Formulas 5 5.1Sequences 5.2Introduction to Functions 5.3Simple Algebraic Fractions Chapter Summary Case Study 5.4Formulas and Substitution 5.5Change of Subject

P. 2 Case Study The following table shows the range of the BMI and the corresponding body condition. 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: For example, if a person’s weight and height are 50 kg and 1.6 m respectively, then the BMI  50   BMICondition Less than 18.5Underweight 18.5 – 22.9Ideal 23 – 24.9Overweight 25 – 29.9Obese 30 or overSeverely obese Let us check your body mass index and see whether you are normal or not. Is my weight normal?

P. 3 We call such a list of numbers a sequence. A. Introduction to Sequences John recorded the weights of 10 classmates. He listed their weights (in kg) in the order of their class numbers. Each number in a sequence is called a term. We usually denote the first term as T 1, the second term as T 2, and so on. In the above sequence, T 1  46, T 2  42, T 3  50,..., etc. 5.1 Sequences

P Sequences B. Some Common Sequences Some sequences have certain patterns, but some do not. 1.Consider the sequence 1, 5, 9, 13,.... 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. We can guess the subsequent terms: 17, 21, 25,... 2.Consider the sequence 8, 16, 32, 64,.... 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. We can guess the subsequent terms: 128, 256, 512, 22 22

P. 5 B. Some Common Sequences 5.1 Sequences 3.We can arrange some dots to form some squares. 4.We can arrange some dots to form some triangles. The number of dots used in each square is called a square number. 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 Sequences C. General Terms For a sequence that shows a certain pattern, we can use T n to represent the nth term. It is a common practice to write the general term of a sequence as an algebraic expression in terms of n. T n is called the general term of the sequence. 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 T n  2n. Once the general term of a sequence is obtained, we can use it to describe any term in a sequence.

P. 7 (a)The sequence 7, 14, 21, 28,... can be written as 7(1), 7(2), 7(3), 7(4), Sequences C. General Terms Example 5.1T Find the general terms of the following sequences. (a)7, 14, 21, 28,... (b)  1,  2,  4,... (c)(d)15, 14, 13, 12,... Solution: (b)The sequence  1,  2,  4,... can be written as  2 1  1,  2 2  1,  2 3  1... ∴ The general term of the sequence is 7n. ∴ The general term of the sequence is  2 n  1.

P Sequences C. General Terms Example 5.1T Solution: Find the general terms of the following sequences. (a)7, 14, 21, 28,... (b)  1,  2,  4,... (c)(d)15, 14, 13, 12,... (c)The sequence can be written as ∴ The general term of the sequence is. (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.

P 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,... Solution: Rewrite the terms of the sequence as expressions in which the order of the terms can be observed. (a)T 1   1 T 2  2 T 3  5 T 4  8 ∴ T n  3n  4  3  4  6  4  9  4  12  4 (b)T 1  1 T 2  4 T 3  9 T 4  16 ∴Tn n2∴Tn n2  1 2  2 2  3 2  4 2 T 9  3(9)  4  23 T 9  9 2   3(1)  4  3(2)  4  3(3)  4  3(4)  4

P Introduction to Functions Consider a sequence with the general term T n  5n  2. The above figure shows an ‘input-process-output’ relationship, which is called a function. In this example, we call T n a function of n. 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. For each input value of n, there is exactly one output value of T n.

P. 11 For every value of x, there is only one corresponding value of y. 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. 5.2 Introduction to Functions The idea of function is common in our daily lives. x12345 y We say that y is a function of x.

P 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. (a) 6(b)  5(c) Solution: Substitute q  6 into the expression 4q  5. (a)When q  6, p  4(6)  5 (b)When q   5, p  4(  5)  5   25 (c)When q , p     19

P. 13 we call such an expression an algebraic fraction. A. Simplification 5.3 Simple Algebraic Fractions We learnt at primary level that numbers in the form, where 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: Note that is not an algebraic fraction because the denominator is a constant.

