Learning Basic Algebra Free

Introduction to learning basic algebra free:

Algebra, a branch of mathematics deals with the expressions of various order, their operations, relations and construction and concepts of expressions.Algebra is the combination of numerical numbers, alpha-numeric variables, expressions and etc., the basic aim of the algebra is balancing the algebraic equations on both sides.In learning basic algebra free, you can learn about the various expressions and order of expressions, with their standard forms.In this article for learning basic algebra free we shall discuss about the basics of algebra. Is this topic Improper Fractions to Mixed Numbers hard for you? Watch out for my coming posts.


Basics of Algebra – learning basic algebra free:


Adding and Subtracting ,
Multiplication of algebra ,
Division of algebraic and
Solving Equations.


Adding and Subtracting Algebraic Expressions:

This shows how to solve the addition and subtraction in algebraic expression.

eg:

Addition:

2a^2 + 2b + 5a^2= 7a^2 +2b.

Subtraction:

15a^2 – 31b -10a^2 = 5a^2 – 31b.

Multiplication of algebra expressions:

This shows how to solve the Multiplication in algebraic expression.

eg:

multiplication:

2(a + 3b) = 2a +6b.

Division of algebraic expressions:

This shows how to solve the Division on in algebraic expression.

Eg:

Division:

12ab2 / 31ab2

Solving Equations:

This shows how to solve the system of algebraic expression.

Eg:

Solve:

5 - (x + 2) = 0.

-5x + 10 = 0

-5x = -10

x = 2

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Example for solving algebra - learning basic algebra free:


Solve the following linear equations: x + 2y + 3z = 14, 3x + y + 2z = 11, 2x + 3y + z = 11.

Solution:

x + 2y + 3z = 14 (1)

3x + y + 2z = 11 (2)

2x + 3y + z = 11 (3)

Consider the equations (1) and (3)

(1) ? x + 2y + 3z = 14

(3) × 3 ? 6x + 9y + 3z = 33 (subtracting)

–5x – 7y = –19

5x + 7y = 19 (4)

Consider the equations (2) and (3)

(2) ? 3x + y + 2z = 11

(3) × 2 ? 4x + 6y + 2z = 22 (subtracting)

–x – 5y = –11

x + 5y = 11 (5)

Consider the equations (4) and (5)

(4) ? 5x + 7y = 19

(5) × 5 ? 5x + 25y = 55 (subtracting)

–18y = –36

? y = 2

Substitute y = 2 in (5) we get

x + 5(2) = 11

x + 10 = 11

x + 10 - 10 = 11 - 10

? x = 1

Substitute x = 1, y = 2 in (3) we get

2(1) + 3(2) + z = 11

2 + 6 + z = 11

8 - 8 + z = 11 - 8

? z = 3

The solution is x = 1, y = 2, z = 3.