Focus On Maths Grade 7

CHAPTER 4 Algebraic Expressions 136 4.4 Division of Algebraic Expressions A Dividing algebraic expressions (I) Quotient of like terms When finding the quotient of algebraic expressions that consist of one term, the number is divided by number and only the same variable can be simplified. EXAMPLE 21 Find the quotient of each of the following. (a) 16p ÷ 2p (b) 32g4 h2 ÷ (–12g4 h 2 ) Solution: (b) 32g4 h 2 ÷ (–12g4 h2 ) = – 32g 4 h2 12g4 h2 = – 8 1 1 1 1 1 1 32 × g × g × g × g × h × h 12 × g × g × g × g × h × h 3 1 1 1 1 1 1 = – 8 3 = –2 2 3 (a) 16p ÷ 2p = 16p 2p = 8 1 16 × p 2 × p 1 1 = 8 (II) Quotient of unlike terms Division of one variable can only be carried out by another same variable to find the quotient of two algebraic terms. EXAMPLE 22 Find the quotient of each of the following. (a) 35xyz 7xy (b) –12pq2 r 2 ÷ 18qr 2 Solution: (a) 35xyz 7xy = 5 1 1 35 × x × y × z 7 × x × y 1 1 1 = 5z (b) –12pq2 r 2 ÷ 18qr 2 = – 12pq2 r2 18qr2 = – 2 1 1 1 12 × p × q × q × r × r 18 × q × r × r 3 1 1 1 = – 2pq 3 ©Praxis Publishing_Focus On Maths

Algebraic Expressions CHAPTER 4 137 EXAMPLE 23 Find the quotient for each of the following. (a) 18x 6 y 3 ÷ 2x 2 y (b) –21abc 5 28ac 4 Solution: (a) 18x 6 y 3 ÷ 2x 2 y = 18x 6 y 3 2x 2 y = 9 18 × x × x × x × x × x × x × y × y × y 1 2 × x × x × y = 9x 4 y 2 (b) –21abc 5 28ac 4 = –21 × a × b × c × c × c × c × c 4 28 × a × c × c × c × c = – 3bc 4 Write in the fraction form. Simplify –3 EXAMPLE 24 Simplify each of the following. (a) – 8a2 b5 × 6bc ÷ 12ab 3 c (b) 10uv 3 ÷ (–15u2 vw) × (–4u4 w) Solution: (a) – 8a2 b 5 × 6bc ÷ 12ab3 c = – 8a2 b5 × 6bc 12ab 3 c = – 8 × a2 × b5 × 6 × b × c 12 × a × b3 × c = – 8 × 6 × a × a × b × b × b × b × b × b × c 12 × a × b × b × b × c = –4 –48 × a × a × b × b × b × b × b × b × c 1 12 × a × b × b × b × c = –4ab 3 Write in the fraction form. Write each expression as multiplication of factors. • y a × y b = y a + b For example: y 5 × y3 = y 5 + 3 = y 8 • x a ÷ x b = x a – b For example: x5 ÷ x3 = x5 ‒ 3 = x2 This law can only be applied for the multiplication and division involving the same variable. ©Praxis Publishing_Focus On Maths

CHAPTER 4 Algebraic Expressions 138 (b) 10uv 3 ÷ (–15u2 vw) × (–4u4 w) = 10uv 3 –15u2 vw × (–4u4 w) = 10 × u × v 3 –15 × u2 × v × w × (–4 × u4 × w) = 10 × u × v × v × v –15 × u × u × v × w × (–4 × u × u × u × u × w) = 2 × v × v –3 × (–4 × u × u × u) = 8 3 u3 v 2 –3 2 The diagram shows a cuboid aquarium. The height of the water level in the aquarium is 11y m. The empty space will be filled when 18x identical marbles are put into the aquarium and 30x2 y m3 of water will overflow at the same time. What is the volume, in m3 , of a marble? 13y m 11y m 6x m 20x m Maths LINK History The word ‘algebra’ comes from the Arabic word ‘al-jabr’. This word appeared in the title of one of the earliest algebra texts, written around the year 825 by al-Khwarizmi. He lived in what is now Uzbekistan. EXAMPLE 25 Simplify –12uv 2 × (–18uvw2 ) 27uv 2 w . Solution: –12uv 2 × (–18uvw2 ) 27uv 2 w = 4 1 1 1 2 1 –12 × u × v × v × (–18) × u × v × w × w 27 × u × v × v × w 3 1 1 1 1 1 = 8uvw ©Praxis Publishing_Focus On Maths

Algebraic Expressions CHAPTER 4 139 EXAMPLE 26 Simplify 5(p + xy) – 1 6px 3 y – 8p2 x 2 2px2 2. Solution: 5(p + xy) – 1 6px 3 y – 8p2 x 2 2px2 2 = 5p + 5xy – 6px 3 y 2px 2 + 8p2 x 2 2px 2 = 5p + 5xy – 3xy + 4p = 5p + 4p + 5xy – 3xy = 9p + 2xy (III) Dividing algebraic expressions by a number In division of an algebraic expression by a number, every term in the expression is divided by the same number. EXAMPLE 27 Simplify the following expression. 12p – 21qr 3 Solution: 12p – 21qr 3 = 12p 3 – 21qr 3 = 4p – 7qr b + c a = b a + c a Practice 4.4 Basic Intermediate Advanced A Find the quotient of each of the following. (a) 12c ÷ (–4c) (b) 60ab2 ÷ 5ab2 (c) (–18p2 q) ÷ 3p2 q (d) 20mnp2 ÷ 4mnp2 B Find the quotient of each of the following. (a) 64kmn 16mn (b) 35efg –7eg (c) –12uv 3 w2 ÷ (–16uv 2 w) (d) 7e2 f 2 g ÷ (–28ef 2 g) C Find the quotient for each of the following. (a) 24k 3 h 8 ÷ 6kh 3 (b) 28a5 bc 2 ÷ (–12a4 b) (c) 5pq6 40pq3 (d) –16x 3 y 4 20y 3 z ©Praxis Publishing_Focus On Maths

CHAPTER 4 Algebraic Expressions 140 D Simplify each of the following. (a) 12p2 q × 9pq2 27p2 q2 (b) –8rs 2 t × (–12r 2 t) –32r 2 t (c) 15pqr × (–20q2 ) ÷ 24pq2 (d) –6y 3 z ÷ 8xy 2 × (–72xz 2 ) E P Q R U T V W S 10xy cm 16xy cm Diagram above shows a rectangle PRTU. Q is the midpoint of edge PR and the ratio of length of RS to ST is 3 : 2. If PQWV is a square, find the area of the shaded region in terms of x and y. 4.5 Simplifying the Algebraic Fractions A Simplification of algebraic fractions by dividing common factors Algebraic fractions are fractions with either its numerator or denominator or both having algebraic expressions. For example, 2x 5 , 3 2y – x , 4x x + 3y . Algebraic fractions can be simplified by: • cancelling out any factor which is common to both the numerator and the denominator, • factorising the numerator or the denominator or both and then cancel out any factor which is common to both the numerator and the denominator. When the highest common factor (HCF) of both the numerator and the denominator has been cancelled out, the remaining fraction is reduced to its simplest form. (a) Simpilfy ad bd . (b) Find the sum of these fractions, a b + c d . (c) Find the sum of these fractions, a b + c d + e f . Critical Thinking EXAMPLE 28 Simplify each of the following algebraic fractions. (a) 24p 2 q 36pq 2 (b) 20xy x(x + y) ©Praxis Publishing_Focus On Maths

Algebraic Expressions CHAPTER 4 141 Solution: (a) 24p2 q 36pq2 = 2 × 2 × 2 × 3 × p × p × q 2 × 2 × 3 × 3 × p × q × q = 2p 3q (b) 20xy x(x + y) = 20xy x(x + y) = 20y (x + y) EXAMPLE 29 Reduce each of the following algebraic expression to its simplest form. (a) 2x 2 – 4x 6xy (b) 4pq – 6pq 2 2pq + 10p2 q2 Solution: (a) 2x 2 – 4x 6xy = 2x(x – 2) 6xy = x – 2 3y (b) 4pq – 6pq 2 2pq + 10p2 q2 = 2pq(2 – 3q) 2pq(1 + 5pq) = 2 – 3q 1 + 5pq For a fraction a b , a is known as the numerator and b is called the denominator, where b ≠ 0. a b Denominator Numerator Divide both the numerator and the denominator by their common factors. 1 3 Factorise the numerator first and then cancel out the common factor. Factorise the numerator and denominator first and then cancel out the common factor. B Addition and subtraction of algebraic fractions (I) Addition and subtraction of algebraic fractions with the same denominator To add and subtract algebraic fractions with the same denominator, rewrite as a single fraction with the denominator remaining unchanged while adding or subtracting the numerators. We need to reduce the single algebraic fraction to its simplest terms if possible. Different denominators Equalise the denominators of each algebraic fraction. Add or subtract the numerators. Same denominator Identify the denominator of each fraction. ©Praxis Publishing_Focus On Maths

