Chapter 10 of 19

53%

Andhra Pradesh (BIEAP)Class 10 Mathematics★ AP Board SSC Mathematics · 8 marks · Paper: 4-mark word problem + 4-mark solve-by-method + 2-mark consistency check

Pair of Linear Equations in Two Variables

Solve PAIRS of linear equations using graphical, substitution, elimination, and cross-multiplication methods. Understand consistent, inconsistent, and dependent systems. AP Board Class 10 Mathematics.

⏱ 12 min readMedium difficulty✓ Free · NCERT-aligned

Worksheet

Podcast

Listen as a conversation

A teacher & a student talk through the chapter — in your language.

Mind Map

The chapter at a glance

Every concept and exam-favourite branch — revisable in 5 minutes.

…

Study Guide

Night-before notes

Key points, glossary and the questions most likely to show up.

…

VocabularyBuild your vocabularyLexiGenome mines the key words and schedules them with spaced repetition.Mine words →

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

1Solve a pair of linear equations by graphical method (nature of solution from intersection/parallel/coincident)
2Solve by substitution, elimination, and cross-multiplication methods
3Determine consistency by comparing ratios a₁/a₂ : b₁/b₂ : c₁/c₂
4Reduce non-standard equations (with 1/x, 1/y) to standard form for solving
5Set up and solve word problems: age, digits, fractions, speed-distance-time, coins

💡

Why this chapter matters

Pair of Linear Equations is one of the highest-scoring algebra chapters in AP SSC — typically 6–8 marks per paper. Students must set up and solve simultaneous equations using elimination, substitution, and cross-multiplication. Word problems (age, money, speed, fractions) that require two equations are near-certain exam questions. The consistency check (comparing a₁/a₂, b₁/b₂, c₁/c₂) is a reliable 2-mark question.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

🔗

Linear equations in one variable (Class 9)

One-variable equations — foundation for substitution method

🔗

Basic algebra and transposition (Class 8)

Rearranging equations — needed in every solving method

Pair of Linear Equations in Two Variables

Standard Form: a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0.

Solving Methods

Method	How
Graphical	Plot both lines. Intersection point = solution.
Substitution	Express one variable in terms of the other. Substitute.
Elimination	Make coefficients equal. Add/subtract to eliminate.
Cross-Multiplication	x/(b₁c₂−b₂c₁) = y/(c₁a₂−c₂a₁) = 1/(a₁b₂−a₂b₁)

Nature of Solutions

Condition	Solution Type	Geometric Meaning
a₁/a₂ ≠ b₁/b₂	UNIQUE solution	Lines INTERSECT
a₁/a₂ = b₁/b₂ ≠ c₁/c₂	NO solution	Lines PARALLEL
a₁/a₂ = b₁/b₂ = c₁/c₂	INFINITE solutions	Lines COINCIDE

Word Problems: Translate into TWO equations. Solve. Check for reasonableness.

Solving Methods — Worked Examples

1. Graphical Method

Example 1: Solve graphically: x + y = 5 and x − y = 1.

Step 1: Write each equation in slope-intercept form (y = mx + c). y = 5 − x. y = x − 1.

Step 2: Create tables of values.

For y = 5 − x: x: 0, 1, 2, 3, 4, 5 y: 5, 4, 3, 2, 1, 0

For y = x − 1: x: 0, 1, 2, 3, 4, 5 y: −1, 0, 1, 2, 3, 4

Step 3: Plot both lines on the same graph. Intersection point = (3, 2). Verify: 3+2=5 ✓, 3−2=1 ✓.

'For the graphical method to be accurate, use graph paper. The intersection point is the solution.'

2. Substitution Method

Example 2: Solve using substitution: 2x + 3y = 11 and x − 2y = −12.

Step 1: Express one variable. From the second equation: x = 2y − 12.

Step 2: Substitute. 2(2y−12) + 3y = 11 → 4y − 24 + 3y = 11 → 7y = 35 → y = 5.

Step 3: Back-substitute. x = 2(5) − 12 = 10 − 12 = −2.

Solution: x = −2, y = 5.

Example 3 — Another Substitution: Solve 3x − 5y = −1 and x − y = −1. x = y − 1. Substitute: 3(y−1) − 5y = −1 → 3y − 3 − 5y = −1 → −2y = 2 → y = −1. x = −1 − 1 = −2. Solution: x = −2, y = −1.

3. Elimination Method

Example 4: Solve using elimination: 3x + 2y = 11 and 2x + 3y = 4.

Step 1: Make coefficients equal. Multiply first equation by 2 and second by 3: 6x + 4y = 22 6x + 9y = 12

Step 2: Subtract. (6x+4y) − (6x+9y) = 22 − 12 → −5y = 10 → y = −2.

