Course Content
Chapter 01 – Sets
A set is a list of objects in no particular order; they could be numbers, letters, or even words. A Venn diagram is a way of representing sets visually.
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Chapter 02 – Rational Numbers
In mathematics, a rational number is a number that can be expressed as the quotient or fraction p/q of two integers, a numerator p, and a non-zero denominator q. In this chapter, we will learn to represent rational numbers on a number line and perform arithmetic operations.
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Chapter 03 – Decimals
Decimals are a set of numbers lying between integers on a number line. They are just another way to represent fractions in mathematics. In this chapter, we will learn about the conversion of decimals to rational numbers, the kinds of decimals, and absolute values.
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Chapter 04 – Exponents
The exponent of a number says how many times to use that number in a multiplication. The laws of exponents simplify the multiplication and division operations and help to solve the problems easily. In this chapter, we are going to discuss the six important laws of exponents.
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Chapter 05 – Square Root of Positive Numbers
Square root, in mathematics, is a factor of a number that, when multiplied by itself, gives the original number. In this chapter, we will learn about what makes perfect squares and will find the roots of positive numbers by considering real-life scenarios.
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Chapter 06 – Direct and Inverse Variation
Variation means change. With direct variation, numbers change proportionately in the same direction, while with inverse variation, they change in opposite directions. In this chapter, we will earn how to solve direct and inverse variation problems, explore their definitions, and work examples to understand the equations and techniques for solving them. Also, we learn to find the continued ratio for two or more ratios.
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Chapter 07 – Financial Arithmetic
Financial mathematics describes the application of mathematics and mathematical modeling to solve financial problems. In this chapter, we will learn about the concept of taxation, profit/markups, zakat & ushr, and how they relate to our daily life.
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Chapter 08 – Algebraic Expressions
Algebraic expressions are the idea of expressing numbers using letters or alphabets without specifying their actual values. The algebraic equations which are valid for all values of variables in them are called algebraic identities. In this chapter, we will learn to perform operations on polynomials and to factorize an algebraic equation by using identities.
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Chapter 09 – Linear Equations
Linear equations are equations having variables with power 1. ax+b = 0 is an example with one variable where x is the variable, and a and b are real numbers. In this chapter, we will learn the definition, type of solutions, and how to solve these equations with one variable and two variables using different methods along with examples.
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Chapter 10 – Fundamentals of Geometry
Geometry is the study of different types of shapes, figures, and sizes in Maths or real life. In geometry, we learn about different angles, transformations, and similarities in the figures. It is important to know and understand some basic concepts. We will learn about working in different numbers of dimensions, and about some of the most fundamental concepts in geometry, including points, lines, and planes.
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Chapter 11 – Practical Geometry
The practical Geometry chapter will teach you about lines and to construct two-dimensional given different kinds of measurements. A quadrilateral is a closed two-dimensional shape that has four sides and four angles. Any four-sided closed shape such as square, rectangle, rhombus, parallelogram, trapezium, etc. And a closed two-dimensional shape that has 3 sides and 3 angles is known as a triangle.
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Chapter 12 – Circumference, Area and Volume
This topic comes under analytical geometry and the formulas for the volume and the surface area of the sphere were first discovered by Archimedes. In this chapter, we will learn about the area and volume of two-dimensional and three-dimensional shapes.
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Chapter 13 – Information Handling
Frequency distribution, in statistics, is a graph or data set organized to show the frequency of occurrence of each possible outcome of a repeatable event observed many times. And, a pie chart is a way of representing data in a circular graph. Pie slices of the chart show the relative size of the data. In this chapter, we will learn to construct the frequency distribution table, some new pie chart vocabulary, and learn to construct the pie chart to represent the data.
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Grade 7 – Mathematics
About Lesson

Algebraic Identities

An algebra identity means that the left-hand side of the equation is identically equal to the right-hand side, for all values of the variables.

Standard Algebraic Identities

What Are Algebraic Identities?

Algebraic identities are equations in algebra where the value of the left-hand side of the equation is identically equal to the value of the right-hand side of the equation. They are satisfied with any values of the variables. Let us consider an example to understand this better.

Consider the equations: 5x – 3 = 12, 10x – 6 = 24, and x2 + 5x + 6 = 0. These equations satisfy only for certain value(s) of x and do not work for any value in general.

Now let us consider an equation x2 – 9 = (x + 3)(x – 3).

Note that this equation is satisfied for any value of x (try to substitute any number for x on both left and right sides, you should be getting the same answer).

These are helpful to work out numerous math problems. The four basic algebra identities are as follows.

Identity-I: (a + b)2 = a2 + 2ab + b2

Identity-II: (a – b)2 = a2 – 2ab + b2

Identity-III: (a + b)(a – b) = a2 – b2

Identity-IV: (x + a)(x + b) = x2 + (a + b) x + ab

Verify: (a + b)2 = a2 + 2ab + b2

Algebraic Verification:

(a + b)2 = a2 + b2

Step 1: Given data

We have to prove the identity-

(a + b)2 = a2 + 2ab + b2

Step 2: Proving the identity

L.H.S = (a + b)2 = (a + b)(a + b)

a(a + b) + b(a + b) = a² + ab + ab + b² = a² + 2ab + b² = R.H.S

L.H.S = R.H.S

Therefore, the algebraic identity (a + b)² = a² + 2ab + b² is verified

Geometrical Verification:

Square of a binomial | (a+b)^2

Examples:

Using the identity , simplify 

Solution

Given –
Using this Identity,



Using the identity , simplify 

Solution:

As We Know from the question, substituting the values
we Get,

=

Evaluate  by using the identity 

Solution:

Using 
We can write 
  
Hence the answer is  

Verify: (a – b)2 = a2 – 2ab + b2

Algebraic Verification:

(a – b)2 = a2 – b2

Step 1: Given data

We have to prove the identity-

(a – b)2 = a2 – 2ab + b2

Step 2: Proving the identity

L.H.S = (a – b)2 = (a – b)(a – b)

a(a – b) – b(a – b) = a² – ab – ab + b² = a² – 2ab + b² = R.H.S

L.H.S = R.H.S

Therefore, the algebraic identity (a – b)² = a² – 2ab – b² is verified

Geometrical Verification:

Algebraic Identities Charts | Printable Formulas

Examples:

Find the value of (x – 2y)by using the (a – b)2 formula.

