- CBSE Class 6 Maths Notes
- CBSE Class 6 Maths Notes
- Chapter 1 - Knowing Our numbers
- Chapter 2 - Whole numbers
- Chapter 3 - Playing with numbers
- Chapter 4 - Basic Geometrical Ideas
- Chapter 5 - Understanding Elementary Shapes
- Chapter 6 - Integers
- Chapter 7 - Fractions
- Chapter 8 - Decimals
- Chapter 9 - Data Handling
- Chapter 10 - Mensuration
- Chapter 11 - Algebra
- Chapter 12 - Ratio and Proportion
- Chapter 13 - Symmetry
- Chapter 14 - Practical Geometry

# Chapter 11 - Algebra

## Introduction to Algebra

### Branches of Mathematics

Mathematics has several branches, but broadly it can be of three types:

**Arithmetic**− It deals with numbers, their types, their properties, operations on numbers, applications of numbers and so on.**Geometry**− It deals with different types of shapes and figures in 1, 2 and 3 dimensions.**Algebra**− It is a branch of mathematics that uses the concept of variables to find the value of unknown quantities in equations.

### What is Algebra?

Algebra is a lot like arithmetic, as it uses all the main rules and operations of arithmetic. However, it introduces a new concept, the idea of an unknown quantity that needs to be found. Algebra uses arithmetic operations like addition, subtraction, multiplication, and division.

### Variables in Algebra

In algebra, we use alphabets or symbols to represent things or objects whose values are unknown. These alphabets or symbols are known as variables.

More than one alphabet or symbols are used when there are multiple unknowns.

## Exploring Variables

Variables are used to establish relationships between different known and unknown quantities. Variables are represented by using alphabets or symbols.

Relationships are written as equations involving variables, numerals, and operations.

**Example**

Question: Sam and Jack are two friends. Jack is older than Sam by 3 years. Express the relationship between their ages. Solution: The relationship between their ages can be expressed as, Sam's age + 3 = Jack's age If Sam's age is taken asX, then Jack's age =X+ 3 If Jack's age is taken asY, then Sam's age =Y− 3

**Example**: In the following matchstick square pattern, find the relationship between the number of squares and the number of matchsticks.

*Solution*:

Analyzing the pattern of squares,

First square has 4 matchsticks = 1 + (3 × 1)

Second square has 7 matchsticks = 1 + (3 × 2)

Third square has 10 matchsticks = 1 + (3 × 3)

If we continue the pattern, then

N^{th} square would have 1 + (3 × N) matchsticks.

**Example**

Question: Jack transfers potatoes from a sack equally into two boxes. He finds that there are 7 potatoes still left out in the sack. Find a relation to express the total number of potatoes. Solution: Let's assume the no. of potatoes in each box = P And, total number of potatoes = T Then, Total no. of potatoes in the sack = potatoes in each box + leftover potatoes T = (2 × potatoes in each box) + 7 T = (2 × P) + 7 T = 2P + 7

## Using Variables in Geometry

Let's see how algebra handles mathematical rules and formulae with its powerful concept of variables.

**Example**: Develop a formula to find the perimeter of a square.

*Solution*: It is known that a square has four equal sides and its perimeter is given as four times the length of its side.

Let's assume,

P = Perimeter of the square, and

L = Length of each side

Then, the perimeter of the square can be written as,

P = 4L

**Example**: Develop a formula to find the perimeter of a regular polygon.

*Solution*: A polygon is a closed figure with at least 3 sides. A regular polygon is one that has equal sides. Examples of regular polygons are equilateral triangle, square, regular pentagon, etc.

Let L = Length of each side

The perimeter of a polygon is the sum of lengths of its sides.

Hence, we have the following formulae,

Perimeter of triangle = 3L

Perimeter of square = 4L

Perimeter of regular pentagon = 5L

Generalizing this pattern,

Perimeter of a regular polygon of N sides = N × length of each side

= N × L = NL

### Perimeter of a Rectangle

In a rectangle, the opposite sides are parallel and equal in length.

Consider a rectangle PQRS with PQ and RS as its lengths and QR and SP as its breadths.

Let's assume,

L = length of the rectangle

B = breadth of the rectangle

P = perimeter of the rectangle

The perimeter of a rectangle is the sum of its two lengths and two breadths. Hence,

P = PQ + QR + RS + SP

= L + B + L + B

= 2L + 2B

= 2(L + B)

Thus the general formula for the perimeter of any rectangle is, P = 2(L + B)

### Area of a Rectangle

Let's assume,

A = Area of the rectangle

The area of a rectangle is the product of its length and breadth. Hence,

Area = Length × Breadth

A = L × B = LB

### Radius and Diameter

The diameter of a circle is twice the length of its radius.

Assume,

D = diameter of the circle

R = radius of the circle

Then,

D = 2 × R = 2R

## Using Variables with Arithmetic

### Commutative property

This property states that changing the order of numbers in an operation does not affect the result. For example,

5 + 4 = 9 and 4 + 5 = 9

Let's put it in the form of an algebraic formula.

If A and B are two variables, then

A + B = B + A

Commutative property holds true for addition and multiplication.

Subtraction and division are NOT commutative.

A − B ≠ B − A

A ÷ B ≠ B ÷ A

### Associative Property

The sum or product of three or more numbers remains the same regardless of how the numbers are grouped. For example,

2 × 5 × 4 = (2 × 5) × 4 = 10 × 4 = 40

2 × 5 × 4 = 2 × (5 × 4) = 2 × 20 = 40

Associativity holds true for addition and multiplication, but not for subtraction and division.