CHAPTER 4 Algebraic Expressions 142 EXAMPLE 30 Simplify each of the following. (a) 4a 9 + 2a 9 (b) 5a 2b + 3 – 3a 2b + 3 Solution: (a) 4a 9 + 2a 9 = 4a + 2a 9 = 6a 9 = 2a 3 (b) 5a 2b + 3 – 3a 2b + 3 = 5a – 3a 2b + 3 = 2a 2b + 3 Rewrite as a single fraction and add the two algebraic terms in the numerators. Then, simplify to its simplest form. Rewrite as a single fraction and subtract 3a from 5a. (II) Addition and subtraction of algebraic fractions with the different denominator (a) Adding and subtracting two algebraic fractions where one denominator is the multiple of the other denominator We use the following steps to add or subtract two algebraic fractions where one denominator is the multiple of the other. 1 Change the algebraic fraction with non-multiple denominator to its equivalent fraction so that both fractions have the same denominator. 2 Write as a single fraction while adding or subtracting the numerators. 3 Simplify the single algebraic fraction to its simplest form if possible. EXAMPLE 31 Express each of the following as a single fraction in its simplest form. (a) p 10 + 3p 5 (b) 2 q 2 – 4 5q 2 Solution: (a) p 10 + 3p 5 = p 10 + 3p × 2 5 × 2 Multiply by 2 to get the same denominator. = p 10 + 6p 10 = p + 6p 10 Write as a single fraction and add the numerators. = 7p 10 • a c + b c = a + b c • a c – b c = a – b c • a c + b d = ad + bc cd • a c – b d = ad – bc cd ©Praxis Publishing_Focus On Maths

Algebraic Expressions CHAPTER 4 143 (b) 2 q 2 – 4 5q 2 = 2 × 5 5 × q2 – 4 5q 2 = 10 – 4 5q 2 = 6 5q2 (b) Adding and subtracting two algebraic fractions with denominators not having any common factors We can find the equivalent algebraic fractions by the following steps to add or subtract two algebraic fractions without any common factors in the denominators. 1 Convert each algebraic fraction to its equivalent fraction. 2 Add or subtract the numerators while keeping the denominator unchanged. 3 Simplify the single algebraic fraction to its simplest form if possible. EXAMPLE 32 Express each of the following as a single algebraic fraction in its simplest form. (a) a 3 + b 4 (b) 2 + c 2a – 3 + d b Solution: (a) The LCM of 3 and 4 = 3 × 4 = 12 a 3 + b 4 = a × 4 3 × 4 + b × 3 4 × 3 = 4a 12 + 3b 12 = 4a + 3b 12 (b) The LCM of 2a and b = 2ab 2 + c 2a – 3 + d b = (2 + c) × b 2a × b – (3 + d) × 2a b × 2a = 2b + bc 2ab – (6a + 2ad) 2ab = 2b + bc – 6a – 2ad 2ab Generally, the expansion or factorisation is performed before the addition and subtraction of algebraic expressions. ©Praxis Publishing_Focus On Maths

CHAPTER 4 Algebraic Expressions 144 (c) Adding and subtracting two algebraic fractions with denominators having a common factor We follow the steps below to add or subtract two algebraic fractions with a common factor in the denominators. 1 Find the least common multiple of the denominators. 2 Convert each algebraic fraction to its equivalent fraction. 3 Add or subtract the numerators while keeping the denominator unchanged. 4 Simplify the single algebraic fraction to its simplest form if possible. EXAMPLE 33 Simplify each of the following and express your answer as a single fraction in its simplest form. (a) 3b 4 + b 6 (b) z + 2 4y – 4y – x 10xy Solution: (a) 3b 4 + b 6 = 3b × 3 4 × 3 + b × 2 6 × 2 2 4 6 2 2 3 3 1 3 1 1 LCM = 2 × 2 × 3 = 12 = 9b 12 + 2b 12 = 9b + 2b 12 = 11b 12 (b) z + 2 4y – 4y – x 10xy 2 4 10 2 2 5 5 1 5 1 1 LCM = 2 × 2 × 5 × x × y = 20xy = (z + 2) × 5x 4y × 5x – (4y – x) × 2 10xy × 2 = 5xz + 10x 20xy – 8y – 2x 20xy = 5xz + 12x – 8y 20xy C Multiplication and division of algebraic fractions Steps to multiply and divide algebraic fractions: Division Multiply numerator by numerator and denominator by denominator. Multiplication Factorise first if it is necessary to simplify algebraic fraction. Convert the division operation to multiplication operation. ©Praxis Publishing_Focus On Maths

Algebraic Expressions CHAPTER 4 145 (I) Multiplication of two algebraic fractions Multiplication of algebraic fractions can be done in the same way as we do for numerical fractions, that is, simplify first if possible and then multiply the numerators and denominators of the algebraic fractions separately. EXAMPLE 34 Simplify each of the following to its simplest form. (a) p 3 × 4q 5r (b) 2b 7d × 21ad 8 (c) 4pq p + 1 × r(p + 1) 6 Solution: (a) p 3 × 4q 5r = p × 4q 3 × 5r = 4pq 15r (c) 4pq p + 1 × r (p + 1) 6 = 4pq p + 1 × r(p + 1) 6 = 2pq 1 × r 3 = 2pqr 3 (II) Division of two algebraic fractions We can divide an algebraic fraction by another, first change the division sign (÷) to the multiplication sign (×), invert the divisor and then proceed with the normal procedure for multiplication. EXAMPLE 35 Simplify each of the following to its simplest form. (a) 2ab 3c ÷ 5d e (b) 2p 9q ÷ 8p 2 15qr (c) 9 2x – 1 ÷ 6y x(2x – 1) (b) 2b 7d × 21ad 8 = 2b 7d × 21ad 8 = b 1 × 3a 4 = 3ab 4 1 3 1 4 2 3 ©Praxis Publishing_Focus On Maths

CHAPTER 4 Algebraic Expressions 146 Solution: (a) 2ab 3c ÷ 5d e = 2ab 3c × e 5d = 2ab × e 3c × 5d = 2abe 15cd (b) 2p 9q ÷ 8p2 15qr = 2p 9q × 15qr 8p2 Change ÷ to × and invert the divisor 8p2 15qr . = 2p 9q × 15qr 8p2 Simplify. = 1 × 5r 3 × 4p = 5r 12p (c) 9 2x – 1 ÷ 6y x(2x – 1) = 9 2x – 1 × x(2x – 1) 6y 3 = 9 2x – 1 × x(2x – 1) 6y 2 = 3 × x 1 × 2y = 3x 2y EXAMPLE 36 Simplify each of the following. (a) (3 + 2x)(3 – 2x) × 5y(y + 7) (b) (4e2 + 16e – 9) ÷ (2ef + 12e – f – 6) Solution: (a) (3 + 2x)(3 – 2x) × 5y(y + 7) = (9 – 4x 2 )(5y 2 + 35y) = 45y 2 + 315y – 20x 2 y 2 – 140x 2 y (b) (4e2 + 16e – 9) ÷ (2ef + 12e – f – 6) = 4e 2 + 16e – 9 2ef + 12e – f – 6 = (2e – 1)(2e + 9) 2e(f + 6) – (f + 6) = (2e – 1)(2e + 9) (2e – 1)(f + 6) = 2e + 9 f + 6 Change ÷ to × and invert the divisor 5d e . Multiply the numerators and the denominators separately. 1 5 43 p Multiply the numerators and the denominators separately. • a b × c d = ac bd • a b ÷ c d = a b × d c = ad bc Use (a + b)(a – b) = a2 – b2 in expansion. Write in the form of fraction. Factorise. 1 1 ©Praxis Publishing_Focus On Maths

Algebraic Expressions CHAPTER 4 147 Practice 4.5 Basic Intermediate Advanced A Simplify each of the following algebraic fractions. (a) 2(a + b) 3(a + b) 2 (b) 4(b + c) 2 8(b + c) (c) 6p(p + q) 15p2 q(p + q) B Reduce each of the following to its simplest form. (a) 3a – 6 3a (b) 2b + 18 2b (c) 4d d – 2d2 (d) 3g2 2g – gh (e) 2p + 8p2 12pq (f) –10x 2 yz 2 5xy – 20x 2 y 2 C Simplify each of the following. (a) 2a + 3 4pr + 5a – 5 4pr (b) 3 + 4b 2a – 1 + 5 – 2b 2a – 1 (c) 3x + 4y 8gh – x + 2y 8gh (d) 2a + 3b 3f – 2 – a – 2b 3f – 2 D Express each of the following as a single fraction in its simplest form. (a) f 5h + h 4f (b) 2m 5n – 3m 2pr (c) 2 3x + 5 yz (d) 3m n – n 2mp (e) 2x + 1 3y + x – 5 5xy (f) 3a + b 2c – b 5d E Simplify each of the following and express it as a single fraction in its simplest form. (a) 2d 9ef + 4e 6f (b) 3p 4n – 5 18mn (c) 2z xy 2 + 5z + 1 yz (d) m + n m2 n – 2 + n 2m (e) 3 – p 4q + 2q – p 10pq (f) 2p – 3r 6pr – q – 4p 10qr F Simplify each of the following to its simplest form. (a) 3 4(a + b) × 14(a + b) 9b (b) 2(d + 2c) 35 × 15(ef) (d + 2c) (c) qr 2q – 5 × 3(2q – 5) 4qr 2 (d) 4x + y 6yz × 15y2 z 2(4x + y) 7 Simplify each of the following to its simplest form. (a) 4 3a ÷ 2b 9 (b) 10d 9 ÷ 5de 12 (c) 5gh 3f ÷ 10g (d) 1– 6 25mn2 2 ÷ 4p 5mn (e) 3r 2 8p ÷ 1– 9qr 4p2 2 (f) 8xy 3 ÷ 4x 2 y 15z 8 Simplify each of the following. (a) (2x + 5)(x + 2) + (7x 2 + 6x – 1) ÷ (x + 1) (b) (y 2 – 4x 2 ) ÷ (y + 2x) – 3x(y + 5) (c) (c 2 + 6c – 16) × 3c c + 8 – 5c(c – 4) (d) 1 – 4y 5x + 2 + (5x – 2) ÷ 25x 2 – 4 y + 7 (e) (3a – b)2 + a(a + 13b) 2a2 – b 2 – ab I A clothing shop sells batik cloth and silk cloth. The price of (n + 2) m batik cloth is $(n + 1) and the price of (3n – 1) m silk cloth is $(2n + 3). Puan Faezah wants to buy 6 m batik cloth and 1 m silk cloth. How much is the total payment? ©Praxis Publishing_Focus On Maths

