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5 Formulas Case Study 5.1 Sequences 5.2 Introduction to Functions 5.3 Simple Algebraic Fractions 5.4 Formulas and Substitution 5.5 Change of Subject Chapter Summary Case Study Body Mass Index (BMI) is frequently used to check whether a person’s body weight and body height are in an appropriate proportion. It can be calculated by using the following formula: Body Mass Index (BMI) Body weight (in kg) Body height 2 (in m 2 ) For example, if a person’s weight and height are 50 kg and 1.6 m respectively, then the BMI 50 1.62 19.5. The following table shows the range of the BMI and the corresponding body condition. BMI Less than 18.5 18.5 – 22.9 23 – 24.9 25 – 29.9 30 or over Condition Underweight Ideal Overweight Obese Severely obese P. 2 Is my weight normal? Let us check your body mass index and see whether you are normal or not. 5.1 Sequences A. Introduction to Sequences John recorded the weights of 10 classmates. He listed their weights (in kg) in the order of their class numbers. We call such a list of numbers a sequence. Each number in a sequence is called a term. We usually denote the first term as T1, the second term as T2, and so on. In the above sequence, T1 46, T2 42, T3 50, ... , etc. P. 3 5.1 Sequences B. Some Common Sequences Some sequences have certain patterns, but some do not. 1. Consider the sequence 1, 5, 9, 13, ... . +4 +4 We can guess the subsequent terms: 17, 21, 25, ... When the difference between any 2 consecutive terms is a constant, such a sequence is called an arithmetic sequence, and the difference is called a common difference. 2. Consider the sequence 8, 16, 32, 64, ... . 2 2 We can guess the subsequent terms: 128, 256, 512, ... When the ratio of each term (except the first term) to the preceding term is a constant, such a sequence is called a geometric sequence, and the ratio is called a common ratio. P. 4 5.1 Sequences B. Some Common Sequences 3. We can arrange some dots to form some squares. The number of dots used in each square is called a square number. 4. We can arrange some dots to form some triangles. The number of dots used in each triangle is called a triangular number. 5. Consider the sequence 1, 1, 2, 3, 5, 8, ... . In this sequence, starting from the third term, each term is the sum of the 2 preceding terms. This sequence is called the Fibonacci sequence. P. 5 5.1 Sequences C. General Terms For a sequence that shows a certain pattern, we can use Tn to represent the nth term. Tn is called the general term of the sequence. It is a common practice to write the general term of a sequence as an algebraic expression in terms of n. For example, since the sequence 2, 4, 6, 8, 10, ... has consecutive even numbers, we can deduce that the general term of this sequence is Tn 2n. Once the general term of a sequence is obtained, we can use it to describe any term in a sequence. P. 6 5.1 Sequences C. General Terms Example 5.1T Find the general terms of the following sequences. (a) 7, 14, 21, 28, ... (b) 1, 2, 4, ... 1 1 1 1 (c) , , , , ... (d) 15, 14, 13, 12, ... 2 4 8 16 Solution: (a) The sequence 7, 14, 21, 28, ... can be written as 7(1), 7(2), 7(3), 7(4), ... ∴ The general term of the sequence is 7n. (b) The sequence 1, 2, 4, ... can be written as 21 1, 22 1, 23 1 ... ∴ The general term of the sequence is 2n 1. P. 7 5.1 Sequences C. General Terms Example 5.1T Find the general terms of the following sequences. (a) 7, 14, 21, 28, ... (b) 1, 2, 4, ... 1 1 1 1 (c) , , , , ... (d) 15, 14, 13, 12, ... 2 4 8 16 Solution: 1 1 1 1 , , , , ... can be written as 2 4 8 16 1 2 3 4 1 , 1 , 1 , 1 , ... n 2 2 2 2 1 ∴ The general term of the sequence is . 2 (d) The sequence 15, 14, 13, 12, ... can be written as 16 1, 16 2, 16 3, 16 4, ... ∴ The general term of the sequence is 16 n. (c) The sequence P. 8 5.1 Sequences C. General Terms Example 5.2T Find the general term and the 9th term of each of the following sequences. (a) 1, 2, 5, 8, ... (b) 1, 4, 9, 16, ... +3 +3 +3 Solution: (a) T1 1 3 4 3(1) 4 T2 2 6 4 3(2) 4 T3 5 9 4 3(3) 4 T4 8 12 4 3(4) 4 ∴ Tn 3n 4 T9 3(9) 4 23 P. 9 (b) T1 1 12 T2 4 2 2 T3 9 3 2 T4 16 42 ∴ Tn n 2 T9 9 2 81 Rewrite the terms of the sequence as expressions in which the order of the terms can be observed. 