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Andhra Pradesh (BIEAP)Class 9 Mathematics★ AP Board (BSEAP) Class 9 Mathematics · ~8 marks · Board: AP Board

Linear Equations in Two Variables

LINEAR EQUATIONS — ax+by+c=0. SOLUTIONS as ordered pairs (x,y) — infinitely many. GRAPHING a linear equation as a STRAIGHT LINE. x=k and y=k as special cases. Applications and worked examples. AP Board Class 9 Mathematics.

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By the end of this chapter you'll be able to…

1Write a linear equation in two variables in standard form ax + by + c = 0
2Find multiple solutions to a linear equation and represent them as ordered pairs
3Graph a linear equation by finding two points (x-intercept and y-intercept method)
4Identify x = k as a vertical line parallel to y-axis and y = k as a horizontal line parallel to x-axis
5Solve word problems by setting up a linear equation in two variables
6Understand that every point on the graph satisfies the equation and vice versa

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Why this chapter matters

Linear Equations in Two Variables is a foundational chapter that bridges algebra and geometry. The key insight — that a linear equation in two variables has infinitely many solutions forming a straight line — is conceptually important and directly tested. Word problems that require setting up a linear equation (age problems, cost problems) appear for 3-4 marks. Graphing the line by finding two points (x-intercept, y-intercept) is a standard 3-4 mark question. Special cases x=k (vertical line) and y=k (horizontal line) are tested in MCQ and short answer. This chapter is directly prerequisite to Pair of Linear Equations in Class 10.

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Linear Equations in Two Variables — Class 9 Mathematics

1. Standard Form: ax + by + c = 0

Where a, b, c are REAL numbers, and a and b are NOT BOTH ZERO.

2. Solutions

A solution is an ORDERED PAIR (x,y) that satisfies the equation. Substitute x and y into the equation → if LHS = RHS, it's a solution. A linear equation in two variables has INFINITELY MANY solutions. 'For any value of x, you can find a corresponding y — and vice versa.'

Finding Solutions

Choose a value for x. Substitute into equation. Solve for y. Write as (x,y). Repeat for other x values.

Example: 2x + y = 7

If x = 0 → y = 7 → (0,7). If x = 1 → 2+y=7 → y=5 → (1,5). If x = 2 → 4+y=7 → y=3 → (2,3). If x = 3.5 → 7+y=7 → y=0 → (3.5,0).

3. Graphing

Plot at least 3 solutions on the Cartesian plane. Join them → STRAIGHT LINE. 'Every point on the line is a solution. Every solution is a point on the line.'

x = k (k is constant)

A VERTICAL line. Parallel to y-axis. Every point on this line has x = k (y can be anything).

y = k

A HORIZONTAL line. Parallel to x-axis. Every point has y = k.

4. Graph of ax + by + c = 0

If a = 0: equation becomes by + c = 0 → y = −c/b → HORIZONTAL line.
If b = 0: equation becomes ax + c = 0 → x = −c/a → VERTICAL line.

5. Real-World Applications

Cost problems: 'Total cost = fixed cost + variable cost × quantity'
Temperature conversion: F = (9/5)C + 32 → 9C − 5F + 160 = 0
Distance-rate-time: d = rt

Common Mistakes

Thinking x=0 is a POINT — it's a LINE (the y-axis). The POINT is (0,y).
Plotting only TWO points — always plot THREE for verification.

More Worked Examples

Example — Finding Solutions: For 2x + 3y = 12, find 4 solutions. If x=0 → 3y=12 → y=4 → (0,4). If x=3 → 6+3y=12 → y=2 → (3,2). If x=6 → 12+3y=12 → y=0 → (6,0). If x=−3 → −6+3y=12 → y=6 → (−3,6).

Example — Graph and Find Area: Graph 4x + 3y = 12. x-intercept (y=0): 4x=12 → x=3 → (3,0). y-intercept (x=0): 3y=12 → y=4 → (0,4). Plot these two points + one more (e.g., x=1.5 → 6+3y=12 → y=2). Draw the line. The line forms a triangle with the axes. Area = ½×3×4 = 6 sq units.

Example — Word Problem: 'The cost of 3 pens and 5 pencils is ₹70. The cost of 5 pens and 3 pencils is ₹90.' Let pen = ₹x, pencil = ₹y. 3x+5y=70. 5x+3y=90. Solve by elimination: Multiply first by 5, second by 3 → 15x+25y=350, 15x+9y=270. Subtract → 16y=80 → y=5. Substitute → 3x+25=70 → x=15. Pen = ₹15, Pencil = ₹5.