Step 3: Back-substitute. 3x + 2(−2) = 11 → 3x − 4 = 11 → 3x = 15 → x = 5.

Solution: x = 5, y = −2.

Example 5 — Addition Instead: Solve 2x − 3y = 7 and 5x + 3y = 28. Add the equations: 7x = 35 → x = 5. Back-substitute: 2(5) − 3y = 7 → 10 − 3y = 7 → y = 1. Solution: x = 5, y = 1.

4. Cross-Multiplication Method

Example 6: Solve by cross-multiplication: 3x − 4y = 10 and 4x + 3y = 5.

Rewrite in standard form: 3x − 4y − 10 = 0 and 4x + 3y − 5 = 0. a₁=3, b₁=−4, c₁=−10; a₂=4, b₂=3, c₂=−5.

x/(b₁c₂−b₂c₁) = [−4×(−5) − 3×(−10)] = 20 + 30 = 50. y/(c₁a₂−c₂a₁) = [(−10)×4 − (−5)×3] = −40 + 15 = −25. 1/(a₁b₂−a₂b₁) = [3×3 − 4×(−4)] = 9 + 16 = 25.

x = 50/25 = 2. y = −25/25 = −1. Solution: x = 2, y = −1.

Nature of Solutions — Detailed Examples

Consistent (Unique): a₁/a₂ ≠ b₁/b₂. Example: 2x+3y=8, 3x+2y=7. 2/3 ≠ 3/2 → unique solution.

Inconsistent (No Solution): a₁/a₂ = b₁/b₂ ≠ c₁/c₂. Example: 2x+3y=8, 4x+6y=15. 2/4 = 3/6 ≠ 8/15 → parallel lines, no solution.

Dependent (Infinite): a₁/a₂ = b₁/b₂ = c₁/c₂. Example: 2x+3y=8, 4x+6y=16. 2/4 = 3/6 = 8/16 → coincident lines, infinite solutions.

Word Problems — With Solutions

Age Problems

Example 7: The sum of the ages of a father and son is 55. Five years ago, the father's age was twice the son's age. Find their present ages. Let father = x, son = y. x + y = 55. (x−5) = 2(y−5) → x−5 = 2y−10 → x−2y = −5. Subtract: (x+y) − (x−2y) = 55 − (−5) → 3y = 60 → y = 20. x = 35. Father is 35, son is 20.

Digit Problems

Example 8: A two-digit number is 4 times the sum of its digits. The number obtained by reversing the digits is 27 more than the original. Find the number. Let tens digit = x, units digit = y. Number = 10x + y. 10x+y = 4(x+y) → 10x+y = 4x+4y → 6x = 3y → y = 2x. Reversed: 10y+x = (10x+y) + 27 → 10(2x)+x = 10x+2x+27 → 21x = 12x+27 → 9x = 27 → x = 3. y = 6. Number = 36.

Fraction Problems

Example 9: The denominator of a fraction exceeds the numerator by 4. If 3 is added to both numerator and denominator, the fraction becomes 4/5. Find the fraction. Let numerator = x, denominator = y. y = x + 4. (x+3)/(y+3) = 4/5. Cross-multiply: 5(x+3) = 4(y+3) → 5x+15 = 4y+12 → 5x−4y = −3. Substitute y = x+4: 5x−4(x+4) = −3 → 5x−4x−16 = −3 → x = 13. y = 17. Fraction = 13/17.

Speed-Distance-Time

Example 10: A boat travels 30 km upstream and 44 km downstream in 10 hours. It travels 40 km upstream and 55 km downstream in 13 hours. Find the speed of the boat in still water and the speed of the stream. Let speed in still water = x km/h, stream speed = y km/h. Upstream = x−y, downstream = x+y. 30/(x−y) + 44/(x+y) = 10. 40/(x−y) + 55/(x+y) = 13. Let u = 1/(x−y), v = 1/(x+y). 30u+44v=10. 40u+55v=13. Solve: Multiply first by 4: 120u+176v=40. Second by 3: 120u+165v=39. Subtract: 11v=1 → v=1/11. Then u=(10−44/11)/30 = (10−4)/30 = 6/30 = 1/5. x+y=11, x−y=5 → x=8, y=3. Boat speed = 8 km/h, stream speed = 3 km/h.

