## Introduction:

Basic algebra is the branch of mathematics where we deal with variables while performing arithmetic operations i.e. use of letters which will allow us to write the rules and formulas in a general way.

So basically in basic algebra, we will be dealing with some unknown values called variables in order the solve the problems easily and quickly.

Now you must be willing to know what kind of problems do algebra gonna help in? You must know that algebra is not confined to your mathematics textbook only, it exists in the real world as well rather all mathematical concepts you learn by time are applicable somewhere in the real world which is their real importance.

To make it simpler for your level, let’s take a real-life example of the use of algebra in daily life.

**Eg: **Your mom asked you to buy 1kg sugar, 2 packets of biscuits and 1 Ltr milk from the grocery shop and gave you Rs.100. How much money shopkeeper must return you so that your all expenses must be correct where sugar costs Rs.40/kg, biscuits costs Rs.5/packet and milk costs Rs.40/ltr?

This is where basic algebra comes to rescue. This is a very simple daily life problem and I think you can solve it with your prior knowledge as well. If not, don’t worry you will be able to do so by the end of this article.

Algebra is a lot more than this example, it is more often used in different fields if engineering, statistics, economics, accountancy and other innumerable subjects. So, algebra basics are a gateway to other fields of your interest and you must learn basics of algebra for a fruitful higher education.

Now, when you know what basic algebra is, shouldn’t we dive deeper into the concepts to know more about the basics of algebra and get more familiar with it?

So, let’s get started.

## Basic Algebra Concepts:

Elementary algebra includes rules and operations such as:

- Addition
- Subtraction
- Multiplication
- Division
- Equation solving techniques
- Functions
- Polynomials
- Algebraic Expressions

Now, we will be learning these basic algebraic operations with the help of some examples.

**Eg: Find x in the equation 54x + 69 = 4x +** **119**

**Solution:**

As you can see we have introduced a term ‘**x’ **here which is known as the variable, which has some unknown value which we are going to find by using elementary algebraic operations like addition, subtraction, multiplication, and division.

1) 54x – 4x = 119 – 69 (Arrange the like together on the same side)

2) 50x = 50 (Perform subtraction operation)

3) x = 1 (Perform division operation)

So, the answer is x = 1

In this way, all other operations can be performed.

Way too simple isn’t it? This was the simpler example for your better understanding, as we move on we will discuss more examples with the increased level of difficulty.

This is the way all operations can be performed over different types of equations.

Now comes different equation solving techniques for that please go through our article on Linear Systems with two variables.

### Basic Algebra Identities:

We often encounter some mathematical problems involving manipulative calculations which become very easy if we use standard algebraic identities.

