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

  • 1Represent a pair of linear equations graphically and interpret the intersection
  • 2Classify a pair as consistent (unique / infinite) or inconsistent using coefficient ratios
  • 3Solve a pair by substitution, elimination and cross-multiplication
  • 4Reduce equations of the form a/x + b/y = c to linear form and solve
  • 5Translate word problems (ages, digits, speed, fractions) into equations and solve them
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Why this chapter matters
A high-yield algebra chapter: it almost always carries a short 'method' question plus a long word problem worth 4–5 marks. The coefficient-ratio consistency test is a standing one-mark question, and the solving methods reappear throughout higher maths.

Before you start — revise these

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

Pair of Linear Equations in Two Variables — RBSE Class 10 (Mathematics)

One equation in two unknowns has endless answers — a whole line of them. Pair it with a second equation and, usually, exactly one point satisfies both: the place where the two lines cross. This chapter is about finding that point cleanly, and about spotting the special cases where the lines never meet or lie exactly on top of each other.


1. What a pair looks like

A linear equation in two variables has the form (with a, b not both zero). A pair is two such equations:

A solution is a pair that satisfies both. Graphically each equation is a straight line, so a solution is a point of intersection.


2. Three geometric outcomes

Compare the ratios of the coefficients:

ConditionLinesSolutionsCalled
intersectingexactly oneconsistent (independent)
parallelnoneinconsistent
coincidentinfinitely manyconsistent (dependent)

This ratio test is a favourite one-mark question — memorise it.


3. Algebraic methods

Substitution

Make one variable the subject in one equation, substitute into the other. Best when a coefficient is already .

Elimination

Multiply the equations to match the coefficient of one variable, then add or subtract to eliminate it. The most reliable all-purpose method.

Example — solve and . Subtract: . Then .

Cross-multiplication

For and : A direct formula — useful when the numbers are awkward, provided .


4. Equations reducible to linear form

Some equations look non-linear but become linear after a substitution. For example, with , put to get — an ordinary linear pair. Solve for u, v, then invert to get x, y.


5. Word problems — the real exam prize

The five-mark questions are almost always word problems: ages, two-digit numbers, speed of boat/stream, fractions, and geometry. The skill is translation:

  1. Name the two unknowns clearly (let the number be , etc.).
  2. Turn each sentence into one equation.
  3. Solve by elimination (safest), then check against the original words.

Example — the sum of a two-digit number and the number formed by reversing its digits is 66; the digits differ by 2. Find the number. Let digits be x (tens) and y (units): ; and . Solving gives or → the number is 42 or 24.


6. Closing thought

Every method here lands at the same place — the crossing point of two lines. Pick elimination as your default, keep the ratio test ready for "how many solutions" questions, and treat word problems as translation exercises. In the RBSE board this chapter reliably carries a short algebra question and a long word problem, so it is worth a large share of your practice time.

Key formulas & results

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

General pair
a₁x+b₁y+c₁=0, a₂x+b₂y+c₂=0
Two lines in the plane.
Unique solution
a₁/a₂ ≠ b₁/b₂
Intersecting lines — consistent, independent.
No solution
a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Parallel lines — inconsistent.
Infinite solutions
a₁/a₂ = b₁/b₂ = c₁/c₂
Coincident lines — consistent, dependent.
Cross-multiplication
x/(b₁c₂−b₂c₁) = y/(c₁a₂−c₂a₁) = 1/(a₁b₂−a₂b₁)
Valid when a₁b₂−a₂b₁ ≠ 0.
⚠️

Common mistakes & fixes

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

WATCH OUT
Confusing the parallel and coincident conditions
Both need a₁/a₂ = b₁/b₂. Parallel (no solution) ALSO needs ≠ c₁/c₂; coincident (infinite) needs = c₁/c₂ as well.
WATCH OUT
Sign errors in cross-multiplication
Write all equations as = 0 first and use the formula exactly; a single sign slip flips the answer. Cross-check by substituting back.
WATCH OUT
Not simplifying reducible equations before solving
For 2/x + 3/y = 13, substitute u = 1/x, v = 1/y FIRST, solve for u, v, then take reciprocals — don't clear fractions blindly.
WATCH OUT
Forgetting to answer the actual question in word problems
After finding x and y, re-read the question — it may ask for the number (10x+y), the sum of ages, etc., not x and y themselves.
WATCH OUT
Reading consistency off the graph carelessly
Plot at least two accurate points per line; a rough sketch can make intersecting lines look parallel.