AP Exam Focus

Topic	Marks	Frequency
Solving by elimination/substitution	4	Very Common
Nature of solutions (consistent/inconsistent)	2-3	Common
Word problems (age, digits, fractions)	5	Very Common
Speed-distance-time word problems	5	Common
Cross-multiplication method	3-4	Moderate

Self-Test Questions

Solve: 2x + y = 10 and x − y = 2. (Answer: x=4, y=2)
Check consistency: x+2y=5, 2x+4y=10. (Answer: 1/2=2/4=5/10 → Dependent, infinite solutions)
Five years ago, A was twice as old as B. Ten years hence, A will be 1.5 times B. Find their ages. (Answer: A=25, B=15)
The sum of digits of a two-digit number is 9. Reversing gives a number 27 less than the original. Find the number. (Answer: 63)
Solve by cross-multiplication: 2x+3y=7, 3x+2y=8. (Answer: x=2, y=1)
A fraction becomes 9/11 if 2 is added to both numerator and denominator. It becomes 5/6 if 3 is added to both. Find the fraction. (Answer: 7/9)
A boat takes 6 hours to go 24 km upstream and 36 km downstream. It takes 7 hours to go 30 km upstream and 40 km downstream. Find speeds. (Answer: boat=10 km/h, stream=2 km/h)

Key formulas & results

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

Standard form

a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

All terms moved to LHS. This form is needed for the consistency check and cross-multiplication.

Consistent (unique solution)

a₁/a₂ ≠ b₁/b₂ → lines INTERSECT

One solution. The MOST common case in exam problems.

Inconsistent (no solution)

a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → lines PARALLEL

No solution. System is contradictory.

Dependent (infinite solutions)

a₁/a₂ = b₁/b₂ = c₁/c₂ → lines COINCIDENT

Infinitely many solutions. One equation is a multiple of the other.

Cross-multiplication method

x/(b₁c₂ − b₂c₁) = y/(c₁a₂ − c₂a₁) = 1/(a₁b₂ − a₂b₁)

From standard form a₁x + b₁y + c₁ = 0. Subscripts follow the cyclic pattern.

Substitution method

Express one variable from one equation; substitute into the other

Best when one equation is simple (e.g., x = 3y − 5).

Elimination method

Multiply equations to make coefficients of one variable equal; add or subtract

Best when coefficients are comparable. Most-used method in AP exams.

⚠️

Common mistakes & fixes

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

WATCH OUT

✗ Sign errors when subtracting equations in elimination

✓ When SUBTRACTING equation 2 from equation 1: EVERY term of equation 2 changes sign. Write the negated equation explicitly before adding. Example: eq1 − eq2 means each term of eq2 is negated.

WATCH OUT

✗ In cross-multiplication, confusing the positions of b₁c₂ − b₂c₁ and c₁a₂ − c₂a₁

✓ Draw the 2×3 matrix with coefficients. Apply the cross-multiplication visual: b₁ × c₂ − b₂ × c₁ for x numerator. Practice the standard form consistently — it eliminates confusion.

WATCH OUT

✗ In word problems, forming only ONE equation instead of two

✓ Every pair-of-equations problem gives TWO pieces of information. Find BOTH constraints before solving. Define TWO variables clearly (e.g., 'Let father's age = x, son's age = y').

WATCH OUT

✗ Forgetting to check the answer makes physical sense (negative age, negative speed)

✓ For word problems: after solving, verify both equations AND check real-world validity (ages must be positive, etc.).

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 Pair of Linear Equations in Two Variables?

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

6 questions~4 min worth ~8 marks in Andhra Pradesh (BIEAP) exams

5-minute revision

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

•Standard form: a₁x + b₁y + c₁ = 0 (all terms to LHS). Required for consistency check and cross-multiplication.
•Consistency check: compare a₁/a₂, b₁/b₂, c₁/c₂. All unequal → unique. First two equal but third different → no solution. All three equal → infinite solutions.
•Elimination: multiply equations to equalise ONE variable's coefficient; add or subtract.
•Substitution: isolate one variable from the SIMPLER equation; substitute into the other.
•Cross-multiplication: x/(b₁c₂−b₂c₁) = y/(c₁a₂−c₂a₁) = 1/(a₁b₂−a₂b₁). Care with signs.
•Word problems: define BOTH variables at the start; write TWO equations from the given conditions; verify both equations AND the physical interpretation.
•Boat problems: let u = 1/(x−y) and v = 1/(x+y) to convert to linear form.
•Age problems: 'n years ago' → subtract n from both ages. 'n years hence' → add n to both ages.
•Digit problems: two-digit number = 10×(tens digit) + units digit. Reversed = 10×(units) + tens.
•Fraction problems: if denominator exceeds numerator by k, write d = n + k; use given condition for second equation.

Andhra Pradesh (BIEAP) marks blueprint

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

Typical chapter weightage: 8 marks

Question type	Marks each	Typical count	What it tests
Very Short Answer (VSA)	1	2	Definition, formula recall
Short Answer (SA)	2	2	Direct application
Long Answer (LA)	4	1	Multi-step problem with full working

Prep strategy

AP SSC board rewards clearly shown working — never skip steps
Memorise all formulas: AP board MCQs test formula recall directly
Practise 4-mark questions with full working under timed conditions
For word problems: identify variables, write equation, solve, state answer with units

Where this shows up in the real world

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

Economics — supply and demand

Market equilibrium price and quantity are found by solving two simultaneous linear equations.