CHAPTER 4 Algebraic Expressions 148 4.6 Expansions of Brackets A Recognise expansion Expansion of two algebraic expressions is the product between an algebraic expression and a term or another algebraic expression. According to the Distributive Law, a × (b + c) = a × b + a × c = ab + ac ab + ac is the product of the term a and the expression (b + c). Therefore, the product ab + ac is known as expansion of the multiplication a(b + c). EXAMPLE 37 Given 5x(x + 2) = 5x 2 + 10x, state the expansion of 5x(x + 2). Solution: From 5x × (x + 2) = 5x 2 + 10x Multiplication of two expressions. Expansion. Therefore, the expansion of 5x(x + 2) is 5x 2 + 10x. The expansion of two algebraic expressions is performed by multiplying each term in the first expression to each term in the second expression inside the brackets. Generally, (i) a(a + b) = a2 + ab Each term in the bracket is multiplied by a. (ii) (a + b)(c + d) = ac + ad + bc + bd An expansion of algebraic expressions can be represented by algebra tiles using the concept of area. For instance, x(x + 2). x x x 2 x + 2 x x 1 1 The rectangle is formed from 1 tile of x 2 and 2 tiles of x. The total area of all tiles is x 2 + 2x. The area of the whole rectangle is equal to the total area of all tiles. The rectangle with the width x and length (x + 2) has the area of x 2 + 2x. Therefore, x(x + 2) = x 2 + 2x. Each term in the first bracket is multiplied by each term in the second bracket respectively. (Expression 1) × (Expression 2) = Expression 3 Multiplication of two expressions. Expansion. ©Praxis Publishing_Focus On Maths

Algebraic Expressions CHAPTER 4 149 B Expansion involving single brackets When a linear algebraic expression is multiplied by an algebraic term (or a number), the expansion can be carried out by multiplying each term in the linear algebraic expression by the algebraic term (or the number). EXAMPLE 38 Expand each of the following. (a) 2(3a – b) (b) 5f(2 f + 7g) (c) –3x(2x + 4y – 3z) Solution: (a) 2(3a – b) = 2 × 3a – 2 × b = 6a – 2b (b) 5f(2f + 7g) = 5f × 2f + 5f × 7g = 10f 2 + 35fg (c) –3x(2x + 4y – 3z) = –3x × 2x + (–3x) × 4y + (–3x) × (–3z) = – 6x2 – 12xy + 9xz EXAMPLE 39 Expand each of the following. (a) x(x + 7) (b) – 3n(n – 5m + 9) Solution: (a) x(x + 7) = x × x + x × 7 = x 2 + 7x (b) –3n(n – 5m + 9) = –3n × n + (–3n) × (–5m) + (–3n) × 9 = –3n 2 + 15mn – 27n ©Praxis Publishing_Focus On Maths

CHAPTER 4 Algebraic Expressions 150 C Expansion involving two brackets When two linear algebraic expressions are multiplied, the expansion can be carried out by multiplying each term in the first expression by every term in the second expression. If there are like terms in the product of the expansion, simplify it by grouping the like terms together. EXAMPLE 40 Expand each of the following. (a) (d + e)(e + f ) (b) (2m – n)(3m – 5n) (c) (p – q)(p + q) (d) (4x – 3y)(x + 2y) Solution: 2 1 4 3 (a) (d + e)(e + f) = d × e + d × f + e × e + e × f = de + df + e 2 + ef Like terms (b) (2m – n)(3m – 5n) = 6m2 – 10mn – 3mn + 5n2 = 6m2 – 13mn + 5n2 (c) (p – q)(p + q) = p2 + pq – pq – q2 = p2 – q2 (d) (4x – 3y)(x + 2y) = 4x 2 + 8xy – 3xy – 6y 2 = 4x 2 + 5xy – 6y 2 EXAMPLE 41 Expand each of the following. (a) (x + 2)(x + 3) (b) (m – 1)(m + 5) (c) (3p + n)(q – 6) Solution: (a) (x + 2)(x + 3) = x × x + x × 3 + 2 × x + 2 × 3 = x2 + 3x + 2x + 6 = x2 + 5x + 6 Each term in the first bracket is multiplied by each term in the second bracket respectively. Simplify the like terms. ©Praxis Publishing_Focus On Maths

Algebraic Expressions CHAPTER 4 151 (b) (m – 1)(m + 5) = m2 + 5m – m– 5 = m2 + 4m – 5 (c) (3p + n)(q – 6) = 3pq – 18p + nq – 6n Simplify the like terms. Algebra tiles 1 Represent with the following tiles. Represents x2 Represents x Represents +1 2 Arrange the tiles in a rectangle based on the expressions being multiplied. x + 3 x + 2 3 The formed rectangle has the length of (x + 3) and width of (x + 2). x + 3 x + 2 1 1 x2 x x x 1 1 1 1 x x The area of rectangle = Total area of all tiles (x + 2)(x + 3) = 1 blue tile of x2 + 5 grey tiles of x + 6 blue tiles of 1 = x2 + 5x + 6 ©Praxis Publishing_Focus On Maths

CHAPTER 4 Algebraic Expressions 152 EXAMPLE 42 Expand each of the following. (a) (m + n)² (b) (p – q)² Solution: (a) (m + n) 2 = (m + n)(m + n) = m2 + mn + mn + n2 = m2 + 2mn + n2 (b) (p – q)2 = (p – q)(p – q) = p2 – pq – pq + q 2 = p2 – 2pq + q2 EXAMPLE 43 3 m 2x m p m 1 m 3 m 1 m Azmah wants to place a rectangular carpet in the middle of her room as shown in the diagram above. Calculate the area, in m2 , of the carpet. Solution: The length of carpet = (2x – 6) m The width of carpet = (p – 2) m The area of carpet = (2x – 6)(p – 2) = 2px – 4x – 6p + 12 • (a + b)2 = a2 + 2ab + b2 is useful in similar expansions such as (3x + 4y)2 = (3x)2 + 2(3x)(4y) + (4y)2 = 9x 2 + 24xy + 16y 2 • (a – b) 2 = a2 – 2ab + b2 is useful in similar expansions such as (5x – 6y)² = (5x)² – 2(5x)(6y) + (6y)² = 25x ² – 60xy + 36y ² Practice 4.6 Basic Intermediate Advanced A Expand each of the following. (a) 3(4s – 5t) (b) 4e(3e – 5f) (c) 5m(2m – n + 3p) (d) – 6p(5p – 3q) (e) 3 2 p(4p – 2q + 6r) ( f) –3x1 – 2 3 x – 2y + 4z2 2 Expand the following. (a) (a + c)(b + d ) (b) (3e – 4g)(2g + f ) (c) (x + 3)(x + 8) (d) (2p + 3)(3p + 7) (e) (3q – 5)(2q – 3) ( f) (5x – 2y)(2x – 5y ) (g) (3x – 4y)(5x + 3y) (h) (5y + 6z)(3y – 4z ) 3 Expand each of the following. (a) (h + k)(h – k) (b) (p + 3)(p – 3) (c) (2d + 1)(2d – 1) (d) (3f – 2g)(3f + 2g) (e) (4x + 5y)(4x – 5y) ( f) (–6y + 5z)(–6y – 5z) 4 Find the expansion for each of the following. (a) (a + 4)2 (b) (d + 6)2 (c) (2m + 5n)2 (d) (5x + 7y)2 (e) (6q + 1 2 r)2 (f) (e – 6)2 ©Praxis Publishing_Focus On Maths

Algebraic Expressions CHAPTER 4 153 (g) (2p – 5)2 (h) (–3t – 6)2 (i) (5g – 3h)2 (j) 16y – 1 4 z2 2 5 (a) In a rectangular marching formation, each row consists of 2x members. If the marching formation has (x + 8) rows and the total number of members is 2x 2 + 16x, determine the expansion of 2x(x + 8). (b) The area of a square with side of (a – 2) cm is (a2 – 4a + 4) cm2 , state the expansion of (a – 2)(a – 2). (c) A machine can print (p + 8) cards per second. The number of cards can be printed by the machine in t seconds is pt + 8t. Determine the expansion of t(p + 8). 4.7 Factorisation of Algebraic Expressions The multiplied terms and algebraic expressions are the factors of a product of the algebraic expressions. For instance,a × (b + c) = ab + ac. Therefore, a and (b + c) are factors of the expression ab + ac. Factorisation of an algebraic expression is the process of expressing the expression as a product of its factors. There are two methods of factorisation: (a) Taking out the common factor. (b) Regrouping and then taking out the common factor. A Factorisation by taking out common factors The general procedure is to take out the highest common factor and place it in front of the brackets which contains the algebraic expression of the remaining terms. Therefore, to factorise 4x + 6xy, we first identify the highest common factor of the two algebraic terms. Thus we have 4x = 2 × 2 × x 6xy = 2 × 3 × x × y 4x + 6xy = 2x(2 + 3y) Notice that 2x is actually the HCF of the algebraic terms 4x and 6xy. The algebraic expression in the brackets is the remaining terms after the common factor has been taken out. When the HCF of the algebraic terms has been taken out, the expression is said to be completely factorised. 2x is taken out and placed in front of the brackets. EXAMPLE 44 Factorise each of the following. (a) 14a + 18ab (b) 6xy – 10x 2 y + 12xy 2 ©Praxis Publishing_Focus On Maths

