Pair of Linear Equations — Class 10 Notes

CBSE 2027 — dates, rules and official papers →

Maths · Ch 3Class 10 · CBSE 2026–27High weightage

Pair of Linear Equations in Two Variables

What the board actually asks: solving a pair by substitution and by elimination, reading a solution off a graph, and classifying a pair as consistent / inconsistent / dependent from the coefficient ratios — the two algebraic methods plus a word-problem application carry almost every mark here.

NCERT VERBATIM sourced, word-for-wordAUTHORED LazyTopper teaching scaffolding

The Big Idea AUTHORED

Two conditions, one point.

A pair of linear equations in two variables is two straight-line equations $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ that you solve together. Geometrically each is a line, so the whole chapter answers one question: where do the two lines meet? They can cross once (one solution), run parallel (no solution), or lie on top of each other (infinitely many). You learn to find that meeting point three ways — by graph, by substitution, and by elimination — and to predict the outcome just from the coefficient ratios.

Formulae

General form

a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0

Unique solution (intersecting)

\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}

No solution (parallel)

\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}

Infinitely many (coincident)

\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}

Definitions the board rewards NCERT VERBATIM

State these in NCERT’s exact words

Consistent pair of linear equations

NCERT verbatim — memorise this exact wording· Ch 3,

“A pair of linear equations in two variables, which has a solution, is called a consistent pair of linear equations.”

What it means: any pair that HAS at least one solution — the lines either cross once (unique) or coincide (infinitely many). “Has a solution” is the whole test.

Inconsistent pair of linear equations

NCERT verbatim — memorise this exact wording· Ch 3,

“A pair of linear equations which has no solution, is called an inconsistent pair of linear equations.”

What it means: the lines are parallel and never meet, so there is NO common

(x, y)

. In working, a false statement like

0 = 9

signals this.

Dependent pair of linear equations

NCERT verbatim — memorise this exact wording· Ch 3,

“A pair of linear equations which are equivalent has infinitely many distinct common solutions. Such a pair is called a dependent pair of linear equations in two variables. Note that a dependent pair of linear equations is always consistent.”

What it means: the two equations are really the SAME line (coincident), so every point on it solves both — infinitely many solutions. A dependent pair is a special case of consistent, never inconsistent.

Substitution method

NCERT verbatim — memorise this exact wording· Ch 3,

“We have substituted the value of one variable by expressing it in terms of the other variable to solve the pair of linear equations. That is why the method is known as the substitution method.”

What it means: isolate one variable in one equation (

x = \dots

), then substitute it into the other so only one variable is left. Fastest when a coefficient is already

1

Elimination method

NCERT verbatim — memorise this exact wording· Ch 3,

“The method used in solving the example above is called the elimination method, because we eliminate one variable first, to get a linear equation in one variable.”

What it means: multiply the equations so one variable’s coefficients match, then add or subtract to cancel that variable. Best when neither variable is already isolated.

Key terms — NCERT verbatim (state word-for-word)

Graph = two lines: The graph of a pair of linear equations in two variables is represented by two lines. (Ch 3, )
Algebraic methods: Algebraic Methods : We have discussed the following methods for finding the solution(s) of a pair of linear equations : (i) Substitution Method (ii) Elimination Method (Ch 3, )

Must-know concepts AUTHORED FRAMING

…and where they’re tested

Graphical method — three cases

Plot both lines. They either intersect at one point (unique solution, consistent), coincide (infinitely many, dependent), or are parallel (no solution, inconsistent). The meeting point read off the graph IS the solution.

NCERT Fig. 3.1The intersecting case — x + 3y = 6 and 2x – 3y = 12 meet at B(6, 0), the unique solution (Example 1).

Graphically solve / find the triangle formed with an axis — 3 marks

Classify from the coefficient ratios

For

a_{1} x + b_{1} y + c_{1} = 0

and

a_{2} x + b_{2} y + c_{2} = 0

\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}

→ intersecting (one solution) ·

\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}

→ parallel (none) ·

\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}

→ coincident (infinitely many). Compare the ratios best on a diagram:

FigureOne pair, three fates — the graph of a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0

Figure asset pending extraction — the sourced original is Figure.

Left: intersecting lines → one solution (consistent). Middle: coincident lines → infinitely many (dependent). Right: parallel lines → no solution (inconsistent).

“Without solving, state whether the pair is consistent” — 1–2 marks

Substitution method

Express one variable from one equation (

x = 3 - 2 y

), substitute into the other to get a single-variable equation, solve it, then back-substitute for the second variable. A no-variable TRUE statement means infinitely many solutions; a FALSE one means no solution.

“Solve by substitution” — 2–3 marks

Elimination method

Multiply the two equations by suitable non-zero constants to make one variable’s coefficients numerically equal, then add or subtract to eliminate it. Solve the resulting one-variable equation and back-substitute. Often quicker than substitution when nothing is pre-isolated.