P. 14 For example: For algebraic fractions, we can simplify them in a similar way, when the common factors are numbers, variables or polynomials. A. Simplification 5.3 Simple Algebraic Fractions We can simplify a numerical fraction, whose numerator and denominator both have common factors, by cancelling the common factors. For example:

P. 15 A. Simplification 5.3 Simple Algebraic Fractions Example 5.4T Simplify the following algebraic fractions. (a) (b) First factorize the numerator and the denominator. Then cancel out the common factors. Solution: We cannot cancel common factors from the terms  7y and  21y only, i.e., the fraction cannot be simplified as. 3

P. 16 A. Simplification 5.3 Simple Algebraic Fractions Example 5.5T Simplify the following algebraic fractions. (a) (b) Solution: First factorize the numerator. You may check if 2k + h (the denominator) is a factor of the numerator.

P. 17 B. Multiplication and Division 5.3 Simple Algebraic Fractions For example: We can perform the multiplication or division of algebraic fractions in a similar way. 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. For example:

P. 18 B. Multiplication and Division 5.3 Simple Algebraic Fractions Example 5.6T Solution: Simplify the following algebraic fractions. (a) (b)

P. 19 C. Addition and Subtraction 5.3 Simple Algebraic Fractions For example: The method used for the addition and subtraction of algebraic fractions is similar to that for numerical fractions. 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:

P. 20 C. Addition and Subtraction 5.3 Simple Algebraic Fractions Example 5.7T Simplify. Solution:

P. 21 C. Addition and Subtraction 5.3 Simple Algebraic Fractions Example 5.8T Solution: Simplify. For any number x  0,

P. 22 C. Addition and Subtraction 5.3 Simple Algebraic Fractions Example 5.9T Simplify. Solution: 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).

P Formulas and Substitution Consider the volume (V) of a cuboid: V  lwhwhere 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 Formulas and Substitution Example 5.10T Consider the formula. Find the value of h if T  121 and r  2.5. Solution: Factorize the expression.

P Change of Subject Given the area (A) of a triangle: where b is the base length and h is the height. 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 convenient to use another formula, with h being the subject. The process of obtaining the formula from is called the change of subject. We can use the method of solving equations to change the subject of a formula.

P Change of Subject Example 5.11T Make u the subject of the formula 3k  4  5u. Solution: First move the other terms to one side such that only the variable u remains on the other side. Rewrite the formula such that only the variable u is on the L.H.S.  4 is transposed to the L.H.S. to become  4.

P Change of Subject Example 5.12T Make m the subject of the formula. Solution: Simplify the fractions. The L.C.M. of r and p is pr. Multiply both sides by mpr.

P Change of Subject Example 5.13T Make m the subject of the formula. Solution: First move all the terms that involve m to one side. Remove the brackets Take out the common factor m.

P. 29 ∴ It takes 6 hours for the temperature of the food to become  3  C. 5.5 Change of Subject Example 5.14T A bag of food is put into a refrigerator. The temperature T (in  C) of the food after time t (in hours) is given by the formula. (a)Make t the subject of the formula. (b)How long will it take for the temperature of the food to become –3  C? Solution: (a)(b)

P. 30 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 Introduction to Functions Chapter Summary A function describes an ‘input-process-output’ relationship between 2 variables. Each input gives only one output.

P Simple Algebraic Fractions Chapter Summary The manipulations of algebraic fractions are similar to those of numerical fractions.

P Formulas and Substitution Chapter Summary 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 Change of Subject Chapter Summary When there is only one variable on one side of a formula, this variable is called the subject of the formula.

5.1 Sequences C. General Terms Follow-up 5.1 Find the general terms of the following sequences. (a)4, 8, 16, 32,... (b)  6,  12,  18,  24,... (c)3, 9, 27, 81,...(d)6, 7, 8, 9,... Solution: (a)The sequence 4, 8, 16, 32,... can be written as 2  2 1, 2  2 2, 2  2 3, 2  2 4,... ∴ The general term of the sequence is 2 n  1. (b)The sequence  6,  12,  18,  24,... can be written as  6(1),  6(2),  6(3),  6(4),... ∴ The general term of the sequence is  6n.