CHAPTER 4 Algebraic Expressions 154 Solution: (a) 14a = 2 × 7 × a 18ab = 2 × 3 × 3 × a × b ∴ HCF = 2 × a = 2a ∴ 14a + 18ab = 2a(7 + 9b) (b) 6xy = 2 × 3 × x × y 10x 2 y = 2 × 5 × x × x × y 12xy 2 = 2 × 2 × 3 × x × y × y ∴ HCF = 2 × x × y = 2xy ∴ 6xy – 10x 2 y + 12xy 2 = 2xy(3 – 5x + 6y) B Factorisation by regrouping When an algebraic expression has four terms and is of the form ac + ad + bc + bd, there is no common factor among them. However if we regroup the terms with a common factor, we will be able to factorise it. EXAMPLE 45 Factorise each of the following. (a) ab + ac + bd + cd (b) 2xy – 2xz – sy + sz Solution: (a) ab + ac + bd + cd = (ab + ac) + (bd + cd) Group two terms with common factor together. = a(b + c) + d(b + c) Take out the common factor from each group. = (b + c)(a + d) Take out the common factor (b + c). (b) 2xy – 2xz – sy + sz = (2xy – 2xz) – (sy – sz) = 2x(y – z) – s(y – z) = (y – z)(2x – s) Factors of an algebraic expression can be determined from its factorisation. For example, Expansion (x + 1)(x + 2) = x 2 + 3x + 2 Written inversely Factorisation x 2 + 3x + 2 = (x + 1)(x + 2) Factors Therefore, x + 1 and x + 2 are the factors while (x + 1)(x + 2) is the factorisation of x 2 + 3x + 2. ©Praxis Publishing_Focus On Maths

Algebraic Expressions CHAPTER 4 155 EXAMPLE 46 Given that 5x(2x + 7) = 10x 2 + 35x, state the factorisation of 10x 2 + 35x. Hence, list the factors of 10x 2 + 35x. Solution: Factorisation = 5x(2x + 7) 5x(2x + 7) = 1 × 5x(2x + 7) = 5 × x(2x + 7) = x × 5(2x + 7) = 5x × (2x + 7) Therefore, the factors of 10x 2 + 35x are 1, 5, x, 5x, 2x + 7, 5(2x + 7), x(2x + 7) and 5x(2x + 7). Practice 4.7 Basic Intermediate Advanced A Factorise completely each of the following. (a) ab – ad (b) 2b2 – 10bd (c) 4c 2 – 6cd (d) 6pq + 28pq2 (e) 15xy2 – 35x 2 y (f) 16xy3 – 24x 2 y 2 2 Factorise completely each of the following. (a) 2a + 2b – 2c (b) 12p – 8q + 20r (c) 4ab + 6ac – 10ad (d) 5pq + 10pq2 – 25p2 q (e) 3xyz – 6xy 2 z + 15x 2 y 2 (f) –12x 2 y 2 z 2 + 16x 3 y 2 z 3 – 36x 2 y 3 z 4 3 (a) Given that 5p(p2 + 2) = 5p3 + 10p, state the factorisation of 5p3 + 10p. Hence, list the factors of 5p3 + 10p. (b) Given that (a – 2)(b – 7) = ab – 7a – 2b + 14, state the factorisation of ab – 7a – 2b + 14. Hence, list the factors of ab – 7a – 2b + 14. (c) Given that 2(n + 4)(n – 1) = 2n2 + 6n – 8, state the factorisation of 2n2 + 6n – 8. Hence, list the factors of 2n2 + 6n – 8. (d) Given that (h + 3k) 2 = h2 + 6hk + 9k2 , state the factorisation of h2 + 6hk + 9k2 . Hence, list the factors of h2 + 6hk + 9k2 . 4 A machine can produce 8n canned food per minute. Find the time taken for the machine to produce (40n2 + 16n) canned food. 5 Teacher Mazana bought (6n – 1) boxes of pens. If the total number of pens bought by her is (6n 2 + 29n – 5), determine the number of pen in each box. 6 Sabariya uses a piece of wire to form a rectangle. If the area of rectangle formed is (3l 2 + 29l + 18) cm2 , find the length of the wire used. 7 The base of a triangle is (2x + 5) cm. If the area of the triangle is (2xy + 18x + 5y + 45) cm2 and the height is more than 30 cm, find the possible range of values of y. • 1 is a factor for all algebraic expressions. • The product of expressions is also the factor of itself. ©Praxis Publishing_Focus On Maths

156 CHAPTER 4 Algebraic Expressions Variables • A variable whose value has not been determined. • Variable can be represented by letters. Algebraic expressions • A combination of numbers, variable and operational signs ‘+’ or ‘–’. For example, 3p + 2q. • Like terms can be simplified by adding or subtracting their coefficients of the same variable. For example, 2p + 7x + 3p – 4x = (2 + 3)p + (7 – 4)x = 5p + 3x Algebraic terms with one variable The product of an variable and a number. For example, 5x. Coefficient Variable Like terms Terms that have the same variable. For example, 3m and –7m. Unlike terms Terms that have the different variable. For example, 4x and 9y. Summary Summary Summary Addition and subtraction To add or subtract two algebraic fractions, we find the lowest common multiple of the denominator, then perform the addition or subtraction operation for the numerator. For example, 5 2x + 1 + 2 x = 5x x(2x + 1) + 2(2x + 1) x(2x + 1) = 5x + 4x + 2 x(2x + 1) = 9x + 2 x(2x + 1) Algebraic fractions Algebraic fractions are fractions with either its numerator or denominator or both having algebraic expressions. For example, x 3 , 4 2x + 1 , 4 – 5x 3x . Simplification Algebraic fractions can be simplified by dividing the denominator and numerator by their common factor. For example, 14p2 q 21pq2 = 2 × 7 1 × p 1 × p × q 1 3 × 7 1 × p 1 × q 1 × q = 2p 3q Multiplication and division • To multiply two algebraic fractions, we simplify the two algebraic fractions, then multiply. For example, 3a 1 4a 1 b 1 × 5 1 b2 b 102 a3 = 3b 8a3 • To divide two algebraic fractions, we change the division operation to multiplication operation, then perform the multiplication operation. For example, 6mp n × 3m2 4n2 = 6 2 m 1 p n × 4n2n 3 1 m m 2 = 8np m Expansion Expanding multiplication involving algebraic expressions. For examples, (a) p(q + r) = pq + pr (b) (p + q)(s + t) = ps + pt + qs + qt (c) (a + b) 2 = a2 + 2ab + b2 (d) (a – b) 2 = a2 – 2ab + b2 (e) (a + b)(a – b) = a2 – b2 Factorisation Algebraic expression can be factorised by • using common factor, for example, 2xy 2 – 4xy = 2xy(y – 2) • expressing algebraic expression in the form of the difference of two squares, for example, p2 – q2 = (p + q)(p – q) • expressing algebraic expression as a product of two algebraic expressions, for example, (a) x 2 + 2xy + y 2 = (x + y)2 (b) ac + ad + bc + bd = (ac + ad) + (bc + bd) = a(c + d) + b(c + d) = (c + d)(a + b) ©Praxis Publishing_Focus On Maths

157 Algebraic Expressions CHAPTER 4 4 Section A 1. “The distance between the hospital and the police station is only a few kilometres.” The variable in the above statement is A the hospital. B the police station. C the distance. D kilometres. 2. “Albert bought n books for $ x from shop k. Each book has m pages.” In the above statement, which letter does not represent a variable? A n C k B x D m 3. Which of the following is an algebraic term with one variable? A 2xy C 5pq B 5 m D 2k 7 4. Given that 8hk 2 n, state the coefficient of the term kn. A 8hk C hkn B 8hn D hn 5. Which of the following algebraic expression has like term as 1 3 x 2 yz? A 1 3 y 2 xz C xyz 3 B 3yz D 19zyx 2 6. 4— cm x3 y x2y 4cm The diagram shows a rectangle. State the coefficient of 4x. A 4xy C x 2 y B 4x 5 y D xy 3 7. State the algebraic expression for the situation below. The area of a rectangle with length of 3ab cm and width of 2a cm. A 6ab cm2 C 3a2 b cm2 B 6a2 b cm2 D 3ab2 cm2 8. There are x goats and y cows in a farm. The cost for rearing a goat and a cow are $20 per month and $30 per month respectively. Determine the total rearing cost incurred by the farm owner in a month. A 2x + 3y C 50(x + y) B 20x + 30y D 30x + 20y 9. The number of terms in “2p – 7 + 8q + 1 + 6q” is A 3 C 5 B 4 D 6 10. Expand 3p(p – 4q). A 3p2 – 4q C 3p2 – 12q B 3p2 – 12pq D 3p – 4q 11. 9m – 6n – 7m – 4 + 5n = A 2m – 11n – 4 C 16m – 10n + 5 B 2m – n – 4 D 16m – n – 4 12. Simplify (2a + 7)(b – 4) + 5a(2b – 3). A 7ab – 14a + 7b – 12 B 10ab + 20a + 5b – 6 C 12ab + 23a – 7b + 16 D 12ab – 23a + 7b – 28 13. Given that (x + 7)(x – 2) = x 2 + kx – 14, where k is a constant. Determine the value of k. A 2 C 7 B 5 D 12 14. Expand 3p(p – 4q). A 3p 2 – 4q C 3p2 – 12q B 3p2 – 12pq D 3p – 4q 15. Factorise 2xy + 12x – y – 6. A (2x – 1)(y + 6) C (x – 1)(2y + 6) B (2x + 1)(y – 6) D (x + 1)(2y – 6) ©Praxis Publishing_Focus On Maths

