This document summarizes the history of algebra, beginning with the contributions of Al-Khwarizmi. It discusses how Al-Khwarizmi synthesized Greek and Hindu knowledge and introduced the concepts of algebra, including the use of zero. It then explains the six standard forms Al-Khwarizmi used to reduce equations. Next, it provides examples and explanations of how to perform basic algebraic operations like addition, subtraction, multiplication, division and exponents on algebraic expressions. Finally, it discusses operations on fractions including simplifying, adding, subtracting, multiplying and dividing fractions with algebraic expressions.

1. [PART 2]
By :
Handayani Sinaga : 8146171034
Nadrah Afiati : 8146171056
Yulia Tiara Tanjung : 8146171090
Mathematic Education – Magister Program
State University of Medan
2014

2. History of Algebra
Algebra by Al-Khwarizmi
 Al-Khwarizmi's book on arithmetic synthesized
Greek and Hindu knowledge and also contained
his own fundamental contribution to
mathematics and science including an
explanation of the use of zero. The reduction is
carried out using the two operations of al-jabr
and al-muqabala.
 Here "al-jabr" means "completion" and is the
process of removing negative terms from an
equation. It is where we get the word algebra
from: Restoration and equivalence
 Having introduced the natural numbers, al-

3. Algebra by Al-
Khwarizmi
He first reduces an equation (linear or quadratic)
to one of six standard forms :
 Squares equal to roots.
 Squares equal to numbers.
 Roots equal to numbers.
 Squares and roots equal to numbers; e.g. x2 +
10 x = 39
 Squares and numbers equal to roots; e.g. x2 +
21 = 10 x.
 Roots and numbers equal to squares; e.g. 3 x
2

4. Operation of Algebraic
Expressions
Addition and
Subtraction Operation
Multiplication and
Division Operation
Exponential Operation

5. Addition and Subtraction
Operation
In algebraic expressions, the terms that
can be added and subtracted are only
the similar terms or like terms. The like
terms contain the same variable with
the same exponent or terms containing
constants. The procedures to run
addition and subtraction of algebraic
expression are as follows.
 Group the like terms
 Operate each group by adding or

6. Examples
1. Calculate (3x2 + 2x + 5) + (6x + 7)
2. Substract 2y2 + 11y from x2 + 8y2 + 7
Solutions :
1. (3x2+2x+5) + (6x+7) = 3x2 + (2x+6x) + (5+7)
= 3x2 + 8x + 12
2. The sentence “Substract 2y2 + 11y from x2 + 8y2
+ 7” means that you must run a substraction
operation of (x2 + 8y2 + 7) – (2y2 + 11y). See that
the like terms are 8y2 and 2y2. Therefore,
(x2 + 8y2 + 7) – (2y2 + 11y) = x2 + 8y2 + 7 – 2y2 - 11y
= x2 + (8y2 -2y2) - 11y + 7

7. Multiplication and Division
Operation
 Multiplications of one terms and two terms fo algebraic
expressions.
These multiplications can be done using distributive
characteristic below.
a(x ± y) = ax ± ay or (x ± y)a = ax ± ay
 Multiplications of two terms and two terms fo algebraic
expressions.
These multiplications can be done using distributive
characteristic or using multiplication scheme.
(x + y) (a + b) = x(a + b) + y (a + b) or (x + y) (a + b) =
ax + bx + ay + by
= ax + bx + ay + by

8. Examples
Expand the following multiplication of algebraic
expressions!
1. 2(x + 3)
2. 3(x + 5) + 5(x - 3)
Solution :
1. 2(x + 3) = 2x + 2 (3)
= 2x + 6
2. 3(x + 5) + 5(x - 3)= 3x + 3(5) + 5x + 5(3)
= 3x + 15 + 5x – 15
= (3 + 5)x + 15 – 15 = 8x

