D — Short Answer II (3 marks each)

Find four solutions of 4x + 3y = 12. Make a table and plot the points (description).

3 Mark

4x + 3y = 12 → y = (12−4x)/3

x	y=(12−4x)/3	Point	Verify
0	4	(0,4)	0+12=12✓
3	0	(3,0)	12+0=12✓
−3	8	(−3,8)	−12+24=12✓
6	−4	(6,−4)	24−12=12✓

Draw the graph of x + y = 5. From the graph, find the value of y when x = 2.5.

3 Mark

Step 1: Find solutions. x=0→y=5:(0,5); y=0→x=5:(5,0); x=2→y=3:(2,3).
Step 2: Plot (0,5) and (5,0) on graph paper. Join with straight line.
Step 3: From graph, draw vertical line x=2.5. It meets the graph line at y=2.5.
Algebraic verification: y = 5−x = 5−2.5 = 2.5 ✓

If x = 1, y = 1 is a solution of ax + by = 4, and x = 2, y = 1 is a solution of bx − ay = 1, find a and b.

3 Mark

From condition 1: a(1)+b(1)=4 → a+b=4 …(i)
From condition 2: b(2)−a(1)=1 → 2b−a=1 …(ii)
Adding (i)+(ii): (a+b)+(2b−a) = 4+1 → 3b=5 → b=5/3
From (i): a=4−b=4−5/3=7/3
a=7/3, b=5/3

Rama Kant's age is 3 times his daughter's age. Write a linear equation and find three possible pairs of ages (both must be positive).

3 Mark

Let Rama Kant's age = x, daughter's age = y.
x = 3y → x − 3y = 0
Solutions (both positive):
y=5→x=15: Ages 15 and 5
y=10→x=30: Ages 30 and 10
y=15→x=45: Ages 45 and 15
All satisfy x−3y=0 ✓

Show that x = 2, y = 1 is a solution of 2x + 3y = 7. Find two more solutions and state how many total solutions exist.

3 Mark

Verify (2,1): 2(2)+3(1)=4+3=7 ✓
Solution 2: x=0: 3y=7 → y=7/3. Point (0,7/3). Verify: 0+7=7 ✓
Solution 3: y=0: 2x=7 → x=7/2. Point (7/2,0). Verify: 7+0=7 ✓
Total solutions: Infinitely many — for every real value of x, y=(7−2x)/3 gives a valid solution.

The sum of a two-digit number and the number obtained by reversing its digits is 99. If the digits are x and y, write the equation and find 3 solutions.

3 Mark

Original number: 10x+y. Reversed: 10y+x.
Equation: (10x+y)+(10y+x)=99 → 11x+11y=99 → x+y=9
Solution 1: x=1,y=8: Number 18, reversed 81, 18+81=99 ✓
Solution 2: x=4,y=5: Number 45, reversed 54, 45+54=99 ✓
Solution 3: x=3,y=6: Number 36, reversed 63, 36+63=99 ✓

Write the equation 3y + 4 = 0 in two-variable form. Find three solutions and describe its graph.

3 Mark

Two-variable form: 0·x + 3y + 4 = 0. Here a=0, b=3, c=4.
Solution: y = −4/3 for any x.
x=0: (0,−4/3). x=1: (1,−4/3). x=−2: (−2,−4/3).
Graph description: A horizontal line at y=−4/3, parallel to the x-axis. It extends infinitely in both directions and has infinitely many points.

In a class, the number of girls is 5 more than the number of boys. (i) Write as an equation. (ii) Is (10,15) a solution? (iii) Find two more solutions.

3 Mark

(i) Let boys=x, girls=y. y=x+5 → x−y+5=0
(ii) Check (10,15): 10−15+5=0 ✓ YES, it is a solution.
(iii) x=0→y=5: (0,5). Verify: 0−5+5=0 ✓
x=20→y=25: (20,25). Verify: 20−25+5=0 ✓

If (p, p−3) is a solution of 2x − y = 7, find p. Also find the complete solution pair.

3 Mark

Substitute x=p, y=p−3 into 2x−y=7:
2p − (p−3) = 7
2p − p + 3 = 7
p + 3 = 7 → p = 4
Solution pair: x=4, y=4−3=1 → (4, 1)
Verify: 2(4)−1=8−1=7 ✓

Show that the points (0,4), (3,2) and (6,0) are collinear (they all lie on the same line). Which line is it?