158 CHAPTER 4 Algebraic Expressions Section B 1. (a) Write an algebraic expression for each of the following situations. Situation Algebraic expression (i) A class has 28 students. Given the number of male students is y, state the number of female students. (ii) The price of two calculators is equal to the price of three dictionaries. If the price of a dictionary is $x, state the price of a calculator. (b) Given the algebraic expression –7p 3 qr, state the coefficient for each of the following terms: (i) pq (ii) 7p2 2. Simplify. (a) 5c + 8d + 3c + d (b) 14p + 11q + 9p – 8q (c) 20g + 10h – 13g – 8h (d) –10x + 4y + 7x – 9y (e) x 2 + 6x + 2x 2 + 3x (f) 7a2 + a2 + a – 4a (g) 9u – 4u2 – u2 + 3u (h) 4mn + 5m – 3mn – 9n 3. Find an algebraic expression for the perimeter of each diagram, in simplest form. (a) 8t (b) (c) 7a 9a + 4 3x + 4 2x 4. Simplify. (a) 1 x + 3 2x (b) 3 5h + 5 9h (c) 2w – 5 12 + w – 1 4 (d) 7e – 1 8 + 2e – 5 3 5. (a) Given that p = 3, q = –8 and r = 4 5 , find the value of each of the following algebraic expressions. (i) 2pq + p2 – 5q (ii) – 1 2 q + 6pr – 7 (b) Kenny bought 5 fluorescent lamps which costs $x each and 2 fans which costs $y each. If he paid with cash $200, express the balance received in terms of x and y. (c) Two years ago, the age of Farhana’s father was thrice of her age. If Farhana’s father is x years old this year, what is the age of Farhana, in terms of x, after five years? 6. (a) Given that (8p – 5pq + 2) – (p + 6pq – 9) + (10p – pq) = ap + bpq + c, where a, b and c are integers. Determine the values of a, b and c. (b) The product of three algebraic expressions that consist of only one algebraic term is 15p2 qr 3 . If two of the algebraic expressions are –8pq3 and 6rs 2 respectively, what is the third algebraic expression? (c) The diagram shows two photo frames which are in the shape of a rectangle and a square respectively. (x + 5) cm (x + 5) cm A B Given that the area of the photo frame A is twice the area of photo frame B. (i) Express the total perimeter of both photo frames. ©Praxis Publishing_Focus On Maths

159 Algebraic Expressions CHAPTER 4 (ii) If x = 8, find the total perimeter of both photo frames. 7. (a) Simplify each of the following. (i) 1 2 5 m + 7n2 – (3m – n + 1) + 1 3 4 m – 92 (ii) (5ab – bc + 2) – (ab + 10bc – 6) (b) In a badminton match, the point obtained by Zhi Ling was twice the point obtained by Fadillah while the point obtained by Everlyn was 5 points less than Zhi Ling. Given that the total points of Zhi Ling and Fadillah was 6mn. Determine the point obtained by Everlyn. (c) The diagram shows the plan of the ground floor of Mr Yazid’s house. Door Reading room (7x – 2y) m (5xy – 2) m (xy + 3x) m x m The shaded region in the diagram represents the plan of a living room. Mr Yazid intends to paste wallpapers throughout his living room. What is the perimeter of the region that is required to be pasted with wallpapers? 8. (a) Expand. (i) –2(p – 1)(p + 1) (ii) (5y – 2x)2 (b) (2x – 5) cm (y – 1) cm (4x + 3) cm The diagram shows a trapezium. Find the area of the trapezium. (c) (4p + 5) m (3p + 1) m (p + 6) m 2 m The diagram above shows the plan of Ayub’s office. Find the area, in square metres, of Ayub’s office in terms of p. 9. (a) Factorise each of the following. (i) x(x + 1) + 3(x – 4) (ii) n(m + 1) + 2(m – 1) + 2m + 6 (b) Simplify 3t(2 – t) + (4t + 3)(t – 2) 7t 2 + 20t – 3 (c) Area of a cardboard is (xy + 3x + 8y + 24) cm2 . Everlyn wants to cut the cardboard into rectangular pieces with length of (x + 5) cm. Justify whether she can cut the cardboard into several rectangular pieces without any remaining cardboard. 10. (a) Simplify –15a2 b –35ab2 c to its simplest form. (b) Find the sum of 4r 7pq and 3r – 7 7pq . Express your answer as a single fraction in its simplest form. (c) Simplify cd – 3d 4h × 20h 3c – 9 to its simplest form. (d) Express 3x – 12 7y 2 ÷ 4 – x 14y as a single fraction in its simplest form. ©Praxis Publishing_Focus On Maths

LINEAR EQUATIONS AND INEQUALITIES IN 5ONE VARIABLE Applications of this chapter The linear equation tells us how much we can afford. Whether we need to bring enough wood to keep the fire burning overnight, calculate our salary, figure out how much paint we need to redecorate the bedroom or buy enough petrol to travel to and from a friend’s house, calculate the cost of any taxi trips, linear equations can provide the answers. Literally, linear systems are everywhere. What other real life situations can we relate to the use of linear equations? 160 ©Praxis Publishing_Focus On Maths

Concept Map Learning Outcomes • Understand the concept of open sentence. • Recognise the equality between two quantities. • Recognise linear algebraic terms and linear algebraic expressions. • Determine linear algebraic equations in one variable. • Solve linear equations in one variable by addition, subtraction, multiplication and division. • Solve problems involving linear equations in one variable. • Understand and use the concept of inequalities. • Solve inequalities by addition, subtraction, multiplication and division. • Solve inequalities involving more than one operation. Linear Equations Inequalities Linear Equations and Inequalities in One Variable Closed Sentence Open Sentence Operations Representation on Number Lines Addition Multiplication Solving Inequalities in Subtraction Division One Variable • Open sentence • Equality • Trial and improvement • Numerical value • Backtracking • Variable • Linear equation • Inequality • Solution • Sentence • Equivalence • Expression • Formula Key Terms Maths History Diophantus of Alexandria was a Greek mathematician known as the Father of Algebra. He had contributed to the knowledge of solving algebraic equations. He was also the author of a series of books called Arithmetica. 161 ©Praxis Publishing_Focus On Maths

162 CHAPTER 5 Linear Equations and Inequalities in One Variable Flashback 1. a(b + c) = a + ca 2. (b – c)a = a – ca 3. 4(2) + 3(2) = (2) 4. 10(5) – 6(5) = (5) 5. Expand a(b + c). 6. Expand (b – c)a. Draw the flow chart whose output number is 6 – 2x. Thinking Find the output number for each of these flow charts. – 1 ×5 (a) x ÷ 8 ×3 (b) x +8 ÷ 5 +3 (c) x ×3 – 7 ÷ 2 (d) x – 4 ×5 +3 (e) x 1 ©Praxis Publishing_Focus On Maths

163 Linear Equations and Inequalities in One Variable CHAPTER 5 5.1 Equality A Open sentences and closed sentences in algebra A mathematical sentence states or compares two expressions. There are many things that can be true about math sentences. It can use words, numbers, variables, or a combination of the three, as in the following examples: • The sum of 2 and 7 is 9. • 2 + 7 = 9 • 2 + n = 9 • 2 add the number n equals 9. A mathematical sentence can be an equation that compares two expressions using the equal sign (=). Mathematical sentences can also be inequalities, which compare two expressions with an inequality symbol, such as greater than or less than (> or <). For example: • 25 × 3 = 75 • 8 × 7  4 × 6 • 47 + 21  9 × 9 A mathematical sentence compares two expressions. It can be true or false, open or closed. A closed sentence is always true or always false. Here are some examples of closed sentences: • A square has four corners always true • 6 is less than 5 always false • −3 is a negative number always true An open sentence can be either true or false depending on what values are used. Here are some examples of opened sentences: • A triangle has n sides Can be true or false (depending on the value of n) • z is a positive number Can be true or false (depending on the value of z) • 3y = 4x + 2 Can be true or false (depending on the values of x and y) • a + b = c + d Can be true or false (depending on the values of a, b, c, d ) B Stating the relationship between two quantities by using the symbol ‘=’ or ‘≠’. Equality is the relationship between two quantities which have the same value. The symbol for equality is ‘=’. In an equation, the quantity on the left-hand side (LHS) equals to the quantity on the right-hand side (RHS). For example, 1 hour = 60 minutes Left-hand side (LHS) = Right-hand side (RHS) ©Praxis Publishing_Focus On Maths

164 CHAPTER 5 Linear Equations and Inequalities in One Variable When two quantities have different values, the inequality symbol ‘≠’ is used. For example, 3 km ≠ 300 m. If a = b, then b = a. For example, if 5 kg = 5000 g, then 5000 g = 5 kg. If a = b and b = c, then a = c. For example, if 0.5 = 1 2 and 1 2 = 2 4 , then 0.5 = 2 4 . The diagram shows 1 kg of copper and 1 kg of cotton. Which object has more mass? Critical Thinking EXAMPLE 1 Write the relationship between the following two quantities using the symbol ‘=’ or ‘≠’. (a) 6 ÷ 8, 1 2 × 1.5 (b) –32 , (–3)2 Solution: (a) 6 ÷ 8 = 0.75 , 1 2 × 1.5 = 0.75 ∴ 6 ÷ 8 = 1 2 × 1.5 (b) –32 = –9 , (–3)2 = 9 ∴ –32 ≠ (–3)2 EXAMPLE 2 Write the following quantities in the form: If a = b, then b = a. (a) 0.5 m, 50 cm (b) 6 – (–2), –2 × (–4) Solution: (a) If 0.5 m = 50 cm, then 50 cm = 0.5 m. • ‘=’ is read as ‘is equal to’. • ‘≠’ is read as ‘is not equal to’. (b) If 6 – (–2) = –2 × (–4), then –2 × (–4) = 6 – (–2). ©Praxis Publishing_Focus On Maths