**1.** (a+b)^{2} = a^{2} + 2ab + b^{2}

**2.** (a-b)^{2} = a^{2} – 2ab + b^{2}

**3.** (a+b) (a-b) = a^{2} -b^{2}

**4.** (x + a)(x + b) = x^{2} + (a + b) x + ab

**5.** (x + a)(x – b) = x^{2} + (a – b) x – ab

**6.** (x – a)(x + b) = x^{2} + (b – a) x – ab

**7.** (x – a)(x – b) = x^{2} – (a + b)x + ab

**8.** (a + b)^{3} = a^{3} + b^{3} +3ab (a + b)

**9.** (a – b)^{3} = a^{3} – b^{3} – 3ab (a – b)

**10.** (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy +2yz + 2xz

**11.** (x + y – z)^{2} = x^{2} + y^{2} + z^{2} + 2xy – 2yz – 2xz

**12.** (x – y + z)^{2} = x^{2} + y^{2} + z^{2} – 2xy – 2yz + 2xz

**13.** (x – y – z)^{2} = x^{2} + y^{2} + z^{2} – 2xy + 2yz – 2xz

**14.** x^{3} + y^{3} + z^{3} – 3xyz = (x + y + z) (x^{2} + y^{2} + z^{2} – xy – yz -xz)

**15.** x^{2 }+ y^{2} = (1/2)[(x+y)^{2} + (x-y)^{2}]

**16.** (x+a) (x+b) (x+c) = x^{3}+ (a + b + c ) x^{2} + (ab + bc +ca) x + abc

**17.** x^{3} + y^{3} = (x + y) ( x^{2} – xy + y^{2})

**18.** x^{3} – y^{3} = (x – y) ( x^{2 }+ xy + y^{2})

**19.** x^{2 }+ y^{2 }+ z^{2 }-xy -yz -zx = (1/2)[(x-y)^{2 }+ (y -z)^{2 }+ (z -x)^{2}]

For more detail, refer to our article on algebraic identities.

## Gauss and sum of n positive integers

Do you know, how Carl Gauss at an early age displayed his genius? So, before moving further with our basics of algebra let’s look into the story of a punishment to Carl Gauss by his teacher in primary school.

He was told to add the numbers from 1 to 100. He was able to compute its sum, which is 5050, in a matter of seconds. But how?

Gauss noticed a relation between the sum of individual numbers that if he adds the first number i.e. 1 and last number i.e. 100, sum will be 101, similarly following the pattern, sum of the second number i.e. 2 and second last number i.e. 99 will also be 101 and same pattern follows for other numbers as well.

In this way, there will be a total 50 sums of 101 and in order to compute the sum of numbers 1 to 100, he multiplied 50 by 101 i.e.** 50*101 = 5050. **Isn’t this a genius work?

Further, to calculate the sum of the first n integers general formula was given which is

Try finding the sum using the formula, it will be the same.

Now you know about basic algebraic operations, different equation solving techniques and basic algebraic identities, so let’s try some numerical problems based on them.

## Solved Examples of basics of algebra:

**Question 1: What is the value of 4 ^{2} – 3^{2}**

Solution: According to the identity * a ^{2} – b^{2} = (a – b) (a + b)*

Here we take the value of a = 4 and b = 3

Hence we get 4^{2} – 3^{2 }= (4 – 3) (4 + 3)

= (1) (7) = 7

**Question 2: Simplify ( p − 1)(c − 1) + 2**

Solution: *p*(*c* − 1) − 1(*c* − 1) + 2

= *pc* − *p* − *c* + 1 + 2

= *pc* − *p* − *c* + 3

**Question 3: Expand (2x – 4y) ^{3 }using standard algebraic identities.**

**Solution: **(2x– 4y)^{3 }is of the form (a – b)^{3} = a^{3} – b^{3} – 3ab (a – b) where a = 2x and b = 4y.

So we have,

(2x – 4y)^{3} = (2x)^{3} – (4y)^{3}– 3(2x)(4y)(2x – 4y) = 8x^{3} – 64y^{3} – 48x^{2}y + 96xy^{2}

**Question 4: Expand 8a ^{3} – 27b^{3 }using standard algebraic identities.**

**Solution: **8a^{3}– 27b^{3 }is of the form x^{3} – y^{3} = (x – y) ( x^{2 }+ xy + y^{2}) where x = 2a and y = 3b.

So we have,

8a^{3}– 27b^{3 }= (2a – 3b) ( 4a^{2 }+ 6ab + 9b^{2}).

**Question 5:** A customer pays Rs. 50 for a coffee after a discount of Rs. 20. What is the original price of the coffee?

**Solution**

Let x be the original price.

x – 20 = 50

x – 20 + 20 = 50 + 20

x + 0 = 70

x = 70

**Question 6:** Half a number plus 5 is 11.What is the number?

**Solution:**

Let x be the number.

(1/2)x + 5 = 11

(1/2)x + 5 – 5 = 11 – 5

(1/2)x = 6

2 × (1/2)x = 6 × 2

x = 12

I hope now you can solve the problem we started with, so try to solve it.

This was all about the basic algebra. Keep following Toppr Bytes for more such informative articles.