Pair of Linear Equations in Two Variables — Detailed Notes

Chapter 3 | NCERT Class 10 Mathematics

Study Diagrams (Visual Revision)

Use these board-style figures while reading the chapter. Open the Diagrams tab for a larger gallery.

Linear Equations — Decision Map

Linear Equations — Must-Write Strip

Linear Pair — Consistency Cards

3.1 Introduction

A situation with two conditions and two unknowns becomes a pair of linear equations, e.g.

$y = \frac{1}{2} x, 3 x + 4 y = 20$

(Akhila’s Giant Wheel & Hoopla example in NCERT). Several methods exist to find common solutions; this chapter develops graphical comparison and two algebraic methods.

3.2 Graphical Method

Language

Term	Meaning
Consistent	At least one common solution
Inconsistent	No common solution
Dependent (consistent)	Infinitely many common solutions (equivalent equations)

Geometry of two lines

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

Condition on ratios	Lines	Solutions
$\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$	Intersecting	Unique (consistent)
$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$	Coincident	Infinitely many (dependent)
$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$	Parallel	None (inconsistent)

Worked idea (Example 1 style): $x + 3 y = 6$ and $2 x - 3 y = 12$ meet at $(6, 0)$ — unique solution, consistent.

Worked idea (Example 2 style): If one equation is a scalar multiple of the other, lines coincide — infinitely many solutions.

Worked idea (Example 3 style): Word problem graphed to an intersection; check the point against the story (e.g. non-negative counts).

Better Method / Strategy — Graphical vs Algebraic

Use graphs to visualise consistency and for integer intersections.

Prefer algebra when the solution has awkward fractions (graphs are error-prone then).

Always start word problems by defining variables clearly.

3.3 Algebraic Methods

3.3.1 Substitution Method

Steps

From the easier equation, express one variable in terms of the other.
Substitute into the second equation; solve for one variable.
Back-substitute to find the other.
If Step 2 gives a true statement with no variable ( $0 = 0$ type) $\Rightarrow$ infinitely many solutions.
If it gives a false statement ( $0 = k$ ) $\Rightarrow$ inconsistent (no solution).

Worked idea (Example 4): $7 x - 15 y = 2$ , $x + 2 y = 3$ .

From (2): $x = 3 - 2 y$ . Substitute: $7 (3 - 2 y) - 15 y = 2 \Rightarrow 21 - 14 y - 15 y = 2 \Rightarrow y = \frac{19}{29}, x = \frac{49}{29}$

Worked idea (Example 6 — infinite): $2 x + 3 y = 9$ , $4 x + 6 y = 18$ are the same line $\Rightarrow$ infinitely many solutions (no unique prices).

Worked idea (Example 7 — none): $x + 2 y - 4 = 0$ , $2 x + 4 y - 12 = 0$ lead to $- 4 = 0$ $\Rightarrow$ rails never meet.

3.3.2 Elimination Method

Steps

Multiply equations by suitable constants so one variable has equal (or opposite) coefficients.
Add or subtract to eliminate that variable.
Solve for the remaining variable; substitute back.
True no-variable identity $\Rightarrow$ infinitely many; false identity $\Rightarrow$ no solution.

Worked idea (Example 8): Incomes $9 x, 7 x$ ; expenditures $4 y, 3 y$ ; each saves $2000$ :

$9 x - 4 y = 2000, 7 x - 3 y = 2000$

Multiply to eliminate $y$ : $x = 2000$ , $y = 4000$ $\Rightarrow$ incomes ₹18 000 and ₹14 000.

Worked idea (Example 9 — none): $2 x + 3 y = 8$ , $4 x + 6 y = 7$ $\Rightarrow$ $0 = 9$ after elimination $\Rightarrow$ inconsistent.

Worked idea (Example 10 — digits): Two-digit number + reverse $= 66$ , digits differ by $2$ $\Rightarrow$ two cases $x - y = \pm 2$ with $x + y = 6$ give $42$ and $24$ .

Better Method / Strategy — Substitution vs Elimination

Prefer substitution when… Prefer elimination when…

One coefficient is already $1$ or $- 1$ Both equations “balanced” with larger coefficients

Easy to isolate $x$ or $y$ Clearing decimals / matching $x$ or $y$ is natural

Checking infinite/none via substitution identity Same check via $0 = 0$ / $0 = k$ after eliminating

(Cross-multiplication is not developed in the rationalised NCERT summary of this chapter — stick to substitution and elimination.)

Word-problem playbook

Define variables in a sentence ( $x =$ …, $y =$ …).
Translate each English condition into one equation.
Choose substitution or elimination.
Solve; verify in both equations.
Interpret: reject negative prices, fractional people (unless meaningful), etc.

Typical story types in NCERT: ages; costs of bats/balls; taxi fixed + per-km; fractions; two-digit numbers; notes of ₹50/₹100; library fixed + daily charge.

Reducing “almost linear” situations

NCERT notes that some situations start non-linear but can be rewritten as a linear pair (e.g. after clearing denominators in fraction problems). Always simplify to $a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0$ before applying ratio tests.

Exam-Oriented Analysis

Aspect	Detail
Weight	High utility chapter; often 4–6 marks across parts
Question types	Ratio consistency; graphical nature; solve by named method; word problems
Common mistakes	Mixing up $\neq =$ vs $=$ in $c$ ratios; arithmetic slips in elimination multipliers; accepting negative ages/costs
Scoring tips	Name the method; show the ratio line; verify both equations; interpret “infinite / none” in words

Question-type table

Style	Typical marks	Key step
Compare ratios / consistent?	1–2	Compute all three ratios carefully
Graph & find intersection	2–3	Two points per line; read common point
Solve by substitution	2–3	Isolate easiest variable
Solve by elimination	2–3	Correct multipliers on both sides
Form equations + solve	3–4	Definitions + method + check

Common mistakes (expanded)

Writing equations as $a x + b y = c$ then forgetting the sign of $c$ when forming $a x + b y + c = 0$ for ratios.
Declaring “inconsistent” whenever coefficients look proportional — must check $c$ ratios.
Multiplying only the $x$ and $y$ terms when eliminating, not the constant.
Stopping at $x$ and forgetting $y$ (or vice versa).
Reporting a graphical answer that does not satisfy both equations.

Scoring tips

Write $\frac{a _{1}}{a _{2}} = \dots, \frac{b _{1}}{b _{2}} = \dots, \frac{c _{1}}{c _{2}} = \dots$ then conclude.
Name the method: “By elimination:”.
End word problems with a one-line interpretation (“Father is 50 years old…”).

Consistency quick card

$\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}} \frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}} \frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}} \Rightarrow unique \Rightarrow infinite \Rightarrow none$

Quick self-test before submitting an answer

Do both original equations hold?
Does the classification match the algebra ( $0 = 0$ / $0 = k$ / unique pair)?
Is the answer meaningful in context?

Substitution vs elimination — one more worked contrast

Solve $3 x + 4 y = 10$ and $2 x - 2 y = 2$ .

Substitution (often shorter here): From the second equation, $x - y = 1$ , so $x = y + 1$ . Substitute into the first: $3 (y + 1) + 4 y = 10 \Rightarrow 7 y = 7 \Rightarrow y = 1, x = 2$

Elimination: Multiply the second equation by $2$ : $4 x - 4 y = 4$ . Add to the first: $(3 x + 4 y) + (4 x - 4 y) = 10 + 4 \Rightarrow 7 x = 14 \Rightarrow x = 2, y = 1$

Same solution. Prefer substitution when a variable is already easy to isolate; prefer elimination when coefficients are multiples and isolating looks messy.