Pair of Linear Equations in Two Variables — Class 10 Mathematics

What CBSE examines here (2026-27). Graphical solution and consistency, the algebraic conditions on the ratios, solving by substitution and by elimination, and situational problems. The rationalised textbook runs to three exercises only — 3.1, 3.2 and 3.3. Both the cross-multiplication method and the section on equations reducible to linear form were removed; both are in the appendix at the end, marked as background.

"Two equations, two unknowns — the foundation of all algebraic problem-solving."

1. About the Chapter

This chapter extends Class 9 linear equations (one variable) to TWO variables. A pair of linear equations forms a SYSTEM that can be solved together.

Standard Form

a₁x + b₁y + c₁ = 0 a₂x + b₂y + c₂ = 0

where a₁, b₁, c₁, a₂, b₂, c₂ are real numbers, and a₁, b₁ not both zero (similarly for second equation).

Why Important

• Foundation for algebra and beyond
• Used in physics, economics, engineering
• Word problems use it everywhere

2. Graphical Representation

Each linear equation in two variables represents a straight line in the coordinate plane.

A PAIR of equations gives TWO LINES. Three cases:

Case 1: Lines Intersect at One Point

• Unique solution (x, y)
• The lines have different slopes
• System is consistent with unique solution

Case 2: Lines are Coincident (overlap)

• Infinitely many solutions
• The lines are EQUAL (same line)
• System is consistent with infinite solutions
• Equations are 'dependent'

Case 3: Lines are Parallel

• No solution
• Lines never meet
• System is inconsistent

3. Algebraic Conditions

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

Unique Solution (Intersecting)

a₁/a₂ ≠ b₁/b₂

Infinite Solutions (Coincident)

a₁/a₂ = b₁/b₂ = c₁/c₂

No Solution (Parallel)

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Example

• 2x + 3y = 7 and 4x + 6y = 14
• a₁/a₂ = 2/4 = 1/2; b₁/b₂ = 3/6 = 1/2; c₁/c₂ = 7/14 = 1/2
• All equal → INFINITE solutions (coincident)

4. Algebraic Methods of Solution

Method 1: Substitution Method

Steps:

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

Example: Solve 2x + 3y = 12 and x + y = 5

• From second: y = 5 − x
• Substitute: 2x + 3(5 − x) = 12
• 2x + 15 − 3x = 12 → −x = −3 → x = 3
• y = 5 − 3 = 2
• Solution: (3, 2)

Method 2: Elimination Method

Steps:

1. Make coefficients of one variable equal (multiply equations)
2. Add or subtract to eliminate that variable
3. Solve for remaining variable
4. Back-substitute

Example: Solve 3x + 4y = 10 and 2x − 2y = 2

• Multiply 2nd by 2: 4x − 4y = 4
• Add to first: 7x = 14 → x = 2
• Substitute: 3(2) + 4y = 10 → 4y = 4 → y = 1
• Solution: (2, 1)

Which method should you pick? Substitution is cleanest when a variable already has coefficient 1. Elimination is cleaner when neither does, especially with fraction-free coefficients. Either is acceptable unless the question names one.

5. Word Problems

Type 1: Age Problems

Example: A father is 4 times as old as his son. After 20 years, he will be twice as old. Find their present ages.

Let son's age = x, father's age = y.

• y = 4x ... (i)
• y + 20 = 2(x + 20) ... (ii)

Substitute: 4x + 20 = 2x + 40 → 2x = 20 → x = 10, y = 40

Son is 10; father is 40.

Type 2: Money Problems

Example: 5 pens and 7 pencils cost ₹250. Later, 7 pens and 5 pencils cost ₹302. Find the price of each.

Let pen = x, pencil = y.

• 5x + 7y = 250 ... (i)
• 7x + 5y = 302 ... (ii)

When the coefficients are swapped like this, adding and subtracting is far quicker than eliminating one variable outright.

Add (i) and (ii): 12x + 12y = 552 → x + y = 46 Subtract (i) from (ii): 2x − 2y = 52 → x − y = 26

Adding those two: 2x = 72 → x = 36, and then y = 46 − 36 = 10.