165 Linear Equations and Inequalities in One Variable CHAPTER 5 EXAMPLE 3 Write the following quantities in the form: If a = b and b = c, then a = c. (a) 3 × 4 = 12 and 12 = 144 ÷ 12 (b) k + k = 2k and 2k = 2 × k Solution: (a) If 3 × 4 = 12 and 12 = 144 ÷ 12, then 3 × 4 = 144 ÷ 12. (b) If k + k = 2k and 2k = 2 × k, then k + k = 2 × k. • If a = b, then b = a. • If a = b and and b = c, then a = c. Discuss with your classmates. (a) If 2 + 3 = 4 + 1, is 4 + 1 = 2 + 3? (b) If 4 + 5 = 2 + 7 and 2 + 7 = 3 + 6, is 4 + 5 = 3 + 6? INTERACTIVE ZONE Practice 5.1 Basic Intermediate Advanced A Write the relationship between the following two quantities using the symbol ‘=’ or ‘≠’. (a) 5 days, 120 hours (b) 0.25 m, 25 mm (c) 6 – (–3), 4 + 5 (d) 42 , 24 B Write the relationship of the following quantities in the form: If a = b, then b = a. (a) 122 , 144 (b) 4 – 9, –(9 – 4) (c) 2.3 cm, 23 mm (d) 25 ÷ 100, 1 4 C Write the following quantities in the form: If a = b and b = c, then a = c. (a) 5 × 2 = 10 and 10 = 15 – 5 (b) 8 15 × 3 4 = 2 5 and 2 5 = 8 20 (c) 15 minutes = 1 4 hour and 1 4 hour = 900 seconds (d) 9m – 5m = 4m and 4m = 4 × m D Determine whether the following is an open sentence. (a) 5 is an odd number. (b) 6 is an odd number. (c) 7 is an odd number. (d) m is an odd number. E Determine whether the following is a closed sentence. (a) A triangle has n corners. (b) b + b = 2 × b (c) x is greater than 9 (d) 4 × y = 12 F Find the value of x makes the sentence x − 4 = 6 true? G Find value of n makes the sentence 6 + n = 4 × n true? H A number divided by 6 is equal to 4. Express the above as an open sentence using a variable y. ©Praxis Publishing_Focus On Maths

166 CHAPTER 5 Linear Equations and Inequalities in One Variable 5.2 Linear Equations in One Variable A Understanding linear equations in one variable (I) Linear algebraic terms A linear algebraic term is a term with only one unknown which is raised to the power of 1. For example, –7x and p are linear algebraic terms. Discuss with your classmates. Discuss why 2x2 – 7 = 0 and 2xy = 15 are not linear equations. INTERACTIVE ZONE EXAMPLE 4 Determine whether each of the following terms is a linear algebraic term. (a) 9x (b) –5y 2 (c) – k 2 (d) 12pq Solution: (a) A linear algebraic term (b) Not a linear algebraic term (c) A linear algebraic term (d) Not a linear algebraic term The unknown y is raised to the power of 2. There are two unknowns, p and q. (I) Linear algebraic expressions A linear algebraic expression is a combination of two or more linear algebraic terms by addition or subtraction. For example, 4x – 7 and 5p + 11q are linear algebraic expressions. EXAMPLE 5 Determine whether each of the following is a linear algebraic expression. (a) 7x – 3 (b) 5x – 4y (c) 6p3 + 5 (d) xy – 1 3 y Solution: (a) A linear algebraic expression (b) A linear algebraic expression (c) Not a linear algebraic expression (d) Not a linear algebraic expression p is raised to the power of 3. xy is not a linear term. ©Praxis Publishing_Focus On Maths

167 Linear Equations and Inequalities in One Variable CHAPTER 5 Is 2x – 3 = 5y is linear equation in one variable? Discuss with your classmates. Critical Thinking Objective : To identify algebraic expressions and equations. Instruction : Do this activity in pairs. 1. Study the cards below that contain mathematical sentences. 6p + 3 5a2 – 10 y + 3 = 6 3 + 2x 6k 3 3h – f = 4 + 7 k 2 – 1 = 24 a + 5a g 2 – 4 = 10 x – 2y 3 = 10 2. Determine and classify the mathematical sentences into expressions and equations. Copy and complete the circle maps below. Expression Equation 3. Compare your answer with your classmates. 1 From the findings of Activity 1, we found that the equations below known as linear equations since the power of the variables is 1. y + 3 = 6 3h – f = 4 g 2 – 4 = 10 x – 2y 3 6p + 3 3 + 2x = 10 As the listed linear equations, we found that the equations below only have one variable and the power of variable also is one. 6p + 3 y + 3 = 6 3 + 2x g 2 – 4 = 10 Hence, these equations are known as linear equations in one variable. A linear equation is an equation which contains one or more linear terms. If a linear equation has only one variable, then the equation is a linear equation in one variable. For example, y + 5 = 19 and 12 = x – 7 are linear equations in one variable. ©Praxis Publishing_Focus On Maths

168 CHAPTER 5 Linear Equations and Inequalities in One Variable EXAMPLE 6 Determine whether each of the following is a linear equation. Hence, determine whether it is a linear equation in one variable. (a) x – 6 = 21 (b) x 2 + 2y = 8 (c) 7m = 3m – 13 (d) p = 1 2 q + 5 (e) 3x = 24 (f) x = 4y Solution: (a) A linear equation in one variable (b) Not a linear equation (c) A linear equation in one variable (d) A linear equation in more than one variable (e) A linear equation in one variable (f) A linear equation in more than one variable x2 is not a linear term. Two variables, p and q Two variables, x and y B Writing linear equation in one variable A linear equation in one unknown can be written based on the information given and vice versa. 2 (a) Write an equation which represented by the diagram above. (b) Show what happens when you take 2 from both sides and write down the new equation. represents an unknown amount represents 1 unit Discuss with your classmates. Give an example of an algebraic expression and of an equation. How are they similar? How are they different? INTERACTIVE ZONE ©Praxis Publishing_Focus On Maths

169 Linear Equations and Inequalities in One Variable CHAPTER 5 EXAMPLE 7 Write a linear equation for each of the following situations. (a) When 9 is subtracted from a number x, the result is 8. (b) Julie bought a book which cost $12 and y pens which cost $3 each. She paid a total of $84. (c) In a Mathematics test, the highest mark, h was 54 points higher than the lowest mark. The sum of the two marks were 138. • Identify the variable for the given situation. • Represent the variable by using a suitable letter if required. • Form a linear equation in one variable by using that letter. Solution: (a) x – 9 = 8 (b) Cost of a book = $12 Cost of y pens = $3 × y = $3y Total cost = $(12 + 3y) = $84 12 + 3y = 84 (c) The highest mark = h The lowest mark = h – 54 The sum of the two marks = h + h – 54 = 138 EXAMPLE 8 Describe the situation in each of the following linear equations. (a) x + 5 = 11 where x represents a number. (b) 8y = 72 where y represents the cost of one badge in $. Solution: (a) The sum of a number x and 5 is 11. (b) The total cost of 8 badges which cost $y each is $72. EXAMPLE 9 Form a linear equation in one variable for each of the following. (a) m marbles are divided equally among 7 people where each of them received 12 marbles. (b) Suzana scored b marks in a test whereas Aliff scored 15 marks less than Suzana. Their total score was 127. (c) When a number is added to 6, the sum is 16. Solution: (a) m ÷ 7 = 12 (b) b + b – 15 = 127 (c) Let x be the number. x + 6 = 16 ©Praxis Publishing_Focus On Maths

170 CHAPTER 5 Linear Equations and Inequalities in One Variable Practice 5.2 Basic Intermediate Advanced A Determine whether each of the following equations is a linear equation. (a) 5x + 6 = 21 (b) x2 – 3y = 9 (c) 4x – 11y = 5 (d) y = 1 2x – 8 B Determine whether each of the following equations is a linear equation in one variable. (a) 2x – 8 = 7 (b) 4y + 7x = 13 (c) 3y – 4 = 15 (d) 3x = 2x – 1 C Write a linear equation for each of the following, using x to represent the variable. (a) The sum of a number and 13 is 15. (b) 8 times a number is 144. (c) 5 is subtracted from a number and then the difference is divided by 3. The result is 18. (d) The difference between x and 4 is 11, given that x is greater than 4. (e) A rectangle has a length of x cm and a width of 5 cm. The perimeter of the rectangle is 32 cm. D Describe the situation represented by each of the following linear equations. (a) x + 12 = 25, given that x is a number. (b) 6p = 420, given that $p is the cost of a book. (c) x 12 = 14, given that x is the number of sweets to be equally divided among 12 students. (d) 3k = 32, given that k cm is the length of one side of an equilateral triangle. 5.3 Solve Linear Equations in One Variable A Determining if a numerical value is a solution of a given linear equation in one variable A numerical value is the solution of a given linear equation in one variable if the value satisfies the equation. The solution of a linear equation is also known as the root of the equation. There are three methods to solve a linear equation in one variable, namely, trial and improvement method, application of equality concept and backtracking method. EXAMPLE 10 Solve the linear equation 4x + 3 = 11 by using (a) trial and improvement method, (b) application of equality concept, (c) backtracking method. Solution: (a) Try x = 1, Left hand side, 4x + 3 = 4(1) + 3 = 4 + 3 = 7 ≠ 11 Thus, x = 1 is not the solution for the given equation. Try x = 2, Left hand side, 4x + 3 = 4(2) + 3 = 8 + 3 = 11 Thus, x = 2 is the solution for the given equation. Start with any number that gives LHS as close as possible to RHS. If the answer obtained on LHS is too small, try a larger number. If the answer is too large, try a smaller number. ©Praxis Publishing_Focus On Maths

