Chapter Notes: Algebra

# Algebra Class 6 Notes Maths Chapter 11

## Introduction to Algebra

There are so many branches of Mathematics

• The study of numbers is called Arithmetic.
• The study of shapes is called Geometry.
• The study to use the letters and symbols in mathematics is called Algebra.

### Algebra

Algebra is a part of mathematics in which the letter and symbols are used to represent numbers in equations. It helps us to study about unknown quantities.

Matchstick Patterns

No. of matchsticks used to make 1st square = 4
No. of matchsticks used to make 2nd square = 7
No. of matchsticks used to make 3rd square = 10
So, the pattern that we observe here is 3n + 1
With this pattern, we can easily find the number of matchsticks required in any number of squares.
Example: How many matchsticks will be used in the 50th figure?
Solution

3n + 1
3 × 50 + 1
= 151 matchsticks

Question for Chapter Notes: Algebra
Try yourself:What is the pattern used to determine the number of matchsticks required to make each square in the given example?

### The Idea of a Variable

Variable refers to the unknown quantities that can change or vary and are represented using the lowercase letter of the English alphabets.
One such example of the same is the rule that we used in the matchstick pattern:
3n + 1
Here the value of n is unknown and it can vary from time to time.
More Examples of Variables

• We can use any letter as a variable, but only lowercase English alphabets.
• Numbers cannot be used for the variable as they have a fixed value.
• They can also help in solving some other problems.

Example: Karuna wanted to buy story books from a bookstall. She wanted to buy 3 books for herself, 2 for her brother and 4 for two of her friends. Each book cost Rs.15. How much money she should pay to the shopkeeper?
Solution:
Cost of 1 book = Rs.15
We need to find the cost of 9 books.

In the current situation, a (it’s a variable) stands for 9
Therefore,
Cost of 9 books = 15 × 9
= 135
Therefore Karan needs to pay Rs.135 to the shopkeeper of the bookstall.
The variable and constant not only multiply with each other but also can be added or subtracted, based on the situation.

Example: Manu has 2 erasers more than Tanu. Form an expression for the statement.
Solution:
1. Erasers that Tanu have can be represented using a variable (x).
Erasers that Manu have = Erasers that Tanu have (x) + 2
Erasers with Manu = x + 2
Solution: 2. Erasers that Manu have can be represented using a variable (y).
Erasers that Tanu has = Erasers that Manu have(y) - 2
Erasers with Tanu = y - 2

Use of Variables in Common Rules (Geometry)

1. Perimeter of Square
The perimeter of a square = Sum of all sides
= 4 × side
= 4s
Thus, p = 4s
Here s is variable, so the perimeter changes as the value of side change.

2. Perimeter of Rectangle
Perimeter of rectangle = 2(length + breadth)
= 2 (l + b) or 2l + 2b
Thus, p + 2 × (l + b) or 2l + 2b
Where, l and b are variable and the value of perimeter changes with the change in l and b.

Use of Variables in Common Rules (Arithmetic)
5 + 4 = 9
4 + 5 = 9
Thus, 5 + 4 = 4 + 5
This is the commutative property of addition of the numbers, in which the result remains the same even if we interchanged the numbers.
a + b = b + a
Here, a and b are different variables.

Example
a = 16 and b = 20
According to commutative property
16 + 20 = 20 + 16
36 = 36

2. Commutativity of Multiplication
8 × 2 = 16
2 × 8 = 16
Thus, 8 × 2 = 2 × 8
This is the commutative property of multiplication, in which the result remains the same even if we interchange the numbers.
a × b = b × a
Here, a and b are different variables.

Example
18 ×12 = 216, 12 ×18 = 216
Thus, 18 × 12 = 12 × 18

3. Distributivity of Numbers
6 × 32
It is a complex sum but there is an easy way to solve it. It is known as the distributivity of multiplication over the addition of numbers.
6 × (30 + 2)
= 180 + 12
= 192
Thus, 6 × 32 = 192
A × (b + c) = a × b + a × c
Here, a, b and c are different variables.