9. Division
In the division of algebraic expressions is known
2 key terms, i.e :
- Divisions with similar terms, such as 2x : x, and
- Divisions with different terms, such as x2 : x
See the following example to know how to
conduct a division of algebraic expressions :
We can use the properties of exponential within
integer to solve division of algebraic
expressions.
𝑥2
+ 2𝑥 : 𝑥 =
𝑥2
+ 2𝑥
𝑥
=
𝑥2
𝑥
+
2𝑥
2
= 𝑥 + 2

10. Exponential Operation
Exponential is a recur multiplication of number. We can
expand pn as follow : pn = p x x p x ... x p
n terms
Exponential properties of integers may apply to algebraic
forms. Exponential of algebraic forms shows the
multiplication of an algebraic form to it self.
Examples :
Determine theexponential result of the algebraic expressions
below.
1. (2x + 4y)2 = (2x)2 + 2(2x)(4y) + (4y)2 = 4x2 + 16xy + 16y2
2. (x – y – z)2 = {(x – y) – z}2
= (x – y)2 + 2(x - y)(-z) + z2

11. Fractions Operation of
Algebraic Expressions
Simplify Algebraic Expressions
in Fraction
Addition and Substraction
Operation in Fraction
Multiplication and Division
Operation in Fraction

12. Simplify Algebraic
Expressions in Fraction
The fraction in algebraic expressions can be
simplified by dividing the nominator and the
denominator in the fraction by the common
factor of the nominator and the denominator.
Example :
Simplify this fraction :
6𝑥2+12𝑥
9𝑥
Solution :
6𝑥2
+ 12𝑥
9𝑥
=
3𝑥 (2𝑥 + 4)
3𝑥(3)
=
2𝑥 + 4
3

13. Addition and Substraction
Operation in Fraction
The algebraic fractions can be added or
subtracted if the fractions have the same
denominator. However, if the fractions that will
be added or subtracted have different
denominators then denominators must be
equalized first.
We can equalize the different fractions
denominators by the following ways.
 Find the Least Common Multiple (LCM) of the
denominators
 Perform the given fractions into equivalent

14. Example
Perform the addition and subtraction for the following fractions.
1.
𝑥
3𝑦
+
4𝑥−𝑦
3𝑦
2.
6𝑚
𝑚−1
−
4
𝑚2−1
Solutions :
1.
𝑥
3𝑦
+
4𝑥−𝑦
3𝑦
=
𝑥+4𝑥−𝑦
3𝑦
=
5𝑥−𝑦
3𝑦
2. The denominators of
6𝑚
𝑚−1
and
4
𝑚2−1
are m-1 and m2
-1 respectively. The LCM of
m-1 and m2
-1 is m2
-1 = (m-1)(m+1), thus
6𝑚
𝑚−1
−
4
𝑚2−1
=
6𝑚(𝑚+1)
𝑚−1 (𝑚+1)
−
4
𝑚−1 (𝑚+1)
=
6𝑚 𝑚+1 − 4
𝑚−1 (𝑚+1)
=
6𝑚2+6𝑚−4
𝑚−1 (𝑚+1)

15. Multiplication and Division
Operation in Fraction
Multiplication with algebraic
fractions can be done by
multiplying both the nominators
and multiplying both the
denominators. Then, the result is
commonly simplified.
To divide fractions, say
𝑥
𝑦
with
𝑎
𝑏
, can be done by multiplying fractions
𝑥
𝑦
with
𝑏
𝑎
. Therefore,
𝑥
𝑦
∶
𝑎
𝑏
=
𝑥
𝑦
𝑥
𝑏
𝑎
=
𝑥𝑏
𝑦𝑎

16. Examples
Perform the multiplication and division of fractions below
1.
7𝑦
2𝑥
𝑥
𝑥𝑦
14𝑦
2.
𝑚
𝑚+1
:
𝑚+3
𝑚+1
Solutions :
1.
7𝑦
2𝑥
𝑥
𝑥𝑦
14𝑦
=
7𝑥𝑦2
28𝑥𝑦
=
𝑦
4
3.
𝑚
𝑚+1
:
𝑚+3
𝑚+1
=
𝑚
𝑚+1
𝑥
𝑚+1
𝑚+3
=
𝑚
𝑚+3

17. - THANK YOU -