A pen costs ₹36 and a pencil ₹10. Check: 5(36) + 7(10) = 250 ✓ and 7(36) + 5(10) = 302 ✓

Type 3: Speed/Distance Problems

Example: A train covers 240 km in some hours. If speed increases by 12 km/h, time reduces by 1 hour. Find original speed.

Let original speed = x km/h, time = y hours.

• xy = 240 ... (i)
• (x+12)(y−1) = 240 ... (ii)

Expand (ii): xy − x + 12y − 12 = 240 → 240 − x + 12y − 12 = 240 → 12y − x = 12

From (i): y = 240/x. Substitute: 12(240/x) − x = 12 → 2880 − x² = 12x → x² + 12x − 2880 = 0

Solve: x = (−12 ± √(144 + 11520))/2 = (−12 ± √11664)/2 = (−12 ± 108)/2

x = 48 (positive), so original speed = 48 km/h, time = 5 hours.

Type 4: Geometric Problems

Example: The length of a rectangle exceeds its breadth by 5 cm. If the length is increased by 5 cm and the breadth decreased by 3 cm, the area is unchanged. Find the dimensions.

Let breadth = y, length = x.

• x = y + 5 ... (i)
• xy = (x + 5)(y − 3) ... (ii)

Expand (ii): xy = xy − 3x + 5y − 15

The xy terms cancel — which is what makes these "area unchanged" problems linear despite starting with a product:

3x = 5y − 15

Substitute x = y + 5: 3y + 15 = 5y − 15 → 2y = 30 → y = 15, so x = 20.

Length 20 cm, breadth 15 cm. Check: original area 20 × 15 = 300 cm²; new area 25 × 12 = 300 cm² ✓

Type 5: Boat in Stream

Example: A boat travels 28 km downstream in 2 hours and 12 km upstream in 2 hours. Find the speed of the boat in still water and the speed of the stream.

Let boat speed in still water = x, stream speed = y.

• Downstream the stream helps, so the effective speed is x + y: 28/(x + y) = 2 → x + y = 14
• Upstream the stream hinders, so the effective speed is x − y: 12/(x − y) = 2 → x − y = 6

Adding: 2x = 20 → x = 10; subtracting: 2y = 8 → y = 4.

Boat = 10 km/h in still water, stream = 4 km/h. Check: 28/14 = 2 h ✓ and 12/6 = 2 h ✓

6. Worked Examples

Example 1: Check Solution

Verify (3, −1) is a solution of x + 2y = 1 and 2x + 3y = 3.

• LHS₁: 3 + 2(−1) = 3 − 2 = 1 ✓
• LHS₂: 2(3) + 3(−1) = 6 − 3 = 3 ✓
• Both verified → (3, −1) IS the solution.

Example 2: Graphical Solution

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

• Plot points for each line:
  • First: (0,5), (5,0), (2,3) — line through these
  • Second: (1,0), (0,−1), (4,3) — line through these
• Lines intersect at (3, 2) → x = 3, y = 2

Example 3: Find Type of Solution

Check: 5x + 4y = 8 and 10x + 8y = 16

• a₁/a₂ = 5/10 = 1/2
• b₁/b₂ = 4/8 = 1/2
• c₁/c₂ = −8/−16 = 1/2 (writing equations as 5x+4y−8=0)
• All equal → INFINITE solutions (coincident lines).

Example 4: Find Type of Solution

Check: 3x + 2y = 8 and 6x + 4y = 17

• a₁/a₂ = 3/6 = 1/2
• b₁/b₂ = 2/4 = 1/2
• c₁/c₂ = −8/−17 = 8/17 ≠ 1/2
• a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → NO solution (parallel lines).

7. Common Mistakes

1. Forgetting to verify

   • Always plug answer back into BOTH original equations.

2. Sign errors

   • Careful with negative signs in elimination.

3. Wrong rearrangement in substitution

   • When isolating variable, watch signs.