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Consistency
For what value of k do the equations 2x + 3y = 4 and (k+2)x + 6y = 3k+2 represent intersecting lines? State the condition.
Show solution
Step 1 — Intersecting lines need a₁/a₂ ≠ b₁/b₂. Step 2 — 2/(k+2) ≠ 3/6 = 1/2 ⇒ (k+2) ≠ 4 ⇒ k ≠ 2. ✦ Answer: any k ≠ 2.
Q2EASY· Method
Solve x + y = 14 and x − y = 4.
Show solution
Step 1 — Add: 2x = 18 ⇒ x = 9. Step 2 — y = 14 − 9 = 5. ✦ Answer: x = 9, y = 5.
Q3EASY· Classify
Are the lines 3x + 2y = 5 and 6x + 4y = 9 consistent?
Show solution
Step 1 — 3/6 = 2/4 = 1/2 but 5/9 ≠ 1/2, so a₁/a₂ = b₁/b₂ ≠ c₁/c₂. Step 2 — Parallel lines ⇒ no solution. ✦ Answer: Inconsistent (no solution).
Q4MEDIUM· Elimination
Solve 2x + 3y = 11 and 2x − 4y = −24 by elimination.
Show solution
Step 1 — Subtract: (3y − (−4y)) = 11 − (−24) ⇒ 7y = 35 ⇒ y = 5. Step 2 — 2x + 15 = 11 ⇒ x = −2. ✦ Answer: x = −2, y = 5.
Q5MEDIUM· Substitution
Solve s − t = 3 and s/3 + t/2 = 6 by substitution.
Show solution
Step 1 — s = t + 3. Step 2 — (t+3)/3 + t/2 = 6 ⇒ multiply by 6: 2(t+3) + 3t = 36 ⇒ 5t + 6 = 36 ⇒ t = 6. Step 3 — s = 9. ✦ Answer: s = 9, t = 6.
Q6MEDIUM· Reducible
Solve 2/x + 3/y = 13 and 5/x − 4/y = −2.
Show solution
Step 1 — Let u = 1/x, v = 1/y: 2u + 3v = 13 and 5u − 4v = −2. Step 2 — Multiply: (×4) 8u + 12v = 52; (×3) 15u − 12v = −6. Add: 23u = 46 ⇒ u = 2. Step 3 — 2(2) + 3v = 13 ⇒ v = 3. Step 4 — x = 1/u = 1/2, y = 1/v = 1/3. ✦ Answer: x = 1/2, y = 1/3.
Q7MEDIUM· Infinite-solution
Find k so that x + 2y = 3 and 5x + ky + 15 = 0 have infinitely many solutions.
Show solution
Step 1 — Write both as = 0: x + 2y − 3 = 0 and 5x + ky + 15 = 0. Step 2 — Need 1/5 = 2/k = −3/15 = −1/5. Step 3 — But 1/5 ≠ −1/5, so there is NO value of k giving infinite solutions. ✦ Answer: no such k exists (the constant ratio contradicts the coefficient ratio).
Q8HARD· Word problem (digits)
The sum of a two-digit number and the number formed by reversing its digits is 66. The digits differ by 2. Find the number(s).
Show solution
Step 1 — Let tens digit x, units y: (10x+y)+(10y+x)=66 ⇒ 11(x+y)=66 ⇒ x+y=6. Step 2 — Digits differ by 2: x − y = ±2. Step 3 — x+y=6, x−y=2 ⇒ x=4, y=2 (number 42); x−y=−2 ⇒ x=2, y=4 (number 24). ✦ Answer: 42 or 24.
Q9HARD· Word problem (speed)
A boat goes 30 km upstream and 44 km downstream in 10 hours. It goes 40 km upstream and 55 km downstream in 13 hours. Find the speed of the boat in still water and of the stream.
Show solution
Step 1 — Let boat speed x, stream y: upstream (x−y), downstream (x+y). Put u = 1/(x−y), v = 1/(x+y). Step 2 — 30u + 44v = 10 and 40u + 55v = 13. Step 3 — Solve: (×4) 120u+176v=40; (×3) 120u+165v=39 ⇒ 11v = 1 ⇒ v = 1/11 ⇒ x+y = 11. Step 4 — 30u + 44/11 = 10 ⇒ 30u = 6 ⇒ u = 1/5 ⇒ x−y = 5. Step 5 — x+y=11, x−y=5 ⇒ x=8, y=3. ✦ Answer: boat 8 km/h, stream 3 km/h.
Q10HARD· Word problem (ages)
Five years ago Nuri was thrice as old as Sonu. Ten years later Nuri will be twice as old as Sonu. Find their present ages.
Show solution
Step 1 — Let present ages Nuri x, Sonu y: (x−5) = 3(y−5) ⇒ x − 3y = −10. Step 2 — (x+10) = 2(y+10) ⇒ x − 2y = 10. Step 3 — Subtract: −y = −20 ⇒ y = 20; then x = 10 + 2(20) = 50. ✦ Answer: Nuri 50 years, Sonu 20 years.