This property states that the result of the numbers added will remain same regardless of their grouping.
(a + b) + c = a + (b + c)

Example
(4 + 2) + 7 = 4 + (2 + 7)
6 + 7 = 4 + 9
13 = 13

Question for Chapter Notes: Algebra
Try yourself:
What is the variable in the expression 3n + 1?

### Expressions

Arithmetic expressions may use numbers and all operations like addition, subtraction, multiplication and division
Example
2 + (9 – 3), (4 × 6) – 8 etc…
(4 × 6) – 8 = 24 – 8
= 16
Expressions with Variable
We can make expressions using variables like
2m, 5 + t etc…..
An expression containing variable/s cannot be analyzed until its value is given.
Example
Find 3x – 12 if x = 6
Solution
(3 × 6) – 12
= 18 – 12
= 5
Thus,
3x – 12 = 5

Formation of Expressions

Practical use of Expressions
Example: 3 boys go to the theatre. The cost of the ticket and popcorn is \$33 and \$15 respectively. What is the cost per person?
Solution:
Let’s say,
x = cost of ticket per person
y = cost of popcorn/person
Total cost of the movie (ticket + popcorn) per person = x + y
Total cost of ticket + popcorn for 3 boys = 3(x + y)
= 3 (33 + 15)
= 3 (48)
= 144
Hence the total cost of movie ticket and popcorn for 3 boys is \$144.

### Equation

If we use the equal sign between two expressions then they form an equation.
An equation satisfies only for a particular value of the variable.
The equal sign says that the LHS is equal to the RHS and the value of a variable which makes them equal is the only solution of that equation.

Example
3 + 2x = 13
5m – 7 = 3
p/6 = 18
If there is the greater then or less than sign instead of the equal sign then that statement is not an equation.

Some examples which are not an equation
23 + 6x  > 8
6f – 3 < 24

The Solution of an Equation

The value of the variable which satisfies the equation is the solution to that equation. To check whether the particular value is the solution or not, we have to check that the LHS must be equal to the RHS with that value of the variable.
Trial and Error Method
To find the solution of the equation, we use the trial and error method.
Example: Find the value of x in the equation 25 – x = 15.
Solution:
Here we have to check for some values which we feel can be the solution by putting the value of the variable x and check for LHS = RHS.
Let’s take x = 5
25 – 5 = 15
20 ≠ 15
So x = 5 is not the solution of that equation.
Let’s take x = 10
25 – 10 = 15
15 = 15
LHS = RHS
Hence, x = 10 is the solution of that equation.

The document Algebra Class 6 Notes Maths Chapter 11 is a part of the Class 6 Course Mathematics (Maths) Class 6.
All you need of Class 6 at this link: Class 6

## Mathematics (Maths) Class 6

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## FAQs on Algebra Class 6 Notes Maths Chapter 11

 1. What are the basic concepts covered in Algebra Class 6?
Ans. Algebra Class 6 covers basic concepts such as understanding variables, expressions, equations, and solving simple algebraic problems using addition, subtraction, multiplication, and division.
 2. How can Algebra Class 6 help in improving mathematical skills?
Ans. Algebra Class 6 helps in improving mathematical skills by strengthening problem-solving abilities, enhancing critical thinking, and providing a foundation for advanced mathematical concepts in higher grades.
 3. What are some common algebraic symbols used in Algebra Class 6?
Ans. Common algebraic symbols used in Algebra Class 6 include variables (like x, y), operations symbols (+, -, *, /), and equality symbols (=).
 4. How can students apply Algebra Class 6 concepts in real-life situations?
Ans. Students can apply Algebra Class 6 concepts in real-life situations by solving everyday problems involving unknown quantities, making budgets, calculating discounts, and understanding patterns in data.
 5. What are some effective study tips for mastering Algebra Class 6?
Ans. Some effective study tips for mastering Algebra Class 6 include practicing regularly, seeking help from teachers or tutors when needed, breaking down complex problems into smaller steps, and using online resources for additional practice.

## Mathematics (Maths) Class 6

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