Handout basic algebra
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1. Basic algebra involves variables, algebraic expressions, and equations. Variables represent unknown values. 2. Algebraic expressions contain variables, numbers, and operators. They can be simplified by combining like terms or using properties of exponents. 3. Equations set two algebraic expressions equal to each other and can be solved algebraically to find the value of variables. There are methods for solving different types of equations like linear, fractional, and simultaneous equations.
Handout basic algebra
- 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