Basic Of Algebra
Basics of Algebra cover the simple operation of mathematics like addition, subtraction,
multiplication, and division involving both constant as well as variables. For example, x+10 =
0. This introduces an important algebraic concept known as equations. The algebraic equation
can be thought of as a scale where the weights are balanced through numbers or constants.
Algebra is the branch of Maths which uses alphabetical letters to find unknown numbers.
These letters are also called variables. The values which are known in the given expression
such as numbers are called constants. Though in higher classes, students will learn the
concept of algebra at the potential level. But when we speak about its basics, it covers the
general algebraic expressions, formulas and identities, which are used to solve many
mathematical problems. Let us learn here the basic concept of algebra with the help of some
terminology, formulas, rules, examples and solved problems.
Basics Of Algebra
In the above-given equation, the letters x and y are the unknown variables which we have to
determine. Whereas 3 and 2 are the numerical values. c denotes the constant term.
Basic Algebra
The algebra for class 6 covers all the basic concepts. Terms related to basic algebra skills are
mentioned below.
Exponent
Expression
Polynomial (Monomial, binomial and trinomial)
Like terms and Unlike terms
Constants
An equation is a statement which implies two same identities separated by “=” sign. Whereas
an expression is a group of different terms separated by ‘+’ or ‘-‘ sign.
Like terms are those terms whose variables and their exponents are same.
Basic Algebra Rules
The basic algebra rules are mentioned below:
The Symmetry rule
The commutative rules
The inverse of adding
Two rules for equation
Basic Algebra Operations
The general arithmetic operations performed in the case of algebra are:
Addition: x + y
Subtraction: x – y
Multiplication: xy
Division: x/y or x ÷ y
where x and y are the variables.
The order of these operations will follow the BODMAS rule, which means the terms inside the
brackets are considered first. Then, roots and exponents are operated on second priority.
Solve all the division and multiplication operations and later addition and subtraction.
Basic Algebra Formula
The general formulas used in algebra to solve algebraic equations and find the values of
unknown variables are given here:
a2 – b2 = (a – b)(a + b)
(a+b)2 = a2 + 2ab + b2
a2 + b2 = (a – b)2 + 2ab
(a – b)2 = a2 – 2ab + b2
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc
(a – b – c)2 = a2 + b2 + c2 – 2ab – 2ac + 2bc
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(a – b)3 = a3 – 3a2b + 3ab2 – b3
These fomulas are used in higher secondary classes. Students can find algebra formulas for
class 8, along with class 9, class 10, class 11 and class 12 here.
Algebra Formulas For Class 9
Algebra Formulas For Class 10
Algebra Formulas For Class 11
Algebra Formulas For Class 12
Some of the basic algebra worksheets are given below containing basic math algebra
questions.
Basic Algebra Examples
Q 1: Find y, when, y + 15 = 30
Solution: y = 30 – 15
y = 15
Q 2 : Find x, when, 9x = 63
Ans. x = 63/9
x=7
Q.3: If x/7 = 21, then find x.
Solution: Given x/7 = 21
or x = 21 x 7
x = 147
Practice Problems
Solve x+12 = 6
Find the value of z, if 23z + 3 = 10
Solve 2y – 8 = 5y
Addition And Subtraction Of Algebraic Expressions
The four basic mathematics operations viz. addition, subtraction, multiplication, and division can also
be performed on algebraic equations or expressions. The addition and subtraction of algebraic
expressions are almost similar to the addition and subtraction of numbers. However, a combination
of variables, constants, and operators constitutes an algebraic expression. But in the case of algebraic
expressions, like terms and the unlike terms must be sorted together.
How to Add and Subtract Algebraic Expressions?
The knowledge of like and unlike terms is crucial while studying addition and subtraction of
algebraic expressions because the operation of addition and subtraction can only be
performed on like terms. The terms whose variables and their exponents are the same are
known as like terms and the terms having different variables are unlike terms.
Example: -5x2 + 12 xy – 3y + 7x2 + xy
In the given algebraic expression, -5x2 and 7x2 are like since both the terms have x2 as the
common variable. Similarly, 12xy and xy are like terms.
Addition of Algebraic Expressions
There are two methods to add and subtract algebraic expressions, namely the horizontal
method and the vertical method. Let us understand these two methods along with the
examples in detailed steps.
Horizontal Method
Consider three algebraic expressions 5xy – 3x2 – 12y +5x, xy – 3x – 12yz + 5x3 and y – 6x2
– zy + 5x3.
Step 1: Write the given algebraic expressions using an additional symbol.
(5xy – 3x2 – 12y +5x) + (xy – 3x – 12yz + 5x3) + (y – 6x2 – zy + 5x3)
Step 2: Open the brackets and multiply the signs.
5xy – 3x2 – 12y + 5x + xy – 3x – 12yz + 5x3 + y – 6x2 – zy + 5x3
Step 3: Now, combine the like terms.
(5xy + xy) + (-3x2 – 6x2) + (-12y + y) + (5x – 3x) + (-12yz – yz) + (5x3 + 5x3)
Step 4: Add the coefficients. Keep the variables and exponents on the variables the same.
6xy – 9x2 – 11y + 2x – 13yz + 10x3
Column Method
In this method, the given expression must be written column-wise one below the other. For
adding two or more algebraic expressions the like terms of both the expressions are grouped
together. The coefficients of like terms are added together using simple addition techniques
and the variable which is common is retained as it is. The unlike terms are retained as it is
and the result obtained is the addition of two or more algebraic expressions.