“Solve by elimination” / solve by BOTH methods — 2–3 marks

Word problems → a pair of equations

Translate the situation (ages, digits, fractions, incomes, fixed-plus-variable charges) into TWO equations in two unknowns, solve by substitution or elimination, and check the answer against the story.

Applied problem — a fixture of Sections C/D, 3–5 marks

Worked examples — real solved problems NCERT VERBATIM

Solved exactly as the source does them

Every problem is a real sourced item, quoted verbatim with its page; the step-by-step decoding and mark-logic are LazyTopper-authored coaching.

Ch 3, Example 1,

“Check graphically whether the pair of equations x + 3y = 6 and 2x – 3y = 12 is consistent. If so, solve them graphically.”

Two solutions of each line.

x + 3 y = 6

gives

(0, 2)

and

(6, 0)

;

2 x - 3 y = 12

gives

(0, - 4)

and

(3, - 2)

(Table 3.2).Concept: make a value table

1 mark

Plot and join.Plot

A (0, 2), B (6, 0), P (0, - 4), Q (3, - 2)

and draw lines

A B

and

P Q

(Fig. 3.1).Concept: draw both lines

1 mark

Read the common point.The lines meet at

B (6, 0)

, so

x = 6, y = 0

; the given pair is consistent.Concept: intersection = solution

1 mark

Mark-scheme logic (authored): 3-mark graphical shape: 1 for a correct table of two solutions per line, 1 for plotting both lines accurately, 1 for reading the common point

(6, 0)

and stating the pair is consistent.

Ch 3, Example 4,

“Solve the following pair of equations by substitution method: 7x – 15y = 2 x + 2y = 3”

Ch 3, Example 8,

“The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save ₹ 2000 per month, find their monthly incomes.”

Ch 3, Example 10,

“The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?”

Problem-solving strategies AUTHORED

One variable already isolated (coefficient 1)?Use substitution — write that variable in terms of the other and plug it in.

Nothing pre-isolated / coefficients line up?Use elimination — scale to match one variable’s coefficients, then add or subtract.

Graph would give ugly / non-integer coordinates?Switch to an algebraic method — NCERT’s own reason for introducing them (p.30).

Asked only to classify (consistent / inconsistent)?Compare

\frac{a _{1}}{a _{2}}, \frac{b _{1}}{b _{2}}, \frac{c _{1}}{c _{2}}

— don’t solve for

x, y

Hit a statement with no variable?TRUE (e.g.

18 = 18

) ⇒ infinitely many solutions; FALSE (e.g.

0 = 9

) ⇒ no solution.

Word problem?Name two variables → form two equations → solve → verify against the story.

Common mistakes that cost marks

Where students lose easy marks

Misreading a graphical intersection — non-integer coordinates are easy to read wrong; if the crossing isn’t on clean grid points, solve algebraically instead.

Swapping the no-solution / infinite-solution verdict — a TRUE no-variable line (

18 = 18

) means infinitely many; a FALSE one (

0 = 9

) means no solution. Students reverse these.

Eliminating without matching coefficients — you must first scale so one variable’s coefficients are numerically equal (and mind the sign: add if opposite, subtract if same).

Confusing the ratio conditions — parallel/inconsistent is

\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}

; coincident/dependent needs ALL three ratios equal.

Stopping at one answer in a two-case word problem — e.g. the two-digit number gives 42 AND 24; report every valid solution.

Formulae

General form

a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0

Unique solution (intersecting)

\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}

No solution (parallel)

\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}

Infinitely many (coincident)

\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}

Mindmap

- - Plot both lines
  - Intersect → unique (consistent)
  - Coincide → infinitely many (dependent)
  - Parallel → none (inconsistent)
- - a₁/a₂ ≠ b₁/b₂ → one solution
  - a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → no solution
  - a₁/a₂ = b₁/b₂ = c₁/c₂ → infinitely many
- - Express one variable
  - Substitute into the other
  - Solve, then back-substitute
- - Match a coefficient
  - Add / subtract to cancel a variable
  - Solve, then back-substitute
- - Two variables → two equations
  - Solve by substitution / elimination
  - Verify against the story
- - TRUE = infinite, FALSE = none
  - Match coefficients before eliminating
  - Report EVERY valid answer
  - Non-integer graph → solve algebraically

Why the coefficient ratios decide the fate of a pair

NCERT establishes the ratio conditions inductively — it tabulates three sample pairs (Table 3.1, p.26), reads off $\frac{a _{1}}{a _{2}}, \frac{b _{1}}{b _{2}}, \frac{c _{1}}{c _{2}}$ for each, and states that “the converse is also true for any pair of lines.” Here is the slope reasoning behind that table, so the three ratio rules are understood rather than memorised. The conditions themselves are exactly NCERT’s (Table 3.1 and Summary point 4).