Key Exam Tip

'For "draw the graph" questions: use GRAPH PAPER. Choose a SCALE that fills most of the paper. Label axes clearly (x and y). Plot points with small crosses (×). Draw a STRAIGHT line with a ruler. Write the equation on the graph.'

Detailed Graphing — Step by Step Method

Equation: 3x + 2y = 6. Express y in terms of x: 2y = 6 − 3x → y = 3 − (3/2)x. Create a table of values:

x	y = 3 − (3/2)x	Point (x,y)
0	3	(0, 3) — y-intercept
2	0	(2, 0) — x-intercept
4	−3	(4, −3)
−2	6	(−2, 6)

Plot these four points on graph paper. Draw a straight line through them. The line extends infinitely in both directions. 'Always find intercepts FIRST — they anchor your graph. Then add 1-2 more points for verification.'

Graph Analysis

x-intercept (where y=0): 3x=6 → x=2 → (2,0). This is where the line crosses the x-axis.
y-intercept (where x=0): 2y=6 → y=3 → (0,3). This is where the line crosses the y-axis.
The line forms a triangle with the coordinate axes. Area of this triangle = ½ × |x-intercept| × |y-intercept| = ½ × 2 × 3 = 3 sq units.

Types of Linear Equations and Their Graphs

Equation Type	Example	Graph Characteristics
ax + by + c = 0 (both a,b ≠ 0)	2x + 3y = 6	STRAIGHT LINE crossing BOTH axes
by + c = 0 (a = 0)	2y + 4 = 0 → y = −2	HORIZONTAL line, parallel to x-axis
ax + c = 0 (b = 0)	3x − 9 = 0 → x = 3	VERTICAL line, parallel to y-axis
ax + by = 0 (c = 0)	2x − y = 0 → y = 2x	Line PASSING THROUGH ORIGIN (0,0)

'When c = 0, the point (0,0) satisfies the equation → the line MUST pass through the origin. This is a favorite 1-mark conceptual question.'

Writing Equations from Given Conditions

Example 1 — Point lies on a line: If (2, −3) lies on 5x − ky = 16, find k. Substitute: 5(2) − k(−3) = 16 → 10 + 3k = 16 → 3k = 6 → k = 2.

Example 2 — Writing an equation from a graph: A line passes through (0,5) and (3,0). Write its equation. The intercepts are given: x-intercept = 3, y-intercept = 5. Equation in intercept form: x/3 + y/5 = 1. Multiply by 15: 5x + 3y = 15 → 5x + 3y − 15 = 0.

Example 3 — Express as ax + by + c = 0: y = 4x − 7. Rearrange: 4x − y − 7 = 0. So a = 4, b = −1, c = −7. The standard form requires ALL terms on LHS, RHS = 0.

Word Problems — From Statement to Equation

Problem 1 — Age relation: 'A father is 5 years more than 3 times his son's age.' Let son's age = x, father's age = y. y = 3x + 5 → 3x − y + 5 = 0.

Problem 2 — Two-digit number: 'The sum of the digits is 9.' Let tens digit = x, ones digit = y. x + y = 9 → x + y − 9 = 0.

Problem 3 — Auto-rickshaw fare: 'Fixed charge ₹25 plus ₹15 per km.' Let km = x, fare = y. y = 15x + 25 → 15x − y + 25 = 0.

Problem 4 — Bookshop: 'Cost of 4 notebooks and 3 pens is ₹145. Cost of 2 notebooks and 5 pens is ₹125.' Let notebook = ₹x, pen = ₹y. Equations: 4x + 3y = 145, 2x + 5y = 125. This is a SYSTEM of two linear equations in two variables — solved using elimination or substitution (Class 10 topic, but introduced here conceptually).

Exam Focus — AP Board (BSEAP)

Question Type	Marks
Check if a point is a solution	1
Find any 3 solutions	2-3
Draw the graph (on graph paper)	4-5
Write equation from word problem	2-3
Identify type of line (horizontal/vertical/through origin)	1-2

Common AP Exam Questions

'Express y = −2x + 5 in the form ax + by + c = 0.' (Answer: 2x + y − 5 = 0.)
'Check whether (2,1) lies on 3x + 2y = 8.' (Answer: 3(2)+2(1)=8 → Yes.)
'Draw the graph of 2x − y = 4.' (Plot intercepts + one more point. Full graph with labels.)
'Write the equation of a line passing through the origin.' (Answer: y = mx or ax + by = 0, where c = 0.)
'If the point (k, 3) lies on 2x + y = 7, find k.' (Answer: 2k+3=7 → k=2.)