5-minute revision

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

  • A solution of a pair is the point where the two lines meet.
  • Unique solution: a₁/a₂ ≠ b₁/b₂ (intersecting).
  • No solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂ (parallel).
  • Infinite solutions: a₁/a₂ = b₁/b₂ = c₁/c₂ (coincident).
  • Methods: substitution, elimination, cross-multiplication — all give the same point.
  • Reducible equations: substitute u = 1/x, v = 1/y, solve, then invert.
  • Word problems: name unknowns, one equation per condition, solve, verify against the words.

Rajasthan (RBSE) marks blueprint

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

Typical chapter weightage: 5–6 marks

Question typeMarks eachTypical countWhat it tests
MCQ / very short11Consistency from coefficient ratios
Short answer2–31Solve by elimination/substitution; reducible forms
Long answer4–51Word problem (age/digit/speed/fraction)
Prep strategy
  • Make elimination your default method — it rarely fails
  • Memorise the three coefficient-ratio cases for the consistency question
  • Drill one word problem of each type: ages, digits, boat-and-stream, fractions
  • Always substitute the answer back into the original equations to catch sign errors

Where this shows up in the real world

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

Cost and break-even

Two cost/revenue lines meet at the break-even quantity — a linear-pair solution.

Mixtures and pricing

Finding quantities of two items given a total and a total cost is a standard pair of equations.

Speed, time and current

Boat-and-stream and aeroplane-and-wind problems model with upstream/downstream speeds.

Supply and demand

Economics finds equilibrium where the supply line crosses the demand line.

Exam strategy

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

  1. Show every step of elimination — method marks are awarded even if arithmetic slips.
  2. For consistency questions, quote the ratio condition explicitly before concluding.
  3. In word problems, clearly define the variables in words before forming equations.
  4. End word problems with the quantity actually asked, in the correct units.
  5. Verify the final (x, y) in BOTH original equations before writing the answer.

Going beyond the textbook

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

  • Systems of three equations in three unknowns and when they are consistent.
  • Determinants (Cramer's rule) as the reason cross-multiplication works.
  • Diophantine (integer-only) solutions of ax + by = c.
  • Geometric meaning: rank of a system and dimension of its solution set.

Where else this chapter is tested

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

RBSE Class 10 Board (BSER Ajmer)High — a method question plus a long word problem almost every year
NTSE / state scholarshipMedium — algebra MCQs
JEE FoundationMedium — linear systems feed into matrices later
Maths Olympiad (IMO/NMTC)Medium — word problems and integer solutions

Questions students ask

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

Yes. RBSE (BSER, Ajmer) prescribes the NCERT Mathematics textbook, so chapters and exercises match the national syllabus; RBSE sets its own exam pattern.

Elimination is the safest default. Use substitution when a coefficient is ±1, and cross-multiplication when the numbers are awkward or the question specifically asks for it.

Write both equations as = 0 and compare a₁/a₂, b₁/b₂, c₁/c₂. Ratios all equal → infinite; first two equal but not the third → none; first two unequal → exactly one.

Equations with 1/x or 1/y terms. Substituting u = 1/x and v = 1/y turns them into an ordinary linear pair; solve, then take reciprocals to get x and y.
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