CBSE Class 10 Maths Chapter 3 Notes — Pair of Linear Equations in Two Variables

General Form and Solutions

A linear equation in two variables has the form: ax + by + c = 0 (where a, b are not both zero).

A pair of linear equations is two such equations considered together:

A solution is a pair (x, y) that satisfies both equations simultaneously.

Graphical Representation

Each linear equation represents a straight line on the coordinate plane. A pair of lines can:

Consistency Check (Without Solving)

Step 1: Compute a₁/a₂, b₁/b₂, c₁/c₂

Step 2: Compare the ratios:

• a₁/a₂ ≠ b₁/b₂ → unique solution (consistent)
• a₁/a₂ = b₁/b₂ = c₁/c₂ → infinitely many solutions (consistent, coincident lines)
• a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → no solution (inconsistent, parallel lines)

Example: 2x + 3y = 8 and 4x + 6y = 16

a₁/a₂ = 2/4 = 1/2; b₁/b₂ = 3/6 = 1/2; c₁/c₂ = 8/16 = 1/2

All ratios equal → infinitely many solutions (coincident lines)

Method 1 — Substitution Method

Steps: 1. From one equation, express one variable in terms of the other 2. Substitute into the second equation 3. Solve for the remaining variable 4. Back-substitute to find the other

Example: Solve: x + y = 14 and x − y = 4

From eq 1: x = 14 − y

Substitute in eq 2: (14 − y) − y = 4 → 14 − 2y = 4 → y = 5

Back-substitute: x = 14 − 5 = 9

Method 2 — Elimination Method

Steps: 1. Multiply equations to make coefficients of one variable equal 2. Add or subtract equations to eliminate that variable 3. Solve for the remaining variable 4. Substitute back

Example: Solve: 3x + 2y = 11 and 2x + 3y = 9

Multiply eq 1 by 3: 9x + 6y = 33 Multiply eq 2 by 2: 4x + 6y = 18

Subtract: 5x = 15 → x = 3

Substitute: 9 + 2y = 11 → y = 1

Method 3 — Cross-Multiplication Method

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

$$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$$

Example: Solve: 2x + 3y − 7 = 0 and 4x + 6y − 14 = 0

Wait — check ratios first: 2/4 = 3/6 = 7/14 = 1/2 → coincident lines, infinitely many solutions.

Proper example: Solve: x + 2y − 4 = 0 and 2x − y − 3 = 0

a₁=1, b₁=2, c₁=−4; a₂=2, b₂=−1, c₂=−3

x = (b₁c₂ − b₂c₁)/(a₁b₂ − a₂b₁) = [(2)(−3) − (−1)(−4)] / [(1)(−1) − (2)(2)] = [−6 − 4] / [−1 − 4] = −10/−5 = 2

y = (c₁a₂ − c₂a₁)/(a₁b₂ − a₂b₁) = [(−4)(2) − (−3)(1)] / [−5] = [−8 + 3] / [−5] = −5/−5 = 1

Reducible Equations

Some equations that are NOT linear can be reduced to a pair of linear equations by substitution.

Example: Solve: 2/x + 3/y = 13 and 5/x − 4/y = −2

Let 1/x = p and 1/y = q:

Solve by elimination: multiply (i) by 4 and (ii) by 3: 8p + 12q = 52 15p − 12q = −6 Add: 23p = 46 → p = 2 → x = 1/2

From (i): 4 + 3q = 13 → q = 3 → y = 1/3

Word Problems — Common Types

Age problems

"The sum of two numbers is 15 and their difference is 3." Let numbers be x and y: x + y = 15; x − y = 3 → x = 9, y = 6

Speed-distance problems

"A boat goes 30 km upstream in 6 hours and 36 km downstream in 6 hours. Find speed of boat and stream." Let boat speed = x, stream speed = y. Upstream: x − y = 5; Downstream: x + y = 6 → x = 5.5, y = 0.5

Fraction problems

"The sum of numerator and denominator is 8. If 3 is added to both, fraction = 3/4." Let fraction = x/y: x + y = 8; (x+3)/(y+3) = 3/4 → 4x+12 = 3y+9 → 4x − 3y = −3 Solve with x + y = 8 → x = 3, y = 5, fraction = 3/5

Board Exam Tips

• 6–8 marks from this chapter — one of the most important in Algebra
• Always check consistency (ratio test) BEFORE attempting to solve — saves time
• For word problems: clearly define variables, set up both equations, then solve
• Substitution is easiest when coefficient of a variable is 1; elimination works for most others
• Learn cross-multiplication formula — it's fastest for straightforward pairs
• Common error: forgetting to bring equation to standard form (ax + by + c = 0) before applying cross-multiplication

📚 RD Sharma Mathematics Class 10 — Linear Equations Chapter

📝 Oswaal CBSE Maths Question Bank — Word Problems Practice