171 Linear Equations and Inequalities in One Variable CHAPTER 5 (b) x x x x = 4x + 3 = 11 x xx x = 4x + 3 – 3 = 11 – 3 = x x x x 4x 4 = 8 4 Thus, x = 2 (c) Let x be the initial value and 11 be the final value. Observe the mathematical operation: x → × 4 → + 3 → = 11 Backtracking method: x ← ÷ 4 ← –3 ← = 11 Therefore, the value of x can be written as 11 – 3 = 8 → 8 ÷ 4 = 2 Thus, x = 2. EXAMPLE 11 Determine whether (a) 8 is a solution of the equation x – 5 = 3, (b) 7 is a solution of the equation 3 + 2x = 19. Solution: (a) Substitute x = 8 into the left-hand side (LHS) of the equation. x – 5 = 8 – 5 = 3 Therefore, 8 is the solution of x – 5 = 3. (b) Substitute x = 7 into the left-hand side (LHS) of the equation. 3 + 2x = 3 + 2(7) = 17 Therefore, 7 is not the solution of 3 + 2x = 19. LHS = RHS = 3 LHS = 17, RHS = 19; LHS ≠ RHS. ©Praxis Publishing_Focus On Maths

172 CHAPTER 5 Linear Equations and Inequalities in One Variable EXAMPLE 12 Solve each of the following linear equations. (a) x + 8 = 16 (b) x – 4 = –9 (c) –7x = 14 (d) x 5 = –2 Solution: (a) x + 8 = 16 x + 8 – 8 = 16 – 8 – 8 x = 8 (b) x – 4 = –9 x – 4 + 4 = –9 + 4 + 4 x = –5 (c) –7x = 14 –7x ÷ (–7) = 14 ÷ (–7) ÷ (–7) x = –2 (d) x 5 = –2 x 5 × 5 = –2 × 5 × 5 x = –10 B Additive and multiplicative inverses The additive inverse is the number with the opposite sign. When a number is added to its additive inverse, we will get 0. 4 is the additive inverse of –4, because (–4) + 4 = 0. –9 is the additive inverse of 9, because 9 + (–9) = 0. EXAMPLE 13 Find the additive inverse of (a) 78, (b) –100. Solution: (a) –78, because –78 + 78 = 0, (b) 100 because 100 + (–100) = 0. The multiplicative inverse of a number is its reciprocal. The product of a number and its multiplicative inverse is 1. For example, 1 8 is the multiplicative inverse of 8, because 8 × 1 8 = 1. 1– 11 2 2 is the multiplicative inverse of – 11 2 , because 1– 11 2 2 × 1– 11 2 2 = 1. ©Praxis Publishing_Focus On Maths

173 Linear Equations and Inequalities in One Variable CHAPTER 5 EXAMPLE 14 Find the multiplicative inverse of (a) –211, (b) 2 5 7 . Solution: (a) – 1 211, because (–211) × 1– 1 2112 = 1. (b) 7 19 , because 19 7 × 7 19 = 1. C Solving equations in the form x + a = b, x – a = b, ax = b and x a = b To solve equations in the form x + a = b, x – a = b, ax = b and x a = b, firstly isolate the term with the unknown by placing that term on one side of the equation using additive inverse and multiplicative inverse when needed. Remember whatever we do to one side of the equation, we must do the same to the other side. EXAMPLE 15 Solve. (a) x + 3 = 11 (b) y – 7 = 12 (c) 5n = –40 (d) k 4 = 3 Solution: (a) x + 3 = 11 x + 3 – 3 = 11 – 3 x = 8 (b) y – 7 = 12 y – 7 + 7 = 12 + 7 y = 19 (c) 5n = –40 5n 5 = – 40 5 n = –8 (d) k 4 = 3 k 4 × 4 = 3 × 4 k = 12 To isolate x, subtract 3 from both sides. To isolate y, add 7 to both sides. To isolate n, divide both sides with 5. To isolate k, multiple both sides with 4. ©Praxis Publishing_Focus On Maths

174 CHAPTER 5 Linear Equations and Inequalities in One Variable D Solving equations in the form ax + b = c EXAMPLE 16 Solve each of the following equations. (a) 4x + 9 = 37 (b) 2x 4 – 3 = 11 Solution: (a) 4x + 9 = 37 4x + 9 – 9 = 37 – 9 4x = 28 4x 4 = 28 4 x = 7 (b) 2x 3 – 3 = 11 2x 3 – 3 + 3 = 11 + 3 2x 3 = 14 2x 3 × 3 2 = 14 × 3 2 x = 21 To isolate 4x, subtract 9 from both sides. To isolate x, divide both sides with 4. To isolate 2x 3 , add 3 to both sides. To isolate x, multiply both sides with 3 2 . E Solving linear equations in one variable (a) Check Lee’s work. Is his solution correct? 8(–x + 3) = 8 8(–x) + 8(3) = 8 – 8x + 24 = 8 – 8x + 24 – 24 = 8 – 24 –8x – 8 = –16 – 8 x = 2 team work (b) If your answer is yes, verify the solution. If your answer is no, describe the error, then correct it. EXAMPLE 17 Solve the equation 5(x + 2) = 3x – 6. ©Praxis Publishing_Focus On Maths

175 Linear Equations and Inequalities in One Variable CHAPTER 5 Solution: 5(x + 2) = 3x – 6 5x + 10 = 3x – 6 5x – 3x = –6 – 10 2x = –16 x = –16 2 = – 8 EXAMPLE 18 Solve each of the following equations. (a) y + 11 = –15 (b) m + 7 5 = m – 5 Solution: (a) y + 11 = –15 y = –15 – 11 = –26 (b) m + 7 5 = m – 5 m + 7 = 5(m – 5) = 5m – 25 m – 5m = – 25 – 7 –4m = – 32 m = – 32 – 4 = 8 Discuss with your classmates. When you solve an equation, how can you be sure that your solution is correct? INTERACTIVE ZONE F Solving problems involving linear equations in one variable 3 The volume of blood in your body (in litres) is approximately equal to the mass in kilograms multiplied by 0.08. (a) Write an equation to represent this situation and state what each pronumeral represents. (b) According to this rule, how many litres of blood does a 60 kg person have? (c) Estimate the mass of a person with 3.8 litres of blood. ©Praxis Publishing_Focus On Maths

176 CHAPTER 5 Linear Equations and Inequalities in One Variable To solve problems involving linear equations in one variable, follow the steps below: 1 Identify the variable in the problem. Use a letter to represent the variable. 2 Form a linear equation using the information given in the problem. 3 Solve the equation to find the value of the variable. EXAMPLE 19 The length of a rectangular field is 40 m longer than its width. Given that the perimeter of the field is 400 m, find the length and width of the field. Solution: Stage 1: Understand the problem List the facts and the question. Facts: Length of rectangular field = 40 m longer than width Perimeter = 400 m Question: Find the length and width of the field. Stage 2: Think of a plan Make a linear equation to calculate the perimeter of the rectangular field. Perimeter = (Length + Length) + (Width + Width) Since Length = 40 + Width, thus, Perimeter = [(40 + Width) + (40 + Width)] + (Width + Width) Stage 3: Carry out the plan Let the width of the field be x m. Then, length of the field = (x + 40) m Perimeter = x + x + (x + 40) + (x + 40) 4x + 80 = 400 4x = 400 – 80 = 320 x = 320 4 = 80 (x + 40) m x m Width = 80 m Length = 80 + 40 = 120 m ©Praxis Publishing_Focus On Maths

177 Linear Equations and Inequalities in One Variable CHAPTER 5 Stage 4: Look back Work backwards to check. Perimeter = 120 m + 120 m + 80 m + 80 m = 400 m When Hazura added three consecutive odd numbers, the result was 87. What is the first odd number? EXAMPLE 20 Marks 2 4 6 8 Number of participants m 10 9 21 The table given shows the marks obtained by a group of participants in a Mathematics quiz. The participants who score more than 4 marks will be given a prize. Given that the number of participants who received a prize is twice the number of participants who did not receive a prize from the organiser. Find the value of m. Solution: Stage 1: Understand the problem List the facts and the question. Facts: Participants who score more than 4 marks will be given a prize. Number of participants who receive a prize is twice the number of participants who did not receive a prize. Question: Find the number of participants who score 2 marks. Stage 2: Think of a plan Get the data from the table. Sum up the number of participants that will receive a prize. Sum up the number of participants that will not receive a prize. ©Praxis Publishing_Focus On Maths

178 CHAPTER 5 Linear Equations and Inequalities in One Variable Stage 3: Carry out the plan Marks 2 4 6 8 Number of participants m 10 9 21 Number of participants who received a prize = 9 + 21 = 30 Number of participants who did not receive a prize = m + 10 Given that the number of participants who received a prize is twice the number of participants who did not receive a prize. Thus, 30 = 2(m + 10) 2(m + 10) = 30 2(m + 10) 2 = 30 2 ÷ 2 m + 10 = 15 m + 10 – 10 = 15 – 10 – 10 m = 5 Stage 4: Look back Work backwards to check. 5 + 10 = 15 9 + 21 = 30 30 15 = 2 Therefore, the number of participants who received a prize is twice the number of participants who did not receive a prize. EXAMPLE 21 A cell phone company offers two plans. Plan A: 120 free minutes, $0.75 per additional minute Plan B: 30 free minutes, $0.25 per additional minute How many minutes of calling will result in the same cost for both plans? ©Praxis Publishing_Focus On Maths

179 Linear Equations and Inequalities in One Variable CHAPTER 5 Stage 3: Carry out the plan 0.75 (t – 120) = 0.25 (t – 30) 0.75(t) + 0.75 (–120) = 0.25(t) + 0.25 (–30) 0.75t – 90 = 0.25t – 7.5 0.75t – 0.75t – 90 = 0.25t – 7.5 0.50t – 90 = – 7.5 0.50t – 90 + 90 = – 7.5 + 90 0.50t = 82.5 0.50t 0.50 = 82.5 0.50 t = 165 The cost is the same for both plans when 165 min of calls are made. Solution: Stage 1: Understand the problem List the facts and the question. Facts: 120 free minutes, $0.75 per additional minute. 30 free minutes, $0.25 per additional minute. Question: How many minutes of calling will result in the same cost for both plans? Stage 2: Think of a plan Let t minutes represent the time for calls. For Plan A, you pay only for the time that is greater than 120 min. So, the time you pay for is (t – 120) min. Each minute costs $0.75, so the cost in dollars is: 0.75 (t – 120) For Plan B, you pay only for the time that is greater than 30 min. So, the time you pay for is (t – 30) min. Each minute costs $0.25, so the cost in dollars is: 0.25 (t – 30) When these two costs are equal, the equation is: 0.75 (t – 120) – 0.25 (t – 30) ©Praxis Publishing_Focus On Maths