Key Reminders for Graph Questions

Graph paper is COMPULSORY for 4-5 mark graph questions. Drawing freehand loses marks.
Mark the scale on both axes: '1 cm = 1 unit' or '1 cm = 2 units' as appropriate.
Plot points as small crosses (×) or dots surrounded by a small circle.
Extend the line beyond the plotted points. Add arrowheads at both ends.
Write the equation ON the line or next to it in the graph.
Title the graph: 'Graph of 3x + 2y = 6.'

Quick Self-Test

Express 2x = 3y + 5 in standard form. (Answer: 2x − 3y − 5 = 0.)
Is (0,0) a solution of x + y = 5? (Answer: No — 0+0 ≠ 5.)
What type of line is y − 3 = 0? (Answer: Horizontal line, parallel to x-axis.)
Find the x-intercept of 4x + 5y = 20. (Answer: set y=0 → x=5 → (5,0).)
Does the point (−1, 2) lie on 3x + y + 1 = 0? (Answer: 3(−1)+2+1=0 → Yes.)

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Linear Equations in Two Variables

STANDARD FORM: ax + by + c = 0, where a, b are not both zero. a, b, c are real numbers. EXAMPLES: 2x + 3y = 6 → 2x + 3y − 6 = 0. x − y + 4 = 0. SOLUTIONS: An ordered pair (x, y) is a solution if it satisfies the equation. A linear equation in TWO variables has INFINITELY MANY solutions. FINDING SOLUTIONS: Fix one variable, solve for the other. For 2x + y = 7: if x=0 → y=7, so (0,7) is a solution. If x=1 → y=5, so (1,5). If x=3 → y=1, so (3,1). GRAPHING: Step 1: Find x-intercept — set y=0, solve for x. Step 2: Find y-intercept — set x=0, solve for y. Step 3: Plot both points. Step 4: Draw a straight line through them. GRAPH IS A STRAIGHT LINE: Every solution is a point on the line. Every point on the line is a solution. SPECIAL CASES: x = k (constant): Vertical line parallel to y-axis. Passes through (k, 0). E.g., x = 3 is a vertical line. y = k (constant): Horizontal line parallel to x-axis. Passes through (0, k). E.g., y = −2 is a horizontal line. x = 0: the y-axis itself. y = 0: the x-axis itself. INTERCEPT FORM: x/a + y/b = 1 (a = x-intercept, b = y-intercept). WORD PROBLEMS SETUP: 'Sum of two numbers is 15' → x + y = 15. 'Cost of 2 pens and 3 books is ₹40' → 2p + 3b = 40. Any single linear equation has infinitely many solutions — need TWO equations for a unique solution (Class 10 topic). CHECKING A SOLUTION: Substitute the ordered pair and verify LHS = RHS. E.g., check (2, 3) for x + y = 5: 2 + 3 = 5 ✓.

AP EXAM KEY TRAPS: (1) A linear equation in ONE variable has exactly one solution. A linear equation in TWO variables has INFINITELY MANY solutions. (2) To graph a line, you only need TWO points — but always find a third to verify. (3) x = 3 is a VERTICAL line (parallel to y-axis). Students often confuse which is vertical/horizontal. Remember: x = constant fixes the x-value, so all points have x=3 → vertical. y = constant fixes y-value → horizontal. (4) The equation y = mx passes through the ORIGIN (0,0). Always check if (0,0) satisfies — if yes, you only need ONE more point. (5) WORD PROBLEM STRATEGY: Define variables, write the equation, find ordered pair solutions. Don't try to find a 'unique answer' — any point on the line is a valid solution.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Saying a linear equation in two variables has one solution, or trying to find a unique answer

✓ A linear equation in TWO variables (like 2x + y = 8) has INFINITELY MANY solutions. Every point on its graph is a solution. For example, (0,8), (1,6), (2,4), (3,2), (4,0) — all are valid solutions. There is NO unique answer. To find a unique solution, you need TWO simultaneous equations (a pair of linear equations) — that is the Class 10 topic. In Class 9, the answer to 'find solutions of 2x + y = 8' is a TABLE of ordered pairs or a GRAPH, not a single pair. Write at least 3 solutions for any graph-based question.

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Linear Equations in Two Variables?