Solution: 

To find: The value of (x – 2y)2.
Let us assume that a = x and b = 2y.
We will substitute these values in (a – b)2 formula:
(a – b)2 = a2 – 2ab + b2
(x-2y)= (x)2 – 2(x)(2y) + (2y)
= x– 4xy + 4y2 

(x – 2y)2 = x2 – 4xy + 4y2

Factorize x2 – 6xy + 9yby using a minus b Whole Square Formula.

Solution: 

To factorize: x2 – 6xy + 9y2.
We can write the given expression as:
x2 – 6xy + 9y2 = (x)2 – 2 (x) (3y) + (3y)2.
Using (a – b)2 formula: 
a2 – 2ab + b2 = (a – b)2 
Substitute a = x and b = 3y in this formula:
(x)2 – 2 (x) (3y) + (3y)2 = (x – 3y)2

 x2 – 6xy + 9y2 = (x – 3y)2

Simplify the following using (a – b)2 formula.

(7x – 4y)2

Solution:

a = 7x and b = 4y
Using formula (a – b)2  =  a2  – 2ab + b2
(7x)2 – 2(7x)(4y) + (4y)2
49x2 – 56xy + 16y2

(7x – 4y)2 = 49x2 – 56xy + 16y2.

Verify: (a + b)(a – b) = a2 – b2

Algebraic Verification:

(a + b)(a – b) = a2 – b2

Step 1: Given data

We have to prove the identity-

(a + b)(a – b) = a2 – b2

Step 2: Proving the identity

L.H.S = (a + b)(a – b)

Expanding the terms, we get

= a(a – b) + b(a – b)

= a² – ab + ab – b²

= a2 + 0 + b2

= a2 – b = R.H.S

L.H.S = R.H.S

Therefore, the algebraic identity (a + b)(a – b) = a2 – b2 is verified

Geometrical Verification:

Difference of squares

Examples:

Using a– b2 formula find the value of 1062 – 62.

Solution: 

To find: 1002 – 62.

Let us assume that a = 100 and b = 6.

We will substitute these in the a– b2 formula.
a– b= (a – b) (a + b)
1062 – 62 = (106 – 6) (106 + 6)
= (100) (112)
= 11200

1062 – 62 = 11200.

Factorize the expression 25x2 – 64.

Solution: 

To factorize: 25x2 – 64.
We will use the a– b2 formula to factorize this.
We can write the given expression as
25x2 – 64 = (5x)2 – 82
We will substitute a = 5x and b = 8 in the formula of a2 – b2
a– b= (a – b) (a + b)
(5x)2 – 82 = (5x – 8) (5x + 8)

25x2 – 64 = (5x – 8) (5x + 8)

Simplify 102 – 52 using a– b2 formula 

Solution: 

To find 102 – 52

Let us assume  a = 10 and b = 5
Using formula a– b= (a – b) (a + b)
102-52 = (10 – 5) (10 + 5)
= 10(10 +5) – 5(10 + 5)
= 10(15) – 5(15)
= 150-75 = 75

102 – 52 = 75.

Verify: (x + a)(x + b) = x2 + (a + b) x + ab

Algebraic Verification:

(x + a)(x + b) = x2 + (a + b) x + ab

Step 1: Given data

We have to prove the identity-

(x + a)(x + b) = x2 + (a + b) x + ab

Step 2: Proving the identity

L.H.S = (x + a)(x + b)

Expanding the terms, we get

= x(x + b) + a(x + b)

= x² + bx + ax + ab

Taking x common from ax and bx

= x² + (a + b) x + ab = R.H.S

L.H.S = R.H.S

Therefore, the algebraic identity (x + a)(x + b) = x2 + (a + b) x + ab is verified

Geometrical Verification:

Product of two binomials | (x+a) (x+b)

Examples:

Use the identity (x + a)(x + b) = x2 + (a + b)x + ab to find the following products

(i) (x + 3)(x + 7)   

(x + 3)(x + 7)

= x2 + (3 + 7)x + (3)(7) 

= x2 + 10x + 21

(ii) (4x + 5)(4x + 1)

(4x + 5)(4x + 1)

= (4x)2 + (5 + 1)(4x) + (5)(1)

= 16x2 + 24x + 5

(iii) (4x – 5)(4x -1)

(4x – 5)(4x – 1)

= (4x)2 + [(-5) + (-1)](4x) + (-5) (-1)

= 16x2 – 24x + 5

(iv) (4x + 5)(4x -1)   

(4x + 5)(4x -1)

= (4x)2 + [(5) + (-1)](4x) + (5)(-1)

= 16x2 + 16x – 5

(v) (2x + 5y)(2x + 3y)

(2x + 5y)(2x + 3y)

= (2x)2 + (5y + 3y)(2x) + (5y)(3y)

= 4x2 + 16xy + 15y2

(vi) (2a2 + 9)(2a2 + 5)   

(2a2 + 9)(2a2 + 5)

= (2a2)2 + (9 + 5)(2a2) + (9)(5)

= 4a4 + 28a2 + 45

Exercise Files
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