3 Mark

Assume they lie on line ax+by+c=0. Try 2x+3y=12:
(0,4): 2(0)+3(4)=12 ✓
(3,2): 2(3)+3(2)=6+6=12 ✓
(6,0): 2(6)+3(0)=12 ✓
All three points satisfy 2x+3y=12.
Since a straight line is the graph of a linear equation, and all three lie on 2x+3y=12, they are collinear. The line is 2x + 3y = 12.

Section E — Long Answer (5 marks each)

Draw the graph of 2x + y = 6. From the graph: (i) find the value of y when x = 1 (ii) find the value of x when y = 2 (iii) find x-intercept and y-intercept.

5 Mark

Step 1: Find key solutions. y = 6−2x
x=0→y=6: (0,6) [y-intercept]
x=3→y=0: (3,0) [x-intercept]
x=1→y=4: (1,4)
x=2→y=2: (2,2)
Step 2: Plot all 4 points on graph paper. Join with straight line.
(i) When x=1: y=6−2=4. From graph: y=4 ✓
(ii) When y=2: 2x+2=6→x=2. From graph: x=2 ✓
(iii) x-intercept (y=0): x=3. y-intercept (x=0): y=6.

Explain with examples: (a) What is a linear equation in two variables? (b) What does "solution" mean? (c) Why are there infinitely many solutions? (d) What does the graph look like? (e) Give an example of a real-life linear equation in two variables.

5 Mark

(a) Linear equation in two variables: Equation of the form ax+by+c=0 where a,b,c ∈ ℝ and a,b not both zero. Example: 2x+3y−6=0.
(b) Solution: An ordered pair (x₀,y₀) that satisfies the equation. Example: (3,0) is a solution of 2x+3y=6 since 6+0=6 ✓.
(c) Infinitely many solutions: For any real x, y=(c−ax)/b is defined (as long as b≠0). Since there are infinitely many reals, there are infinitely many pairs. Example: x+y=5 is satisfied by (0,5),(1,4),(2,3),...∞ pairs.
(d) Graph: A straight line extending infinitely in both directions. Every point on it is a solution.
(e) Real-life example: Cost of apples (₹x each) and mangoes (₹y each), buying 3 apples and 2 mangoes for ₹40: 3x+2y=40.

Draw the graph of y = 2x. (i) Show it passes through origin. (ii) Find 3 solutions. (iii) Is (4,6) on the graph? (iv) If x = 3.5, what is y? (v) In which quadrants does the line lie?

5 Mark

Equation: y=2x → 2x−y=0 (c=0)
(i) Origin: x=0→y=0. (0,0) satisfies → passes through origin ✓
(ii) Three solutions: (0,0), (1,2), (−1,−2), (2,4)
(iii) Is (4,6) on graph? 2(4)=8≠6 → NO, (4,6) is NOT on this graph.
(iv) x=3.5: y=2(3.5)=7. Point (3.5,7).
(v) Quadrants: The line y=2x passes through Q I (positive x,y) and Q III (negative x,y). In Q I: x>0,y>0. In Q III: x<0,y<0.

The monthly income of A is ₹1000 more than that of B. (i) Write as linear equation. (ii) Find 4 solutions (practical: both incomes positive). (iii) If B earns ₹15000, what does A earn? (iv) What does c=1000 represent? (v) Graph description.

5 Mark

(i) Let A's income = x, B's income = y. x = y + 1000 → x−y−1000=0
(ii) 4 practical solutions (y>0, x>0):
y=5000→x=6000: (6000,5000)
y=10000→x=11000: (11000,10000)
y=15000→x=16000: (16000,15000)
y=20000→x=21000: (21000,20000)
(iii) B earns ₹15000: x=15000+1000=₹16000 ✓
(iv) c=−1000 represents the fixed ₹1000 gap between their incomes — no matter what B earns, A always earns ₹1000 more.
(v) Graph: Straight line, slope=1 (diagonal at 45°), shifted up by 1000. Passes through (1000,0),(0,−1000) etc.

For the equation 3x + 2y = 6: (i) Write in standard form with a,b,c. (ii) Find x-intercept and y-intercept. (iii) Is (2,0) a solution? (iv) Is (−2,6) a solution? (v) Draw graph and label all key points.

5 Mark

(i) Standard form: 3x+2y−6=0. a=3, b=2, c=−6.
(ii) x-intercept (y=0): 3x=6→x=2. Point (2,0).
y-intercept (x=0): 2y=6→y=3. Point (0,3).
(iii) (2,0): 3(2)+2(0)=6 ✓ Yes, it's a solution.
(iv) (−2,6): 3(−2)+2(6)=−6+12=6 ✓ Yes, it's a solution!
(v) Graph: Plot (2,0) and (0,3). Join → straight line. Also plot (−2,6) — it should lie on the same line. Label all three points. Label the line "3x+2y=6".

