Study Algebra Edition

Introduction to study algebra edition:

Algebra comes from the Arabic word al-gebra. It represents as the “reunion of the broken words”. In mathematics, Letters like a, b, c, x are been used to represent the numbers, values, complex numbers, integers and these unite with the general operations to form the algebraic expressions. If two algebraic expressions are equated we get the algebraic equations.


Learn study algebra edition:


The algebra edition mainly divides in to two topics they are Algebra1, Algebra2.The algebra 1 further classified into exponents, graphing linear equations, and quadratic equations and factoring. The algebra 2 consists of probability, inequality, trigonometry, parabolas and logarithmic functions. Algebra edition includes all the topics mentioned above.

Algebra 1:

Algebra1 is one of the most important branches in mathematics. Algebraic expression is the collection or of algebraic terms like constants and variables linked by mathematical symbols .In algebra 1 the unknown variables are represented by using letters.

Algebra 2:

Algebra2 get online help of solving equations, graphs and function, quadratic equations, slope of a line, trigonometric identities, etc.

I have recently faced lot of problem while learning Definition of Commutative Property, But thank to online resources of math which helped me to learn myself easily on net.

Example problems for study algebra edition:


Problem 1:

Solve the 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.


Problem 2:

Solve the equation 7x – 12 = 3x – 24

Solution:

7x – 12 = 3x – 24                    (Adding 12 on both sides)

7x – 12 + 12= 3x – 24 + 12

7x = 3x – 12                                     (Subtract 3x on both sides)

7x – 3x = 3x – 3x – 12

4x = -12                                   (Divide by 4 on both sides)

`4x / 4 = - 12 / 4`

x = -3

So, the answer for x is -3.