Addition And Subtraction
We can arrange the terms of the given expressions in the same order to represent the result.
Read more:
Addition and subtraction of Decimals
Addition and Subtraction of fractions
Addition
Subtraction
Subtraction of Algebraic Expressions
Similar to the addition of algebraic expressions, we can subtract algebraic expressions in two
methods. Let’s understand these two methods along with the example.
Example:
Subtract x2y – 2x2 – zy + 5 and –3x2 + 3x3 from y3 + 3x2y – 6x2 – 6zy + 7x3
Solution:
Let’s solve this problem in two ways.
Horizontal Method:
Step 1: Write the given algebraic expressions using an additional symbol.
(y3 + 3x2y – 6x2 – 6zy + 7x3) – (x2y – 2x2 – zy + 5) – (–3x2 + 3x3)
Step 2: Open the brackets and multiply the signs.
y3 + 3x2y – 6x2 – 6zy + 7x3 – x2y + 2x2 + zy – 5 + 3x2 – 3x3
Step 3: Now, combine the like terms.
y3 + (3x2y – x2y) + (-6x2 + 2x2 + 3x2) + (-6zy + zy) + (7x3 – 3x3) + (-5)
Step 4: Add the coefficients. Keep the variables and exponents on the variables the same.
y3 + 2x2y – x2 – 5zy + 4x3 – 5
Column Method:
For subtracting two or more algebraic expressions, it’s a better practice to write the
expressions to be subtracted below the expression from which it is to be subtracted. Like terms
are placed below each other. The sign of each term which is to be subtracted is reversed and
then the resulting expression is added normally.
Addition And Subtraction
In this way, we can add and subtract two or more expressions in either of the methods.
Practise Problems
Add 8a3b3 + 4ab3 – 2a2b3 and 8a2 – a2b3 + 4ab3
Subtract 44x4 + 17x3 + x – 18 from 54x4 – 33x3 – 28x2 + 7x – 6.
Simplify: (10x2 + 14) + (9x2 + 3) – (8x2 – 6)
Simplify: (29x3 – 11x + 30) – (17x3 – 8x – 22) + (37x2 + 4)
Simplify: 5x2 – 9x – 2y2 + 3x2 + 5x + 20x – 5 + 30
Multiplication of Algebraic Expressions
In Maths, Algebra is one of the important branches. The concept of algebra is used to find the
unknown variables or unknown quantity. The multiplication of algebraic expressions is a
method of multiplying two given expressions consisting of variables and constants. Algebraic
expression is an expression that is built by the combination of integer constants and variables.
For example, 4xy + 9, in this expression x and y are variables whereas 4 and 9 are constants.
The value of an algebraic expression changes according to the value chosen for the variables
of the expressions.
If there are brackets given in any expression, then it should be simplified first. When there is
no bracket present, then the algebraic expressions can also be solved by applying division
and multiplication, and then addition and subtraction, similar to BODMAS rule.
For a clear idea on this let us take an expression 2x+1. Now if x = 1, the value of expression
would be 3. If x = 2 the value will be 5 and so on. The value of the expression is dependent
on the value of the variable. And if we have 2(x+1), then we will not get the same answer as
we have got earlier. Here, if we put x =1, then it gives 2(1+1) = 4. So, we can see the difference
when we use brackets in expressions.
Terms Used in Algebra
Variable: The unknown quantity used in algebraic expressions, such as x, y, z, a, b, etc.
Coefficient: The value attached with the variable is the coefficient of it. For example, for 3x,
the coefficient is 3
Monomial: An expression with one term. For example, 4x, 5y, 6z
Binomial: An expression with two terms. For example, 4x+4, 5y+y, 6z+z
Trinomial: An expression with three terms. For example, 4x2+4x+1
Solved Examples
Let us solve some problems here based on the multiplication of different types of algebraic
expressions.
Illustration 1: Multiply 5x with 21y and 32z
Solution: 5x × 21y × 32z = 105xy × 32z = 3360xyz
We multiply the first two monomials and then the resulting monomial to the third monomial.
Illustration 2: Find the volume of a cuboid whose length is 5ax, breadth is 3by and height is
10cz.
Solution:
Volume = length × breadth × height
Therefore, volume = 5ax × 3by × 10cz = 5 × 3 × 10 × (ax) × (by) × (cz) = 150axbycz
Illustration 3: Multiply (2a2 + 9a + 10) by 4a.
Solution:
4a × (2a2 + 9a + 10)
= (4a × 2a2) + (4a × 9a) + (4a × 10)
= 8a3 + 36a2 + 40a
Illustration 4: Simplify the below algebraic expression and obtain its value for x = 3.
x(x − 2) + 5
Solution: Given, x(x − 2) + 5, x = 3.
On simplifying the given expression we get:
x2-2x+5
Now putting x = 3, we get;
= 32-2(3)+5
=9–6+5
=8
Illustration 5: Simplify the below algebraic expression and obtain its value for y = −1
4y(2y − 6) – 3(y − 2) + 20
Solution: 4y(2y − 6) − 3(y − 2) + 20 for y = −1
Substituting the value of y = −1.
4 × −1((2 × −1) – 6) – 3(−1 − 2) + 20
= −4 (−2 − 6) − 3(−3) + 20
= 32 + 9 + 20 = 61.
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