5.2 Introduction to Functions In the previous section, we learnt how to find the values of the terms in a sequence from the general term by substituting different values of n in the general term. Consider a sequence with the general term Tn 5n 2. The above figure shows an ‘input-process-output’ relationship, which is called a function. For each input value of In this example, we call Tn a function of n. n, there is exactly one output value of Tn. P. 10 5.2 Introduction to Functions The idea of function is common in our daily lives. Suppose that each can of cola costs $5. Let x be the number of cans of cola, and $y be the corresponding total cost. Since the total cost of x cans of cola is $5x, the equation y 5x represents the relationship between x and y. The following table shows some values of x and the corresponding values of y. x y 1 5 2 10 3 15 4 20 5 25 For every value of x, there is only one corresponding value of y. We say that y is a function of x. P. 11 5.2 Introduction to Functions Example 5.3T If p is a function of q such that p 4q 5, find the values of p for the following values of q. 2 (a) 6 (b) 5 (c) 5 Solution: (a) When q 6, p 4(6) 5 19 (b) When q 5, p 4(5) 5 25 2 2 (c) When q , p 4 5 5 5 8 5 5 17 5 P. 12 Substitute q 6 into the expression 4q 5. 5.3 Simple Algebraic Fractions A. Simplification a We learnt at primary level that numbers in the form , where b a and b are integers and b 0, are called fractions. When both the numerator and the denominator of a fractional expression are polynomials, where the denominator is not a constant, such as: 3 5b 6x2 , or , 4b 2b 1 ( x 1)( x 2) we call such an expression an algebraic fraction. P. 13 3x 4 y is 5 not an algebraic fraction because the denominator is a constant. Note that 5.3 Simple Algebraic Fractions A. Simplification We can simplify a numerical fraction, whose numerator and denominator both have common factors, by cancelling the common factors. 24 12 2 For example: 36 12 3 2 3 For algebraic fractions, we can simplify them in a similar way, when the common factors are numbers, variables or polynomials. For example: 4a 2a(2) 6a 3 2a(3a 2 ) 2 2 3a P. 14 5.3 Simple Algebraic Fractions A. Simplification Example 5.4T Simplify the following algebraic fractions. 5ua 10uc 14 y 2 7 y (a) (b) 2ua 4uc 21y Solution: 5ua 10uc 5u (a 2c) 2ua 4uc 2u (a 2c) 5 2 14 y 2 7 y 7 y (2 y 1) (b) 7 y (3) 21y 2 y 1 1 2 y or 3 3 (a) P. 15 First factorize the numerator and the denominator. Then cancel out the common factors. We cannot cancel common factors from the terms 7y and 21y only, i.e., the fraction cannot be simplified as 14 y 2 7 y 14 y 2 . 21y 3 3 5.3 Simple Algebraic Fractions A. Simplification Example 5.5T Simplify the following algebraic fractions. mh 2nh 2km 4kn 3 x (a) (b) 2k h (6 2 x ) 2 Solution: (a) mh 2nh 2km 4kn 2k h h( m 2n) 2k ( m 2n) 2k h (h 2k )(m 2n) 2k h m 2n P. 16 (b) 3 x (6 2 x ) 2 3 x [2(3 x)]2 3 x 4(3 x) 2 1 4(3 x) First factorize the numerator. You may check if 2k + h (the denominator) is a factor of the numerator. 5.3 Simple Algebraic Fractions B. Multiplication and Division When multiplying or dividing a fraction, we usually try to cancel out all common factors before multiplying the numerator and the denominator separately to get the final result. 3 25 3 5(5) 10 9 2(5) 3(3) 5 6 We can perform the multiplication or division of algebraic fractions in a similar way. 3a 25b3 3a 5b 2 (5b) 2 For example: 10b 9a 2(5b) 3a(3a) 5b 2 6a For example: P. 17 5.3 Simple Algebraic Fractions B. Multiplication and Division Example 5.6T Simplify the following algebraic fractions. 6q 24q 2 2k 2 8t (a) (b) 3 2 n 4 5n 20 64t 8t k Solution: 2k 2 8t 2k 2 8t (a) 64t 8t 2 k 3 8t (8 t ) k 3 1 4tk 6q 24q 2 6q 5(n 4) (b) n 4 5n 20 n 4 24q 2 5 4q P. 18 5.3 Simple Algebraic Fractions C. Addition and Subtraction The method used for the addition and subtraction of algebraic fractions is similar to that for numerical fractions. 2 5 25 For example: 9a 9a 9a 7 9a When the denominators of algebraic fractions are not equal, first we have to find the lowest common multiple (L.C.M.) of the denominators. For example: 1 2 1(3) 2(2) 6a 9a 18a 7 18a P. 19 5.3 Simple Algebraic Fractions C. Addition and Subtraction Example 5.7T Simplify 7 3 . 