1 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

1 questions~2 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•STANDARD FORM: A linear equation in two variables has the form ax + by + c = 0, where a, b, c are real numbers and a, b are not both zero. The 'linear' name comes from the fact that the graph is a straight LINE.
•INFINITELY MANY SOLUTIONS: A single linear equation in TWO variables (like 2x + y = 8) has INFINITELY MANY solutions. Every ordered pair (x, y) satisfying the equation is a solution. Geometrically, every point on the line represents a solution.
•FINDING SOLUTIONS: Fix one variable's value, solve for the other. For 2x + y = 7: x=0 → y=7 → (0,7); x=1 → y=5 → (1,5); x=3 → y=1 → (3,1). Make a table to organise solutions.
•GRAPH IS A STRAIGHT LINE: The graph of any linear equation ax+by+c=0 is a STRAIGHT LINE. Two points determine a line, so finding any two solutions and plotting them gives the entire line. A third point should be plotted to verify.
•FINDING INTERCEPTS (graphing method): X-INTERCEPT: Set y=0 in the equation, solve for x. The line crosses the x-axis at (x, 0). Y-INTERCEPT: Set x=0, solve for y. Line crosses y-axis at (0, y). Plot these two intercept points + one verification point.
•SPECIAL CASE — VERTICAL LINE: x = k (a constant). All points on the line have x-coordinate k. The line is PARALLEL to the y-axis. Examples: x = 3 (vertical line through (3,0)). x = 0 is the y-axis itself.
•SPECIAL CASE — HORIZONTAL LINE: y = k. All points on the line have y-coordinate k. Line is PARALLEL to the x-axis. Examples: y = 4 (horizontal line through (0,4)). y = 0 is the x-axis itself.
•WORD PROBLEMS: Define variables clearly. Translate the problem to an equation. Example: 'Sum of two numbers is 15' → x + y = 15. 'Cost of 2 pens and 3 books is ₹40' → 2p + 3b = 40. 'Father's age is twice his son's age' → x = 2y or x − 2y = 0.
•CHECKING A SOLUTION: Substitute the ordered pair (x, y) into the equation. If LHS = RHS, the pair is a solution. Example: check (2, 3) for x + y = 5: 2 + 3 = 5 ✓. Check (1, 2) for 2x + y = 7: 2(1)+2 = 4 ≠ 7. Not a solution.
•EVERY POINT IS A SOLUTION (AND VICE VERSA): Every point on the graph satisfies the equation. Every (x, y) pair satisfying the equation lies on the graph. This bi-directional correspondence between algebra (equation) and geometry (line) is the central insight of coordinate geometry.

Andhra Pradesh (BIEAP) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Budget and expense planning

Real-world budgets often follow linear equations. If a family budgets ₹500 for groceries, with rice at ₹40/kg and dal at ₹100/kg: 40r + 100d = 500 (where r = kg of rice, d = kg of dal). This linear equation has multiple solutions — they can buy 5 kg rice + 3 kg dal, or 10 kg rice + 1 kg dal, etc. Understanding multiple solutions to a budget equation helps in flexible planning.

Mixing problems and chemistry

Mixing two solutions of different concentrations to get a desired concentration follows a linear equation. If solution A is 10% acid and solution B is 30% acid, mixing x litres of A and y litres of B to get 5 litres of 20% solution: 0.1x + 0.3y = 0.2(5) = 1.0, and x + y = 5. These are two linear equations in two variables — exactly the setup that leads to Class 10's Pair of Linear Equations. AP's pharma and chemical industries use this mixing math daily.

Distance-time linear relationships

If two trains travel at different speeds and you want to know when they meet or how far each has gone, the relationships are LINEAR equations in distance and time. d = v × t. Plotting distance vs time gives a straight line whose slope is velocity. AP's Railways scheduling and route planning use exactly these linear relationships to ensure trains don't collide and arrive on time.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Graphing (3-4 marks): use the INTERCEPT method — find x-intercept and y-intercept, plus one verification point. Make a small table showing x, y values. Draw axes carefully, plot points, draw the line. Label the intercepts on the graph with their coordinates.
Finding solutions (2 marks): list at least 3-4 ordered pairs as a table. Format: 'When x = 0, y = ___; when x = 1, y = ___; when x = 2, y = ___.' Show the substitution to verify at least one.
Word problem setup (3-4 marks): explicitly DEFINE variables ('Let x = ..., y = ...'). Then write the equation showing the translation. Give 2-3 valid solutions. State that infinitely many solutions exist.
Special cases (1-2 marks): x = k is VERTICAL line parallel to y-axis. y = k is HORIZONTAL line parallel to x-axis. Always state both pieces — type (vertical/horizontal) and orientation (parallel to which axis).
Verification (1 mark): always check by substitution. 'LHS = ___, RHS = ___, so (x, y) IS / IS NOT a solution.' Don't just state the conclusion without showing the LHS = RHS check.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Research SLOPE-INTERCEPT form y = mx + c — an alternative way to write linear equations. m = slope (rise/run), c = y-intercept. Parallel lines have equal slopes; perpendicular lines have slopes whose product is −1 (m₁ × m₂ = −1). This form makes it easy to read line properties. Research the connection between slope-intercept form and standard form.
Investigate LINEAR PROGRAMMING — using systems of linear inequalities to solve optimisation problems (maximise profit, minimise cost, subject to constraints). LP is widely used in operations research, factory scheduling, transportation planning, and resource allocation. The Class 9 single linear equation is the building block — LP combines many of them.
Explore LINEAR ALGEBRA and matrices — when you have multiple linear equations in multiple variables, matrices provide an efficient way to solve them. Gaussian elimination, matrix inverses, and determinants are all techniques for handling systems of linear equations. These topics appear in Class 12 mathematics and engineering courses.
Research how DEEP LEARNING uses linear equations — every artificial neural network layer performs a linear transformation y = Wx + b (W is a matrix of weights, x is input, b is bias, y is output) followed by a non-linear activation. Modern AI is built on millions of linear equations being optimised simultaneously. The conceptual foundation begins with the Class 9 understanding of linear equations.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

