1. BASIC ALGEBRA
Key Terms :
•variable : variable ( nilai yang selalu berubah-ubah )
•algebraic form : bentuk aljabar
•like terms : suku sejenis
•coefficient : koefifien
•factor : pembagi
• polynomials : suku banyak
• Constant term : suku konstan
1. ALGEBRAIC EXPRESSIONS : is a collection of letter ( known as variable with or without
coefficient ) and real numbers ( known as constant )
example :
2a − 7b + 5 → variable : a and b ,
3 term : 2a, − 7b and 5
constant term
coefficient is -7
coefficient is 2
2. Simplification rules for algebraic expression
a. Like term can be combined into single term by addition and substraction.
Example : 3a + b − 2a − 5b = ( 3a − 2a ) + ( b − 5b )
b. In multiplication and division, the coefficient and variables multipled or divided.
Example : 2a × 5a = 10a 2
2m
− 2m ÷ 4n = −
4n
m
=−
2n
c. When brackets occur in algebraic expression simplify expression within the brackets and/
or remove the brackets.
Note that : (+) x (-) = (-)
(-) x (+) =(-)
(-) x (-) = (+)
Example : 7 a − { c − 2( a − c )} = 7a − { c − 2a + 2c}
= 7 a − ( − 2a + 3c )
= 7 a + 2a − 3c
= 9a − 3c
3. Expansion and factorisation of algebraic expressions
a. Some useful formula
a ( b + c ) = ab + ac
( a + b ) 2 = a 2 + 2ab + b 2
( a − b ) 2 = a 2 − 2ab + b 2
( a + b )( a − b ) = a 2 − b 2
( a + b )( c + d ) = ac + ad + bc + bd
b. Some techniques for factorization
 Find common factors
Example : 2a 2 − 6ab = 2a ( a − 3b )
 Collect and regroup the items
Example : ab + 3b − 2a − 6 = ( ab − 2a ) + (3b − 6)
= a( b − 2 ) + 3( b − 2 )

2. = ( b − 2 )( a + 3)
 Use the identity a 2 − b 2 = ( a + b )( a − b )
Example : 4m − 9n = ( 2m + 3n )( 2m − 3n )
2 2
 Cross-multiplication method
Example : 2 x 2 = 5 x − 12 = ( 2 x − 3)( x + 4 )
2x -3 − 3x
x 4 8x
2x 2 -12 5x
4. Algebraic fractions
X
a. Algebraic fractions are algebraic expressions written in the form , where X and Y are
Y
a
algebraic fractions, example :
2ab
b. Rules govering numeral fractions also apply to algebraic fractions
 When simplifying algebraic fractions, cancel common factors
12a 2b 4a
Example : =
3ab 2 b
Note: cancellation can be done only after both of numerator and denominator have
been completely factorised
2a 2 − 4a 2a ( a − 2 ) 2a
Example : = = √
a −4
2
( a + 2)( a − 2) a + 2
2 a 2 − 4a
=2−a x
a2 − 4
 Addition/substraction : First find the LCM of the denominator.
1 x 1 x
Example : + = −
x − 2 6 − 3 x x − 2 3( x − 2 )
3− x
=
3( x − 2 )
Note : As shown, you may need to factorise the denominator before determining the
LCM
 Multiplication : cancel any common factors then multiply numerators and both of
denominators.
3a 2 2b 2 2ab
Example : x =
5b 15a 25
 Division : Multiply by its reciprocal
2 xy 4 y 2 2 xy z
Example : ÷ = x 2
3z z 3z 4 y
x
=
6y
5. Solving linear equations
It involves getting the unknown terms on one side ot the equation and all other terms on the
other side of it.
Example : 3( x + 2 ) − 4(1 − x ) = 2 − ( x + 5)
3x + 6 − 4 + 4 x = 2 − x − 5
7 x + 2 = −3 − x
7 x + x = −3 − 2
8 x = −5
5
x=−
8

3. 6. Solving Fractional equations
It involves rewriting the equation into one without fractions. One way is to multiply each term of
the equation by the LCM of denominators.
x x +1 1
Example : − =
2 3 4
 x  x + 1 1
12  − 12  = 12
2  3  4
6 x − 4( x + 1) = 3
6x − 4x − 4 = 3
2x − 4 = 3
2x = 7
7
x=
2
7. Solving simultaneous linear equations
In general to solve for 2 unknown, 2 equations are needed.
a. Method I → Elimination method
It involves getting rid of one of the unknown either by addition or substrction.
b. Method II → Substitution method
It involves selecting one of the equations and the expressing one unknown in terms of the
other before substituting into the second equation.
Example : Find the value of x and y from
3 x − 2 y = 4 …………….(1)
x + 5 y = 7 …………….(2)
Method I : 3 x − 2 y = 4 x 1 3x − 2 y = 4
x + 5 y = 7 x 3 3 x + 15 y = 21 -
− 17 y = −17
y =1
Substitute y=1 into (2) → x=2
Method II : from (2) , x = 7 − 5 y ………………(3)
Substitute (3) into (1) 3( 7 − 5 y ) − 2 y = 4
21 − 15 y − 2 y = 4
− 17 y = −17
y =1
Substitute y =1 into (3) x = 7 − 5y
=7-5(1) =2