Write the pair in general form — each equation is a straight line.

a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0

Rearranged to slope–intercept form, a line has slope

- a / b

and

y

-intercept

- c / b

(for

b \neq = 0

y = - \frac{a}{b} x - \frac{c}{b}

Two lines cross exactly once iff their slopes differ, are parallel iff slopes match but intercepts differ, and are the same line iff both match.

Different slopes ⇒ the lines intersect at one point ⇒ a unique solution (consistent).

\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}

NCERT: intersecting lines give exactly one solution (Table 3.1, p.26; Summary 4(i), p.37).

Equal slopes but different intercepts ⇒ parallel lines that never meet ⇒ no solution (inconsistent).

\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}

No common point, so the pair has no solution (p.26; Summary 4(ii), p.37).

All three ratios equal ⇒ the two equations describe the SAME line ⇒ infinitely many solutions (dependent, hence consistent).

\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}

Every point of the line satisfies both equations (p.26; Summary 4(iii), p.37).

3 competency questions from the CBSE board pattern, with the full step-marked solution for each — the marks are shown per step, as the official scheme awards them.

5 marksSection D

The present age of a father is three times the present age of his son. Twelve years from now, the father's age will be twice the son's age. Find their present ages.

Step-marked solution

[1 mark] Let the father's present age be x years and the son's present age be y years. 'Father is three times the son': x = 3y …(1).
[1 mark] Twelve years hence, father = x + 12 and son = y + 12, with father twice the son: x + 12 = 2(y + 12) → x − 2y = 12 …(2).
[1 mark] Substitute (1) into (2): 3y − 2y = 12 → y = 12.
[1 mark] From (1): x = 3(12) = 36.
[1 mark] Hence father is 36 years and son is 12 years. (Check: in 12 years, 48 = 2 × 24 ✓.)

5 marksSection D

The sum of the present ages of a mother and her daughter is 60 years. Six years ago, the mother's age was five times the age of the daughter at that time. Find their present ages.

Step-marked solution

[1 mark] Let mother's present age be x years and daughter's be y years. Sum condition: x + y = 60 …(1).
[1 mark] Six years ago: mother = x − 6, daughter = y − 6, with x − 6 = 5(y − 6) → x − 5y = −24 …(2).
[1 mark] Subtract (2) from (1): (x + y) − (x − 5y) = 60 − (−24) → 6y = 84 → y = 14.
[1 mark] From (1): x = 60 − 14 = 46.
[1 mark] Hence mother is 46 years and daughter is 14 years. (Check: 6 years ago, 40 = 5 × 8 ✓.)

5 marksSection D

Ten years ago, a man was thrice as old as his son. Ten years from now, the sum of their ages will be 76 years. Find their present ages.

Step-marked solution

[1 mark] Let man's present age be x years and son's be y years. Ten years ago: x − 10 = 3(y − 10) → x − 3y = −20 …(1).
[1 mark] Ten years hence: (x + 10) + (y + 10) = 76 → x + y = 56 …(2).
[1 mark] Subtract (1) from (2): (x + y) − (x − 3y) = 56 − (−20) → 4y = 76 → y = 19.
[1 mark] From (2): x = 56 − 19 = 37.
[1 mark] Hence the man is 37 years and the son is 19 years. (Check: 10 years ago, 27 = 3 × 9 ✓.)

Practise Pair of Linear Equations in Two Variables competency questions

Official NCERT chapter PDF (jemh103) — ncert.nic.in

Sources: NCERT 2026-27 (tap to view)

Source ledger PROVENANCE

Source edition: NCERT Reprint 2026-27, Class 10 Mathematics, Ch 3 (jemh103.pdf, printed pages 24–37). One row per sourced item — every verbatim, example and NCERT figure traces to its page.

Item	Type	Source
Consistent pair of linear equations	verbatim	NCERT 2026-27 Ch3
Inconsistent pair of linear equations	verbatim	NCERT 2026-27 Ch3
Dependent pair of linear equations	verbatim	NCERT 2026-27 Ch3
Substitution method	verbatim	NCERT 2026-27 Ch3 (Remark, Section 3.3.1)
Elimination method	verbatim	NCERT 2026-27 Ch3 (Remark 1, Section 3.3.2)
Graph = two lines	verbatim	NCERT 2026-27 Ch3 (Summary 2)
Algebraic methods	verbatim	NCERT 2026-27 Ch3 (Summary 3)
Graphical consistency check · NCERT Example 1	example	NCERT 2026-27 Ch3 Example 1 -27
Substitution method · NCERT Example 4	example	NCERT 2026-27 Ch3 Example 4
Elimination method · NCERT Example 8	example	NCERT 2026-27 Ch3 Example 8
Word problem · NCERT Example 10	example	NCERT 2026-27 Ch3 Example 10 -36
NCERT Fig. 3.1 (intersecting lines)	figure	NCERT 2026-27 Ch3 Fig. 3.1

Pair of Linear Equations in Two Variables · spec-rendered note · NCERT Reprint 2026-27, Class 10 Mathematics, Ch 3 (jemh103.pdf, printed pages 24–37)

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