5.1 Sequences C. General Terms Follow-up 5.1 Find the general terms of the following sequences. (a)4, 8, 16, 32,... (b)  6,  12,  18,  24,... (c)3, 9, 27, 81,...(d)6, 7, 8, 9,... Solution: (c)The sequence 3, 9, 27, 81,... can be written as 3 1, 3 2, 3 3, 3 4,... ∴ The general term of the sequence is 3 n. (d)The sequence 6, 7, 8, 9,... can be written as 5  1, 5  2, 5  3, 5  4,... ∴ The general term of the sequence is 5  n.

5.1 Sequences C. General Terms Follow-up 5.2 Solution: Find the general term and the 30th term of each of the following sequences. (a) 15, 24, 33, 42,... (b) (a)T 1  15 T 2  24 T 3  33 T 4  42 ∴ T n  9n  6  9  6  18  6  27  6  36  6 (b)Consider the denominators 2, 5, 8, 11,.... ∴Tn∴Tn T 9  9(30)  6  276 T 30  +9  9(1)  6  9(2)  6  9(3)  6  9(4)  6 +3 Rewrite the denominators as 3  1, 6  1, 9  1, 12  1,....

5.2 Introduction to Functions Follow-up 5.3 If y is a function of x such that y  3x  2, find the values of y for the following values of x. (a) 7(b)  5(c) Solution: Substitute x  7 into the expression 3x  2. (a)When x  7, y  3(7)  2 (b)When x   5, y  3(  5)  2  23   13 (c)When x , y   2  2  4

A. Simplification 5.3 Simple Algebraic Fractions Follow-up 5.4 Solution: Simplify the following algebraic fractions. (a) (b)

A. Simplification 5.3 Simple Algebraic Fractions Follow-up 5.5 Solution: Simplify the following algebraic fractions. (a) (b)

Follow-up 5.6 B. Multiplication and Division 5.3 Simple Algebraic Fractions Solution: Simplify the following algebraic fractions. (a) (b) Note that (1  x)   (x  1).

Follow-up 5.7 C. Addition and Subtraction 5.3 Simple Algebraic Fractions Solution: Simplify the following expressions. (a) (b)

Follow-up 5.8 C. Addition and Subtraction 5.3 Simple Algebraic Fractions Solution: Simplify the following expressions. (a) (b)

Follow-up 5.9 C. Addition and Subtraction 5.3 Simple Algebraic Fractions Solution: Simplify the following expressions. (a) (b)

5.4 Formulas and Substitution Follow-up 5.10 (a)Consider the formula. If A  120, b  6 and h  15, find a. (b)Consider the formula. If u  6, d  15, find t. Solution: (a) 240  15(a  6) 16  a  6 a  10 (b) t  18 Factorize the expression.

5.5 Change of Subject Follow-up 5.11 Make each of the variables below the subject of the formula y  9z  8x. (a)z(b)x Solution: (a) (b) Rewrite the formula such that only the variable z is on the L.H.S.  8x is transposed to the L.H.S. to become  8x.  y is transposed to the R.H.S. to become  y.

5.5 Change of Subject Follow-up 5.12 Make the letter in brackets the subject in each of the following formulas. (a)Tr  mg  mv [m] (b) [g] Solution: (a) (b) Take out the common factor m.

5.5 Change of Subject Follow-up 5.13 Make the letter in brackets the subject in each of the following formulas. (a) [y] (b) [u] Solution: (a) (b)

5.5 Change of Subject Follow-up 5.14 A tap is turned on and water starts to flow into a tank. The volume V of water (in cm 3 ) inside the tank after time t (in seconds) is given by the following formula V  15t  170. (a)Make t the subject of the formula. (b)How long will it take for 500 cm 3 of water to fill the tank? Solution: (a) (b) ∴ It takes 22 seconds to fill the tank.