4. Reading word problems wrong

   • 'Age 5 years later' means add 5, not subtract.

5. Standard form confusion

   • All terms on one side; a₁x + b₁y + c₁ = 0 (constants moved to LHS).

8. Tips for Mastery

For Algebraic Solving

• Master both methods (substitution and elimination) and know when each is quicker
• Clear fractions and decimals BEFORE you start solving
• Rewrite in standard form before applying any ratio test
• Always substitute your answer back into the original equations

For Word Problems

• Define variables CLEARLY
• Translate sentences to equations
• Solve
• Interpret answer in context
• Verify with original problem

For Graphical Method

• Find 2-3 points per line
• Use ruler for straight lines
• Mark intersection point clearly

9. Conclusion

Linear equations in two variables are USED EVERYWHERE — physics formulas, economics, engineering, everyday calculations. This chapter gives you two reliable algebraic methods plus a graphical one, and the ratio test that tells you in advance whether a solution exists at all.

Master the methods. Practice word problems. These skills feed into:

• Class 11-12 algebra
• Calculus
• Engineering mathematics
• Real-life problem-solving

Two equations. Two unknowns. The foundation of mathematical thinking.

Appendix — beyond the current syllabus

Not examinable in CBSE 2026-27. Two things left this chapter during rationalisation: the cross-multiplication method, and the section on equations reducible to a pair of linear equations. The rationalised textbook now runs 3.1 Introduction → 3.2 Graphical Method → 3.3 Algebraic Methods (substitution, elimination) → 3.4 Summary, with exercises 3.1, 3.2 and 3.3 only. Both topics are kept below because older guidebooks and solution sites still lead with them, and students meet them and wonder whether they have missed something. You have not.

The cross-multiplication method

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

$\frac{x}{b_1{c}_2-b_2{c}_1}=\frac{y}{c_1{a}_2-c_2{a}_1}=\frac{1}{a_1{b}_2-a_2{b}_1}$

Example — solve 2x + y = 5 and 3x + 2y = 8.

In standard form: a₁=2, b₁=1, c₁=−5 and a₂=3, b₂=2, c₂=−8.

• x-denominator: b₁c₂ − b₂c₁ = (1)(−8) − (2)(−5) = −8 + 10 = 2
• y-denominator: c₁a₂ − c₂a₁ = (−5)(3) − (−8)(2) = −15 + 16 = 1
• last denominator: a₁b₂ − a₂b₁ = (2)(2) − (3)(1) = 4 − 3 = 1

So x/2 = y/1 = 1/1, giving x = 2 and y = 1 — the same answer elimination gives in two lines.

Equations reducible to a pair of linear equations

The old textbook had a section (and an exercise) on equations that are not linear as written but become linear after a substitution. The section and its exercise are gone, though the chapter's summary still ends with a sentence gesturing at the idea — a leftover the editors did not remove.

The technique is worth seeing once, because the same move reappears in Class 11 and in entrance papers.

The idea. If the unknowns appear only as 1/x and 1/y, rename them.

Example — solve 2/x + 3/y = 13 and 5/x − 4/y = −2.

Let u = 1/x and v = 1/y. The pair becomes genuinely linear:

• 2u + 3v = 13
• 5u − 4v = −2

Eliminate v. Multiply the first by 4 and the second by 3:

• 8u + 12v = 52
• 15u − 12v = −6

Add: 23u = 46, so u = 2. Substituting back into 2u + 3v = 13 gives 4 + 3v = 13, so v = 3.

Finally undo the substitution: x = 1/u = 1/2 and y = 1/v = 1/3.

Check: 2 ÷ (1/2) + 3 ÷ (1/3) = 4 + 9 = 13 ✓ and 5 ÷ (1/2) − 4 ÷ (1/3) = 10 − 12 = −2 ✓

A caution about the exam. Problems whose equations start out non-linear are no longer set as their own type here. But a situational problem can still lead you to a product — the train problem in section 5 is one — and you handle that with ordinary substitution, not with this renaming trick.