2 h 8h Solution: 7 3 7(4) 3 2h 8h 8h 8h 28 3 8h 25 8h P. 20 5.3 Simple Algebraic Fractions C. Addition and Subtraction Example 5.8T Simplify 5 3 . 2u 6v 3v u Solution: 5 3 5 3 2u 6v 3v u 2(u 3v) (u 3v) 5 3(2) 2(u 3v) 1 2(u 3v) 1 2(3v u ) P. 21 For any number x 0, 1 1 1 x x x 5.3 Simple Algebraic Fractions C. Addition and Subtraction Example 5.9T Simplify 4 3 . 3p 9 p Solution: 4 3 4 3 3 p 9 p 3( p 3) p 4 p 3 3( p 3) 3 p ( p 3) 4 p 9 p 27 3 p ( p 3) 5 p 27 3 p( p 3) P. 22 Since 3(p 3) and p have no common factors other than 1, the L.C.M. of 3(p 3) and p is 3p(p 3). 5.4 Formulas and Substitution Consider the volume (V) of a cuboid: V lwh where l is the length, w is the width and h is the height. If the values of the variables l, w and h are already known, we can find V by the method of substitution. Similarly, if the values of V, l and w are known, we can find h. Actually, in any given formula, we can find any one of the variables by the method of substitution if all the others are known. P. 23 5.4 Formulas and Substitution Example 5.10T Consider the formula T r 2 h and r 2.5. 3h . Find the value of h if T 121 4 Solution: 3h 4 3 T h r 2 4 3 121 h 2.52 4 T r 2h 5 2 3 121 h 2 4 P. 24 Factorize the expression. 25 3 121 h 4 4 11h 121 2 121 2 h 11 h 22 5.5 Change of Subject Given the area (A) of a triangle: bh A where b is the base length and h is the height. 2 In the formula, A is the only variable on the left-hand side. We call A the subject of the formula. This formula can be used if we want to find A, when b and h are known. In another case, if we need to find h when A and b are known, it is more 2A convenient to use another formula h , with h being the subject. b 2A bh The process of obtaining the formula h from A is called the b 2 change of subject. We can use the method of solving equations to change the subject of a formula. P. 25 5.5 Change of Subject Example 5.11T Make u the subject of the formula 3k 4 5u. Solution: 3k 4 5u 3k 4 5u 5u 3k 4 3k 4 u 5 4 is transposed to the L.H.S. to become 4. Rewrite the formula such that only the variable u is on the L.H.S. P. 26 First move the other terms to one side such that only the variable u remains on the other side. 5.5 Change of Subject Example 5.12T Make m the subject of the formula 1 2 3 . p m r Solution: 1 2 3 p m r 2 3 1 m r p 2 3p r m pr 2 pr m(3 p r ) m(3 p r ) 2 pr 2 pr m 3p r P. 27 Simplify the fractions. The L.C.M. of r and p is pr. Multiply both sides by mpr. 5.5 Change of Subject Example 5.13T Make m the subject of the formula k h 2m . 3 2m Solution: h 2m 3 2m k (3 2m) h 2m 3k 2mk h 2m 3k h 2m 2mk 3k h 2m(1 k ) 2m(1 k ) 3k h 3k h m 2(1 k ) k P. 28 Remove the brackets Take out the common factor m. First move all the terms that involve m to one side. 5.5 Change of Subject Example 5.14T A bag of food is put into a refrigerator. The temperature T (in C) of 9t the food after time t (in hours) is given by the formula T 24 . 2 (a) Make t the subject of the formula. (b) How long will it take for the temperature of the food to become –3C? Solution: (a) T 24 9t 2 9t 24 T 2 2 t (24 T ) 9 P. 29 (b) 2 t [24 (3)] 9 2 (27) 9 6 ∴ It takes 6 hours for the temperature of the food to become 3C. Chapter Summary 5.1 Sequences A list of numbers arranged in an order is called a sequence. Each number in a sequence is called a term. For a sequence with a certain pattern, we can represent the sequence by its general term. P. 30 Chapter Summary 5.2 Introduction to Functions A function describes an ‘input-process-output’ relationship between 2 variables. Each input gives only one output. P. 31 Chapter Summary 5.3 Simple Algebraic Fractions The manipulations of algebraic fractions are similar to those of numerical fractions. P. 32 Chapter Summary 5.4 Formulas and Substitution A formula is any equation that describes the relationship of 2 or more variables. By the method of substitution, we can find the value of a variable in a formula when the other variables are known. P. 33 Chapter Summary 5.5 Change of Subject When there is only one variable on one side of a formula, this variable is called the subject of the formula. P. 34