Navigation — GPS triangulation

Finding a location from two or more signal sources involves solving simultaneous equations.

Chemistry — mixture problems

Mixing solutions of different concentrations uses two-variable linear equations to find required volumes.

Profit and loss analysis

Breakeven analysis in business uses pairs of linear equations: revenue = cost.

Exam strategy

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

Elimination is the fastest method for clean numbers. Practice it until you can solve in under 2 minutes.
Always VERIFY your answer in BOTH original equations before writing the final answer. This earns full marks and catches calculation errors.
Word problems: write the definition 'Let x = ..., y = ...' first. Then write both equations LABELLED as (1) and (2). Examiners follow your working this way.
Consistency check shortcut: form the 2×3 matrix [a₁ b₁ c₁; a₂ b₂ c₂]. If rank 2 → unique. Rank 1 → infinite. Rank mismatch → no solution.
For boat/stream or pipe problems: ALWAYS substitute u = 1/(x−y), v = 1/(x+y) first. This converts a rational system to a standard linear one.

Going beyond the textbook

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

A system of two linear equations in two unknowns has a unique solution iff the coefficient matrix has determinant ≠ 0 (i.e., a₁b₂ − a₂b₁ ≠ 0) — this is Cramer's Rule.
For n equations in n unknowns: Gaussian elimination generalises the AP SSC methods to larger systems.
Non-linear systems (like pairs involving xy terms) can sometimes be converted to linear systems via substitution — a technique used in EAMCET and JEE.

Where else this chapter is tested

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

AP Board SSC6-8 marks guaranteed — solving methods and word problems appear in every paper

EAMCETSystem of equations; simultaneous linear equations appear in EAMCET Maths

JEE FoundationLinear systems as part of algebra; builds toward matrices in Class 11-12

NTSEWord problems involving pairs of linear equations are common in SAT section

AI helpers for this chapter

Generate more practice on the fly, or have the chapter explained at a different level.

⚡

Generate 5 more questions

Fresh MCQs + short-answer items based on this chapter, plus answer keys.

🧠

Explain at a different level

Re-explain the chapter at the level you actually want.

📝 My notes

Saved on this device — sign in to sync · how this works

0/600

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Elimination: when coefficients are simple numbers easily made equal. Substitution: when one equation expresses one variable simply (like x = 2y + 3). Cross-multiplication: when asked specifically, or when you want a systematic formula-based approach.

When you try to solve parallel lines by elimination, you get a contradiction like 0 = 5. This is the algebraic signature of an inconsistent system.

The two equations represent the SAME line — one is a scalar multiple of the other. Any point on that line is a solution. Algebraically, you end up with 0 = 0 after elimination.

Write your TWO equations, then verify they are INDEPENDENT (not multiples of each other). Then substitute the solution BACK into the original word problem description to confirm it makes sense.

If a₁b₂ − a₂b₁ = 0, the denominator is zero — the lines are either parallel or coincident. In this case, use the consistency check to decide which.

Practice more, ask doubts, find a tutor

Andhra Pradesh (BIEAP) Class 10 mock tests

Andhra Pradesh (BIEAP) Class 10 PYQs

Ask in Q&A

Mathematics flashcards

Related chapters

Students who read Pair of Linear Equations in Two Variables usually read these next.

Mathematics — Quadratics, Progressions, Trigonometry & Mensuration (SSC)

Complete AP SSC Mathematics: QUADRATIC EQUATIONS (formula, discriminant, nature of roots). ARITHMETIC and GEOMETRIC PROGRESSIONS. TRIGONOMETRY (ratios, identities, heights and distances). COORDINATE GEOMETRY. MENSURATION of 3D solids. STATISTICS (mean, median, mode, ogive). PROBABILITY. BSEAP Class 10 Mathematics.

Triangles

Complete TRIANGLES for AP Class 10 — SIMILARITY of triangles (AAA, SSS, SAS criteria). Basic Proportionality Theorem (Thales Theorem) and its converse. Pythagoras Theorem proof and applications. Areas of similar triangles. AP Board Class 10 Mathematics — New 2025-26 NCERT-aligned syllabus.

Coordinate Geometry

Complete COORDINATE GEOMETRY for AP Class 10 — DISTANCE FORMULA derivation and applications. SECTION FORMULA (internal division, midpoint). Area of a TRIANGLE from coordinates. Checking COLLINEARITY. New 2025-26 NCERT-aligned syllabus. AP Board Class 10 Mathematics.

Quadratic Equations

Master QUADRATIC EQUATIONS — the quadratic formula, discriminant, nature of roots, and solving word problems. The most important algebra topic in AP SSC Mathematics.

🧑‍🏫