Pair of Linear Equations in Two Variables – Class 10 CBSE Math Notes

Introduction to Linear Equations in Two Variables

A linear equation in two variables is a rational and integral equation of the first degree involving two variables, typically represented as x and y. The general form of such an equation is:

ax + by + c = 0

where:

• a, b, c ∈ ℝ (real numbers)
• a ≠ 0 and b ≠ 0
• x and y are variables

Characteristics:

1. Neither variable appears under a radical sign
2. Neither variable is in the denominator
3. The exponent of x and y in each term is one
4. The graph of a linear equation in two variables is always a straight line

Examples:

• 3x + 2y = 7 ✓
• 2x - √3y = √5 ✓
• y - 4x = √3 ✓

System of Simultaneous Linear Equations

A pair of linear equations in two variables forms a system of simultaneous linear equations.

General Form:

a₁x + b₁y + c₁ = 0 ...(1) a₂x + b₂y + c₂ = 0 ...(2)

where:

• a₁, a₂, b₁, b₂, c₁, c₂ are real numbers
• a₁² + b₁² ≠ 0 and a₂² + b₂² ≠ 0

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

Types of Solutions: How to Determine Consistency

The nature of solutions depends on the ratios of coefficients:

Example Analysis:

Example 1: Check if these equations are consistent or inconsistent:

(i) 3x + 5y = 6, 6x + 2y = 12

Solution:

• a₁/a₂ = 3/6 = 1/2
• b₁/b₂ = 5/2
• Since 1/2 ≠ 5/2, the equations have a unique solution (consistent)

Methods to Solve Pair of Linear Equations

There are three main algebraic methods to solve simultaneous linear equations:

1. Substitution Method

This method involves expressing one variable in terms of the other and substituting it into the second equation.

Steps:

1. Choose one equation and express y in terms of x (or x in terms of y)
2. Substitute this expression into the other equation
3. Solve the resulting single-variable equation
4. Substitute back to find the other variable
5. Verify the solution

Example with Detailed Solution:

Solve: 4x + 3y = 24, 3y - 2x = 6

Step 1: From equation (1): 4x + 3y = 24

3y = 24 - 4x y = (24 - 4x)/3 ...(3)

Step 2: Substitute (3) into equation (2):

3[(24 - 4x)/3] - 2x = 6 24 - 4x - 2x = 6 24 - 6x = 6 -6x = -18 x = 3

Step 3: Substitute x = 3 into (3):

y = (24 - 12)/3 = 12/3 = 4

Solution: x = 3, y = 4 ✓

2. Elimination Method

This method eliminates one variable by making coefficients equal and then adding or subtracting equations.

Steps:

1. Multiply equations by suitable numbers 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 to find the eliminated variable
5. Verify the solution

Example with Detailed Solution:

Solve: 3x + 4y = 10, 2x - 2y = 2

Step 1: Original equations:

3x + 4y = 10 ...(1) 2x - 2y = 2 ...(2)

Step 2: Multiply (2) by 2:

4x - 4y = 4 ...(3)

Step 3: Add (1) and (3) to eliminate y:

3x + 4y = 10 4x - 4y = 4 ____________ 7x = 14 x = 2

Step 4: Substitute x = 2 in equation (2):

2(2) - 2y = 2 4 - 2y = 2 2y = 2 y = 1

Solution: x = 2, y = 1 ✓

3. Cross-Multiplication Method ✖️

This is a direct formula-based method derived from elimination.

Formula:

For equations:

a₁x + b₁y + c₁ = 0 a₂x + b₂y + c₂ = 0

The solution is given by:

x/(b₁c₂ - b₂c₁) = y/(c₁a₂ - c₂a₁) = 1/(a₁b₂ - a₂b₁)

Visual Representation:

b₁ c₁ c₁ a₁ a₁ b₁ × × × × × × b₂ c₂ c₂ a₂ a₂ b₂

Example:

Solve: x + 2y - 1 = 0, 2x - 3y - 12 = 0

Using cross-multiplication:

x/(2×(-12) - (-3)×(-1)) = y/((-1)×2 - (-12)×1) = 1/(1×(-3) - 2×2) x/(-24 - 3) = y/(-2 + 12) = 1/(-3 - 4) x/(-27) = y/10 = 1/(-7) x = 27/7 y = -10/7

Solution: x = 27/7, y = -10/7

4. Graphical Method

This method involves plotting both equations on a graph and finding their point of intersection.

