ALGEBRAIC 2EXPRESSIONS Applications of this chapter Algebra is like a universally useful tool that can help solve everyday problems. Here are some simple examples that demonstrate the relevance of algebra in the real world: Doubling or halving recipes when cooking, interior and landscape design, business and financial management, sport, finding tax obligations, computer programming and astrological calculations. Everyday life problems can be represented in a simple mathematical model by writing the relationship between quantities involved in algebra language. Discuss the problems of daily life which involves unknown values. 36
37 Learning Outcomes • Distinguish between monomials and polynomials. • Determine the degree of a polynomial. • Model polynomials using algebra tiles. • Perform addition and subtraction involving polynomials. • Perform multiplication and division involving polynomials. • Determine the values of algebraic expressions. • Apply the knowledge of algebraic expressions to solve mathematical problems and real-life problems. Algebraic Expressions Simplify Algebraic Expressions Determine the Value of Algebraic Expressions Apply Algebraic Expressions Monomials and Polynomials Degree of a Polynomial Modelling Polynomials Adding and Subtracting Polynomials Multiplying and Dividing a Polynomial by a Number Multiplying and Dividing a Polynomial by a Monomial Make Generalisation About Patterns Algebraic formulae • Algebraic expression • Term • Monomial • Polynomial • Coefficient • Variable • Constant • Exponent • Degree of a term • Degree of a polynomial • Pattern • Formula Maths History Key Terms The word ‘algebra’ comes from the Arabic word ‘al jabr’, which means ‘reunion of broken parts’. Muhammad ibn Musa al-Khwarizmi was a 9th-century Persian mathematician, geographer, and astronomer known as the ‘Father of Algebra’. Concept Map
38 CHAPTER 2 Algebraic Expressions Flashback 1. (a) List the first six multiples of 2 and 3. (b) Identify the common multiples of 2 and 3. 2. Simplify each of the following. (a) 5 9 + 2 9 (b) 9 17 – 3 17 Farmers Fendi and Adam are transplanting rice seedlings in a paddy field. They transplant rice seedlings bound in the north and south side of their rectangular paddy field. Fendi started first and has transplanted 1 m on the south side but Adam told him that he should start from the north side. So, Fendi moves to the north. Adam finished transplanting the south side and continue working on the north side where he transplanted 2 m. Both farmers completed their work. (a) Which farmer transplanted more of the rice seedlings? (b) How much more? 1 How many types of rice are there? Critical Thinking (c) 2 3 + 1 4 (d) 4 3 – 2 5 3. Simplify each of the following. (a) 3 × 5 6 (b) 3 5 × 10 21 (c) 6 ÷ 5 3 (d) 3 28 ÷ 6 7
39 Algebraic Expressions CHAPTER 2 2.1 Simplifying Algebraic Expressions A Monomials and polynomials Algebraic expression is an expression that contains variables, numbers and operations. Some of its examples include: 9x + 5 x – 2y + 3 4x 3 + xy + 2 Expressions are made up of terms. For instance, 9x + 5 There are two terms, 9x and 5. From the expression 9x + 5, 9 is the coefficient of variable x and 5 is a constant. An algebraic expression made up of terms connected by the operations of addition and subtraction is called polynomial. A polynomial consists of two or more terms. If a polynomial has only one term, then it is a monomial. Monomials Polynomials m 10x + 1 pq 2x + 2y 9xy a + b + c 6m2 3x 2 – 2x + 1 p3q 7p3 + 4p2 – 3p + 2 EXAMPLE 1 State whether each of the following expressions is a monomial or a polynomial. (a) 3ab + 4 (b) 8x 2 – 3xy + y2 (c) 5m3 (d) p2 + 5q2 + r 2 – p – 7 (e) – 2k 3 + k 2 + 2k – 1 (f) 11 Solution: (a) Polynomial (b) Polynomial (c) Monomial (d) Polynomial (e) Polynomial (f) Monomial Terms Some words use ‘poly’ and ‘poly’ means more than one. So, a polynomial is an expression with several terms. Discuss with your friends how the names monomial, binomial and trinomial relate to the number of terms in an expression. INTERACTIVE ZONE
