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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 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 1^{st} square = 4
No. of matchsticks used to make 2^{nd} square = 7
No. of matchsticks used to make 3^{rd} 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 50^{th} figure?
Solution
3n + 1
3 × 50 + 1
= 151 matchsticks
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
Here the value of n is unknown and it can vary from time to time.
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: 1
Karan wanted to buy story books from a bookstall. She wanted to buy 3 books for herself, 2 for her brother and 4 for 2 of her friends. Each book cost Rs.15.how much money she should pay to the shopkeeper?
Solution: 1
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: 2
Manu has 2 erasers more than Tanu. Form an expression for the statement.
Solution: 2.1
Erasers that Tanu have can be represented using a variable (x)
Erasers that Manu have = erasers that Tanu have + 2
Erasers with Manu = x + 2
Solution: 2.2
Erasers that Manu have can be represented using a variable (y)
Erasers that Tanu has = erasers that Manu have - 2
Erasers with Tanu = y - 2
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.
1. Commutativity of Addition
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.
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
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.
4. Associativity of Addition
This property states that the result of the numbers added will remain same regardless of their grouping.
(a + b) + c = a + (b + c)
(4 + 2) + 7 = 4 + (2 + 7)
6 + 7 = 4 + 9
13 = 13
Arithmetic expressions may use numbers and all operations like addition, subtraction, multiplication and division
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.
Find 3x – 12 if x = 6
(3 × 6) – 12
= 18 – 12
= 5
Thus,
3x – 12 = 5
y subtracted from 12
x multiplied by 6
t Multiplied by 4, and then subtract 5 from the product.
3 boys go to the theatre. The cost of the ticket and popcorn is $33 and $15 respectively. What is the cost per person?
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.
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.
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 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.
Find the value of x in the equation 25 – x = 15.
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.
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