AP Board SSC (Class 10) — Pair of Linear EquationsVery High — Class 9 single linear equation directly extends to Class 10 pair of equations (substitution, elimination, graphical methods)

JEE Main and AdvancedVery High — straight lines and systems of equations are core JEE Mathematics topics

NTSE (Mathematics)High — linear equations appear in NTSE Stage I and II

AP EAPCET (Mathematics)Very High — straight lines (slope, equations, intercepts) is a major chapter in EAPCET

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A linear equation in ONE variable (like 2x = 6) has only ONE unknown — solving gives one specific value (x = 3). Once x is determined, there's nothing else to find. A linear equation in TWO variables (like 2x + y = 8) has TWO unknowns. The equation constrains the relationship between them — but doesn't determine each individually. For any value you choose for x, you can solve for y. Choose x=0 → y=8; x=1 → y=6; x=2 → y=4; x=−1 → y=10; and so on, infinitely. Each (x, y) pair satisfying 2x+y=8 is a solution. To get a UNIQUE solution, you need a SECOND equation that adds another constraint — that's the Pair of Linear Equations chapter in Class 10.

The equation x = 3 means: the x-coordinate is always 3. The y-coordinate can be ANYTHING. So (3, 0), (3, 5), (3, −2), (3, 100), and infinitely many others all satisfy x = 3. When you plot ALL these points, you get a VERTICAL line passing through x = 3 on the x-axis. The y values span the entire range (−∞ to +∞). Similarly y = 4 is a horizontal line — x can be anything, y is always 4. KEY: a single equation in two variables fixes ONE constraint, leaving one degree of freedom — hence the line. A pair of equations would intersect at a single point.

USE THE INTERCEPT METHOD — fastest for graphing. Step 1: Find x-intercept by setting y=0. 3x + 0 = 12 → x = 4. So (4, 0). Step 2: Find y-intercept by setting x=0. 0 + 4y = 12 → y = 3. So (0, 3). Step 3: Plot both points on the graph. Step 4: Draw a straight line through them. Step 5 (RECOMMENDED): Find a third point to verify — try x=4, get y by formula. Or x=−4: 3(−4)+4y=12 → 4y=24 → y=6 → (−4, 6). Plot — should be on the line. If all three points are collinear, your line is correct. INTERCEPT METHOD is faster than picking random x-values because intercepts are guaranteed to be 'nice' axis points.

FOUR STEPS: (1) READ CAREFULLY and identify what is asked. (2) DEFINE VARIABLES with clear meaning. 'Let x = age of son, y = age of father.' (3) TRANSLATE the verbal relationship into an equation. 'Father is twice the son's age' → y = 2x. 'Sum of their ages is 60' → x + y = 60. (4) SOLVE or list solutions. Example: 'A two-digit number is such that the digit in the tens place is 3 more than the units digit.' Let units = u, tens = t. Translate: t = u + 3. This is a linear equation. Solutions: (0,3), (1,4), (2,5), ... → numbers 30, 41, 52, ... — multiple valid numbers.

There is a ONE-TO-ONE correspondence — a bi-directional bridge between algebra and geometry. (1) Every SOLUTION (x, y) to the equation corresponds to a POINT (x, y) on the graph. (2) Every POINT (x, y) on the graph corresponds to a SOLUTION (x, y) of the equation. So: the graph is the GEOMETRIC PICTURE of all solutions. If you have a graph and want a solution, just read off a point. If you have a solution and want to verify graphically, just plot the point and check it lies on the line. This bridge — algebra ↔ geometry — is the central insight that Descartes introduced and that underlies all modern mathematics.

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