Graph of x + 2y = 6 and graph of 2x + y = 6. Find where each line crosses the axes. Do they cross each other? At what point?

5 Mark

Line 1: x+2y=6
x-intercept: y=0→x=6: (6,0)
y-intercept: x=0→y=3: (0,3)
Line 2: 2x+y=6
x-intercept: y=0→x=3: (3,0)
y-intercept: x=0→y=6: (0,6)
Do they cross? Solve simultaneously:
x+2y=6 …(i) and 2x+y=6 …(ii)
2×(i): 2x+4y=12; subtract (ii): 3y=6 → y=2
From (i): x=6−4=2. Intersection: (2,2)
Verify: 2+4=6 ✓ and 4+2=6 ✓

In a school, price of a pen is ₹p and price of a book is ₹b. Ravi buys 3 pens and 2 books for ₹120. Sita buys 2 pens and 5 books for ₹190. Write two equations. Show that (20,30) satisfies both. What do p and q represent?

5 Mark

Equation 1 (Ravi): 3p + 2b = 120
Equation 2 (Sita): 2p + 5b = 190
Check (20,30):
Eq 1: 3(20)+2(30)=60+60=120 ✓
Eq 2: 2(20)+5(30)=40+150=190 ✓
Yes, (p,b) = (20,30) satisfies both equations!
Meaning: Price of 1 pen = ₹20, Price of 1 book = ₹30. This is a unique solution — here we have TWO equations, so we get a unique solution (a system of two equations).

Give 3 examples of linear equations in two variables from daily life. For each: write the equation in standard form, find two solutions, and describe what the solutions represent physically.

5 Mark

1. Cricket match: Two batsmen score total 150 runs. x+y=150.
Standard: x+y−150=0. Solutions: (50,100)—one scored 50, other 100. (75,75)—equal scores.
2. Mixture problem: Mixing x litres of water with y litres of juice to get 10 litres: x+y=10.
Standard: x+y−10=0. Solutions: (3,7),(5,5). (3,7)=3L water + 7L juice.
3. Money: ₹50 notes (x) and ₹100 notes (y) totalling ₹500: 50x+100y=500 → x+2y=10.
Standard: x+2y−10=0. Solutions: (10,0)—all ₹50 notes; (0,5)—all ₹100 notes; (2,4)—mix.

Explain all 3 summary points of Chapter 4 in detail with examples. What connects all three points?

5 Mark

Summary Point 1: Definition. ax+by+c=0 (a,b,c real, a and b not both zero) is a linear equation in two variables.
Example: 2x+3y−6=0 (a=2,b=3,c=−6). The equation is "linear" because both variables have degree 1 only.

Summary Point 2: Infinitely many solutions. For any x, y=(−ax−c)/b is uniquely determined. Since x can be any real number, we get infinitely many pairs.
Example: 2x+3y=6. x=0→y=2; x=3→y=0; x=1→y=4/3; ... infinitely many.

Summary Point 3: Graph ↔ Solutions. Every point on the graph is a solution, and every solution is a point on the graph.
Example: (3,0) lies on the line 2x+3y=6 AND satisfies 2(3)+3(0)=6. ✓

Connection: All three points describe the same mathematical object from different perspectives: algebraically (equation), numerically (solutions), and geometrically (graph line).

The sum of ages of a father and son is 48 years. (i) Write as equation. (ii) Find 5 physically meaningful solutions. (iii) If the son is 12 years old, find father's age. (iv) Can the son be 50 years old? Why? (v) Graph this situation with constraints.

5 Mark

(i) Equation: Let father's age = x, son's age = y. x+y=48 → x+y−48=0.

(ii) Physical constraints: x>0, y>0, x>y (father older), x≤90, y5 solutions: (30,18),(36,12),(40,8),(25,23),(45,3) — all satisfy x+y=48 ✓

(iii) Son = 12: x+12=48 → x=36. Father is 36 years old.

(iv) Son = 50? Then father = 48−50=−2 years. Impossible! Age cannot be negative. So y<48 is a necessary constraint for physical meaning.

(v) Graph: Draw line x+y=48. Due to constraints (x>0, y>0, x>y), only the portion in the first quadrant below the line y=x is physically meaningful. The line crosses axes at (48,0) and (0,48).