Let's generalize the formula using A, B, and C as variables,

(A + B) + C = A + (B + C)

(A × B) × C = A × (B × C)

### Distributive Property

Distributive property helps us solve difficult multiplications. For example,

7 × 45

= 7 × (40 + 5)

= (7 × 40) + (7 × 5)

= 280 + 35 = 315

Here, we have distributed the multiplication to each of the numbers being added.

Generalizing the distributive property using variables,

A (B + C) = (A × B) + (A × C)

where A, B, and C are variables representing any three distinct numbers.

## Algebraic Expressions with Variables

A mathematical expression is defined as two or more numbers that are either added, subtracted, multiplied, or divided with one another.

For example, 2 + 3 is a mathematical expression.

Expressions are of two types:

- Numerical expressions
- Algebraic expressions

### Numerical Expressions

Mathematical expressions that contain only numbers are called numerical expressions. For example, (5 + 9) or (7 − 3) are numerical expressions.

### Algebraic Expressions

Mathematical expressions that contain algebraic variables are called algebraic expressions. For example, (x + 2) or (y − 3) are called algebraic expressions because they contain variables like x and y.

### The BODMAS Rule

We use the BODMAS rule to simplify numerical and algebraic expressions. the BODMAS rule is used. This rule lists the priority of execution of different operations like addition, subtraction, multiplication and division.

Brackets >> Order >> Division >> Multiplication >> Addition >> Subtraction

**Example**

Question: Simplify 18 + (6 × 3) + (5 − 6) Solution: First priority is brackets. Simplifying the numbers inside the brackets first, 18 + (6 × 3) + (5 − 6) = 18 + 18 − (−1) = 18 + 18 − 1 Next priority is addition, 18 + 18 − 1 = 36 − 1 Last priority is subtraction, 36 − 1 = 35 So, we have, 18 + (6 × 3) + (5 − 6) = 35

**Example**

Question: Simplify (y × 7) + (2(3 − y)) Solution: Using BODMAS rule, (y × 7) + (2(3 − y)) = 7y + (6 − 2y) = 7y + 6 − 2y = 5y + 6 So, (y × 7) + (2(3 − y)) = 5y + 6

**Example**

Question: Jack has 5 more chocolates than Jill and Sam has twice as many chocolates as Jack. Develop an algebraic expression for the number of chocolates with Sam. Solution: Assume that Jill has y chocolates. Number of chocolates with Jack = y + 5 Number of chocolates with Sam = 2 (y + 5) = 2y + 10

**Example**: A car is travelling between two cities which are D km apart. The car travels for 5 hours and is at a distance of 150 km from its destination. Develop an algebraic expression for the speed of the car.

*Solution*:

The distance between the cities = D km

Distance travelled by the car = D − 150 km

Time of travel of car = 5 hours

Therefore, we have,

Speed of car = ${Distance \: travelled}/{Time \: of \: travel}$ = ${D - 150}/{5}$ km/hr

## Guess the Number Game

### Algebraic Magic

Ask your friend to think of a number. It can be any natural number of her choice.

Then, ask her to perform the following chain of arithmetic operations on it.

Your friend can think of any number, but the final output at the end of the above operations would be 10!

Surprised! Let's find out how it works.

Assume the number that your friend thought of is = N

Adding 2, it becomes,

N + 2

Multiplying 2,

2(N + 2) = 2N + 4

Adding 5,

2N + 4 + 5 = 2N + 9

Multiplying 5,

5 (2N + 9) = 10N + 45

Subtracting 45,

10N + 45 − 45 = 10N

Dividing by N,

$${10 \: N}/{N} = 10$$

Hence, N can be any number, but our output remains unchanged.

### Another Example

Ask another friend to think of a natural number and perform the following arithmetic operations step by step.

Regardless of the number your friend thinks, the final output of this sequence of operations would be 6!

## What is an Equation?

An equation is a statement with an 'equal to' sign (=) showing that the values of expressions on either side are equal.

### Numerical Equation

If the expressions on either side of an equation have only numbers, then such an equation is a numerical expression. For example,

5 − 3 = 2

10 + 7 × 2 = 24

### Algebraic Equation

If the expressions on either side of an equation contain one or more variables, then such an equation is an algebraic equation. For example,

x + y = 12

6n = 72

We solve algebraic equations to find the values of unknown variables in the equations. For example,

4 + x = 6

Subtracting 4 from both sides of the equation isolates the unknown variable and gives its value.

4 + x − 4 = 6 − 4

⇒ x = 2

Similarly, solving

4y = 60

⇒ ${4y}/{4}$ = ${60}/{4}$

⇒ y = 15

**Example**: The letter 'W' can formed using 4 matchsticks. We have the equation M = 4N where M is the total number of matchsticks and N is the number of 'W's formed. Find the number of 'W's formed if M = 20

*Solution*:

The given equation is,

M = 4N

M = 20, so the equation becomes,

20 = 4N

Solving for N,

N = ${20}/{4}$ = 5

Hence, 5 'W's can be formed using 20 matchsticks.

**Example**

Question: Jack has 3 more chocolates than Jill and Jill has 5 chocolates in all. How many chocolates does Jack have? Solution: Let the number of chocolates with Jack = Y. Jill has 3 chocolates less than Jack, which is = Y − 3 Given that Jill has 5 chocolates in all. So, we have the equation, Y − 3 = 5 Adding 3 on both sides, x − 3 + 3 = 5 + 3 ⇒ x = 8 Therefore, Jack has 8 chocolates.