180 CHAPTER 5 Linear Equations and Inequalities in One Variable Stage 4: Look back Work backwards to check. For Plan A, you pay for: 165 – 120, or 45 min The cost is: 45 × $0.75 = $33.75 For Plan B, you pay for: 165 – 30, or 135 min The cost is: 135 × $0.25 = $33.75 These costs are equal, so the solution is correct. Maths LINK Science Ohm’s Law relates the resistance, R ohms, in an electrical circuit to the voltage, V volts, across the circuit and the current, I amperes, through the circuit: R = V t . For a light bulb, when the voltage is 120 V and the resistance is 192Ω, the current in amperes can be determined by solving this equation: 192 = 120 t . Practice 5.3 Basic Intermediate Advanced A Write a situation based on each of the following equations. (a) y + 5 = 6, where y is an integer. (b) t + 3 = 158, where t is Henry’s height, in cm, in the previous year. (c) 6p = 48, where p is the length of a rectangle, in cm. (d) 7 + 3m = –8, where m is an integer. B Solve the linear equation 3x + 2 = 11 by using (a) trial and improvement method, (b) application of equality concept, (c) backtracking method. C Determine whether the number in brackets is the solution for the given equation by using trial and improvement method. (a) 2x – 4 = 32 (x = 13) (b) 7 – x = 3 (x = 4) (c) –3x + 12 = –6 (x = –6) D Solve each of the following equations. (a) 3x + 7 = 28 (b) 3x + 8 = 14 (c) 5x – 3 = 37 (d) 4x – 11 = 17 (e) –2x – 9 = 13 (f) –6x + 11 = –25 E Solve each of the following equations. (a) 3x 5 – 5 = 10 (b) 2x 7 + 3 = 19 (c) 11 + 2x 3 = 9 (d) 13 – 5x 8 = –17 ©Praxis Publishing_Focus On Maths

181 Linear Equations and Inequalities in One Variable CHAPTER 5 F Solve each of the following linear equations. (a) 5w + 3 = 23 (b) 1 – 3w = –5 (c) 2 3 x + 1 = –1 (d) – 1 2 x + 7 = – 4 (e) 2(y + 2) = 16 (f) 4(3 + 2y) = – 4 (g) 2z + 3 3 = 7 (h) z + 4 5 = z – 2 G The perimeter of a rectangle with length of p cm and width of 5 cm is 28 cm. Form a linear equation in one variable in terms of p. Hence, find the value of p. 8 Linda is 5 years younger than her sister. Their total age is 47 years. Calculate Linda’s age. 9 Age Number of visitors 45 – 49 10 50 – 54 17 55 – 59 15 60 – 64 n 65 – 69 6 [45 – 49 means 45 years old until 49 years old.] The table above shows the number of visitors at a recreation park on a certain day. The entrance fee for senior citizens will be given certain discount. Given the number of visitors who did not get discount is 3 times the number of visitors who get discount. Calculate the value of n. [The senior citizens are defined as those who aged 60 and above.] 5.4 Inequalities in One Variable A Understand and use the concept of inequalities When two quantities are not equal, this state of unequal condition can be related by using symbols such as , , < or >. These symbols are known as inequality symbols. For example, John has 6 marbles and Harry has 10 marbles. To show that 10 marbles is more than 6 marbles, we write 10 . 6. We can also write 6 , 10 to indicate that 6 is less than 10. Therefore, an inequality is a relationship between two unequal quantities. The table below shows the meaning of four inequality symbols. Inequality symbol Meaning and read as . is greater than , is less than > is greater than or equal to < is less than or equal to ©Praxis Publishing_Focus On Maths

182 CHAPTER 5 Linear Equations and Inequalities in One Variable 4 (a) Define a variable and write an inequality for each situations. (b) Compare your inequalities with those of your classmates. (c) If the inequalities are different, how can you find out which is correct? P 30 min MON - FRI 9 AM - 6 PM EXAMPLE 22 Use an inequality symbol to represent the relationship involving the quantities given in the following statements. (a) 12 is less than 25. (b) y is greater than 6. (c) The temperature of the chamber, T is less than or equal to 35°. (d) p is greater than or equal to 2r. Solution: (a) 12 , 25 (b) y . 6 (c) T < 35° (d) p > 2r EXAMPLE 23 Write the relationship between the following pairs of numbers using the symbols , or .. (a) 3.25, 3 1 2 (b) 1 4 7 , 1 3 5 Solution: (a) 3 1 2 = 3.5 (b) 1 4 7 = 1 4 × 5 7 × 5 = 1 20 35 \ 3.25 , 3.5 1 3 5 = 1 3 × 7 5 × 7 = 1 21 35 \ 1 3 5 . 1 4 7 ©Praxis Publishing_Focus On Maths

183 Linear Equations and Inequalities in One Variable CHAPTER 5 B Concept of linear inequalities in one variable Consider the inequality relation y . 6 in Example 22(b), there is a variable y in the inequality and its power is one. We call this type of relation a linear inequality in one variable. Notice that if y is an integer, there are many values of y such as 7, 8, 9, 10, … which satisfy this inequality. EXAMPLE 24 Determine the possible solutions for each of the following inequalities, given that x and y are integers. (a) x , 14 (b) y . 4 (c) y > –6 Solution: (a) x , 14 If x is an integer, the possible solutions for x that are less than 14 are 13, 12, 11, 10, 9, … (b) y . 4 If y is an integer, the possible solutions for y that are greater than 4 are 5, 6, 7, 8, 9, … (c) y > –6 If y is an integer, the possible solutions for y that are greater than or equal to –6 are –6, –5, –4, –3, –2, … –6 is included. C Representing linear inequalities on a number line In Chapter 1, we have learnt how to compare two values based on their positions on a number line. In this section, we use the same concept to compare the values of a number with a variable based on their positions on the number line. For example, –4 x x 7 x . – 4 x , 7 A linear inequality can be represented on a number line. For example, (a) x . a (b) x > b a x a ' ' means not including the value a. b x b ' ' means including the value b. (c) x  c (d) x < d x c c x d d ©Praxis Publishing_Focus On Maths

184 CHAPTER 5 Linear Equations and Inequalities in One Variable EXAMPLE 25 Describe the relationship between the number and the variable, x, in each of the following number lines. Hence, form an algebraic inequality for the relationship. (a) x –5 (b) 6 x Solution: (a) x is less than –5. (b) x is greater than 6. x , –5 x . 6 EXAMPLE 26 Represent each of the following inequalities on a number line. (a) x . 3 (b) x > –2 (c) x , – 1 2 (d) x < 4.5 (e) 3 , x < 7 Solution: (a) x 3 2 3 4 5 (b) –3 –2 –1 0 x –2 (c) x – – 1 2 –3 –2 –1 0 – –1 2 (d) x 4.5 2 3 4 4.5 5 (e) 2 3 4 5 6 7 8 3 x 7 When given a number line, a linear inequality can be written by observing (a) the direction of the arrow, (b) the symbol ‘ ’ and ‘ ’. For example, (i) a Arrow towards the right (→) and the symbol ‘ ’. (ii) a Arrow towards the right (→) and the symbol ‘ ’. x . a x > a (iii) a Arrow towards the left (←) and the symbol ‘ ’. (iv) a Arrow towards the left (←) and the symbol ‘ ’. x , a x < a ©Praxis Publishing_Focus On Maths

185 Linear Equations and Inequalities in One Variable CHAPTER 5 EXAMPLE 27 Write an inequality for each of the following representations on the number lines. (a) 9 10 11 12 (b) –7 –6 –5 –4 (c) 0 1 22.3 3 (d) –1 0 1 2 – –1 4 (e) 2 3 4 5 6 7 Solution: (a) x > 10 (b) x , –5 (c) x < 2.3 (d) x . – 1 4 (e) 3 < x , 6 ‘¡’ on the number line represents . or , whereas ‘l’ on the number line represents > or <. Practice 5.4 Basic Intermediate Advanced A Rewrite each of the following statements in terms of one of the inequality symbols ,, ., < and >. (a) –5 is less than –3. (b) The length of the cuboid (l) is less than or equal to 20 cm. (c) The weight of the watermelon (w) is greater than or equal to 3 kg. (d) The number of audience (N) at the concert is greater than 500. B Write the relationship between the two numbers in the following sets of numbers using the symbols . or ,. (a) 3 1 3 , 3.25 (b) – 8, –5 (c) 29 4 , 7 3 8 (d) – 2 3 7 , –2 4 9 C Determine the possible solutions for each of the following inequalities, given that x and y are integers. (a) x . –2 (b) y < 4 (c) p > 3 (d) q , 6 (e) m > –6 ( f) x , 10 D Represent each of the following inequalities on a number line. (a) x . –4 (b) x , 6 (c) x > 2 1 2 (d) y < –5 1 4 (e) 3.5 , y < 9 (f) –3 < y , 2 E Write an inequality for each of the following representation on the number line. (a) 1 2 3 4 5 (b) 765 98 (c) –4 –3 –2 –1 0 –1.5 (d) –8 –7 –6 –5 –4 –6 –1 3 (e) –2 –1 0 1 2 3 4 5 (f) 3.5 432 5 6 7 8 9 10 ©Praxis Publishing_Focus On Maths