Steps:

1. Convert each equation to the form y = mx + c
2. Create a table of values for x and y for each equation
3. Plot points on graph paper
4. Draw straight lines through the points
5. The point of intersection is the solution

Example:

Solve graphically: 2x + y = 8, 3x + 2y = 12

For equation (1): 2x + y = 8 ⟹ y = 8 - 2x

For equation (2): 3x + 2y = 12 ⟹ y = (12 - 3x)/2

Plot these points and draw lines. The lines intersect at (4, 0).

Solution: x = 4, y = 0 ✓

Formula

Special Cases and Important Concepts

Homogeneous System of Equations

A system is homogeneous if all constant terms are zero:

a₁x + b₁y = 0 a₂x + b₂y = 0

Properties:

• Always passes through the origin (0, 0)
• Always consistent
• If a₁/a₂ ≠ b₁/b₂: unique solution (only x = 0, y = 0)
• If a₁/a₂ = b₁/b₂: infinitely many solutions

Equations Reducible to Linear Form

Some non-linear equations can be converted to linear form through substitution:

Type I: Reciprocal Form

a/x + b/y = c a'/x + b'/y = c'

Substitution: Let 1/x = u and 1/y = v

Then solve: au + bv = c, a'u + b'v = c'

Type II: More Complex Forms

a/(lx + my) + b/(cx + dy) = k a'/(lx + my) + b'/(cx + dy) = k'

Substitution: Let 1/(lx + my) = p and 1/(cx + dy) = q

Word Problems: Real-World Applications

Linear equations are extensively used in solving real-world problems involving:

1. Age Problems

When relationships between ages of two or more people are given at different times.

Example: Five years hence, father's age will be three times the age of his son. Five years ago, father was seven times as old as his son. Find their present ages.

Solution: Let present ages be: Father = x years, Son = y years

After 5 years: (x + 5) = 3(y + 5) 5 years ago: (x - 5) = 7(y - 5)

Solving: x = 40 years, y = 10 years

2. Speed, Distance, and Time Problems

Problems involving upstream/downstream, train speeds, or walking speeds.

Example: A boat covers 32 km upstream and 36 km downstream in 7 hours. Also, it covers 40 km upstream and 48 km downstream in 9 hours. Find the speed of the boat in still water and the speed of the stream.

Solution: Let speed in still water = u km/h, speed of stream = v km/h

Setting up equations using time = distance/speed:

• 32/(u-v) + 36/(u+v) = 7
• 40/(u-v) + 48/(u+v) = 9

Solving: u = 10 km/h, v = 2 km/h

3. Number Problems

Problems about two-digit or three-digit numbers and digit manipulation.

4. Cost and Mixture Problems

Problems involving prices, quantities, and mixtures of items.

Common Mistakes to Avoid

1. Sign errors when multiplying equations in elimination method
2. Division by zero – always check if denominators become zero
3. Forgetting to verify solutions by substituting back
4. Incorrect graphing – always plot at least 3 points for accuracy
5. Misreading ratio conditions for determining consistency

Summary

The chapter on Pair of Linear Equations in Two Variables is fundamental to algebra and has extensive real-world applications. Understanding the three solution methods substitution, elimination, and cross-multiplication along with the ability to determine consistency of systems, forms the core of this topic.

Note:

• Linear equations represent straight lines
• Systems can have one, none, or infinitely many solutions
• Coefficient ratios determine the nature of solutions
• Multiple methods exist; choose based on equation structure
• Word problems require careful translation into mathematical form

Author Important Massage for the Students: This guide is prepared based on the official CBSE Class 10 Mathematics curriculum, incorporating best practices from experienced mathematics educators and aligned with NCERT guidelines. All methods and examples have been verified for mathematical accuracy and pedagogical effectiveness.