SUBJECT: MATHEMATICS
CLASS: SS 2
DATE:
TERM: 2nd TERM
REFERENCE BOOKS
TOPIC: ALGEBRAIC FRACTIONS CONTENT -Substitution in Fractions. -Undefined Fractions. SUBSTITUTION IN FRACTIONS Example 1 Given that x:y = 9:4, evaluate 8x-3yx- 34 If x:y = 9:4, then xy= 94 Divide numerator and denominator of 8x-3yx- 34yby y. 8x-3yx- 34y = 8xy- 3xy- 34 Substitute 94 for xy in the expression. Value of expression = 8 × 94- 394- 34 = 18-3 112= 15112 = 15 ÷ 32=15 × 23=10 Example 2 If x = 2a+33a-2 , express x-12x+1 in terms of a. Substitute 2a+33a-2 for x in the given expression. x -12x +1= 2a +33a -2- 12 × 2a +33a -2+ 1 Multiply the numerator and denominator by (3a – 2). x-12x+1= 2a +3- (3a-2)22a +3+ (3a -2) = 2a+3-3a+24a+6+3a -2 = -a+57a+4 or 5 -a4 +7a Example 3 Solve the equation 13a -1= 2a+1- 38 The LCM of the denominators is 8(3a – 1)(a + 1). To clear fractions, multiply the terms on both sides of the equation by 8(3a – 1)(a + 1). If 13a -1= 2a+1- 38 Then 13a -1 ×83a-1(a+1) = 2a -1=83a-1a+1 = -38 ×83a-1a+1 8(a + 1) = 16(3a – 1) – 3(3a – 1)(a + 1) 8a + 8 = 48a – 16 – 3(3a2 + 2a – 1) 8a + 8 = 48a – 16 – 9a2 – 6a + 3 8a + 8 – 48a + 16 + 9a2 + 6a – 3 = 0 9a2 – 34a + 21 = 0 (a – 3)(9a – 7) = 0 a = 3 or 9a = 7 a = 3 or 7/9 Check: if a = 3, 13a-1= 19-1= 18 and 2a+ 1- 38= 24- 38= 12- 38= 18 if a = 79, 13a-1= 173-1= 134= 34 and 2a+1- 38= 2179- 38 = 1816- 38 = 98- 38= 34 EVALUATION UNDEFINED FRACTIONS If the denominator of a fraction has the value zero, the fraction will be undefined. If an expression contains an undefined fraction, the whole expression is undefined. Example 1 Find the values of x for which the following frxactions are not defined. if x + 2 = 0 then x = -2 the fraction is not defined when x = -2. If 3x – 12 = 0 Then 3x = 12 x = 4 Example 2 Find the values of x for which the expression ax- bx2+6x-7is not defined. ax- bx2+6x-7= ax- bx-1(x+7) The expression is not defined if any of the fractions has a denominator of 0. axis undefined when x = 0. (x – 1)(x + 7) = 0 If (x – 1)(x + 7) = 0 Then either (x – 1) = 0 or (x + 7) = 0 i.e. either x = 1 or x = -7 The expression is not defined When x = 0, 1 or -7 Example 3 Solution i.i. when x – 5 = 0 x = 5 multiply both sides by x – 5 x2 + 15x + 50 = 0 (x + 5)(x + 10) = 0 Either x + 5 = 0 or x + 10 = 0 i.e. either x = -5 or x = -10 The expression is zero when x = -5 or x = -10. EVALUATION For what value(x) of x are the following expressions (i) undefined (ii) equal to zero? GENERAL EVALUATION/ REVISION QUESTIONS e.)3c+2- 22c-3= 17 WEEKEND ASSIGNMENT Objectives 1.For what values of x is the expression7x2x+1(x-1) not defined? A.1, B. -1,-1 C. -1,1 D. 2,1 2.For what values of x is the expression1x2-3x+2 not defined? A. 1,2 B. -1,2 C. -1,-2 D. 1,-2 3.Solve3+xx = 0 A. 1 B. 3 C. -3 D. -1 Theory READING ASSIGNMENT New General Mathematics SSS2, pages 195-201, exercise 17f and 17g.
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