40 CHAPTER 2 Algebraic Expressions B Degree of a polynomial What is an exponent? 6m2 p 3 q1 The sum of the exponents on the variables in a single term is known as the degree of the term. For example, 6m2 has a degree of 2 while p3 q has a degree of 4. The degree of the highest-degree term in a polynomial is known as the degree of the polynomial. For example, 3x 2 – 2x + 1 is a polynomial with one variable. The degree of the first term is 2, the second term is 1 and the third term is 0. The highest degree is 2, so the degree of the polynomial is 2. x 3 + 5x 2y 4 + y 2 is a polynomial with two variables. The degree of the first term is 3, the second term is 6 and the third term is 2. The highest degree is 6, so the degree of the polynomial is 6. The polynomials with one variable can be classified into different types based on their degrees. Polynomial Degree Example Constant polynomial 0 7 Linear polynomial 1 9x + 5 Quadratic polynomial 2 3x 2 – 2x + 1 Cubic polynomial 3 x3 – 12x 2 + 48x – 64 Quartic polynomial 4 6x4 + 3x 3 + 3x 2 + 2x + 1 EXAMPLE 2 State the degree of each of the following polynomials. (a) 4a3 + 2 (b) 6x 2 – 2xy + y 2 (c) 3pq + q3 – 5 Solution: (a) The polynomial has two terms. The term with the highest degree is 4a3 . Its degree is 3. So, the degree of the polynomial is 3. (b) The polynomial has three terms. Each of the three terms has a degree of 2. So, the degree of the polynomial is 2. (c) The polynomial has three terms. The term with the highest degree is q3 . Its degree is 3. So, the degree of the polynomial is 3. Exponent Exponent Discuss with your friends whether each of the following algebraic expressions is a polynomial. (a) 3 n + n – 4 (b) 6k + √k INTERACTIVE ZONE
41 Algebraic Expressions CHAPTER 2 C Modelling polynomials In arithmetic, we use Base Ten Blocks to model whole numbers. For example, the number 123 can be modelled as 100 10 3 In algebra, we can use algebra tiles to model a polynomial. For example, yellow represents positive tiles and red represents negative tiles. An x 2 -tile has a dimension of x units by x units. An x -tile has a dimension of 1 unit by x units. A 1-tile has a dimension of 1 unit by 1 unit. EXAMPLE 3 Model each of the following polynomials. (a) 2x + 1 (b) –x 2 – 3 (c) 3x 2 + 2 – x (d) 2p2 + p – 4 Solution: (a) 2x + 1 (b) –x2 – 3 • We can use any variable to write a polynomial and to describe the tiles that model it. For example, the tiles that model the polynomial 3x 2 + 2 – x also model the polynomial 3m2 + 2 – m. • A polynomial is usually written in descending order, that is, the exponents of the variable decrease from left to right. For example, the polynomial 3x 2 + 2 – x is usually written as 3x 2 – x + 2. (a) Which polynomial does each group of algebra title represent? (b) Which of the polynomials in part (a) are equivalent? 2 Model A Model B Model C x 2 –x 2 x –x 1 –1
42 CHAPTER 2 Algebraic Expressions D Adding and subtracting polynomials We have learnt that like terms are terms with the same variable while unlike term are terms with different variables. An algebraic expression is in simplest form if it has no like terms and no parentheses. (c) 3x 2 + 2 – x (d) 2p2 + p – 4 Tiles can be either horizontal or vertical Maths LINK A polynomial can be used to model projectile motion. When a golf ball is hit with a golf club, the distance the ball travels in metres, in terms of the time t seconds that it is in the air, may be modelled by the polynomial –4.9t 2 + 22.8t. EXAMPLE 4 Determine the sum of 3x 2 – x + 2 and x 2 + 4x – 3. Give your answer in the simplest form. Solution: (3x 2 – x + 2) + (–x 2 + 4x – 3) = 3x 2 – x + 2 – x 2 + 4x – 3 = 3x 2 – x 2 – x + 4x + 2 – 3 = 2x 2 + 3x – 1 Group like terms Method 1: Use algebra tiles. Model each polynomial and combine them. 2 Group like terms and remove zero pairs. 0 0 0 (a) Write a polynomial to represent the perimeter of each rectangle. (i) (ii) x x x x x x x 1 1 (b) Each polynomial represents the perimeter of a rectangle. Use algebra tiles to make the rectangle. (i) 4a + 2 (ii) 10b 3
43 Algebraic Expressions CHAPTER 2 EXAMPLE 6 Simplify (3p 2 + p – 2q – 6pq + 2q 2) + (–3q2 + 2pq + 5q – 4p + 6p 2). Solution: (3p 2 + p – 2q – 6pq + 2q 2) + (– 3q2 + 2pq + 5q – 4p + 6p 2) = 3p 2 + p – 2q – 6pq + 2q 2 – 3q 2 + 2pq + 5q – 4p + 6p 2 = 3p 2 + 6p 2 + p – 4p – 2q + 5q – 6pq + 2pq + 2q2 – 3q2 = 9p 2 – 3p + 3q – 4pq – q 2 EXAMPLE 7 Subtract 3x 2 – x – 1 from 2x 2 + 3x + 2. Give your answer in the simplest form. Solution: (2x 2 + 3x + 2) – (3x 2 – x – 1) = 2x 2 + 3x + 2 – (3x 2) – (–x) – (–1) = 2x 2 + 3x + 2 – 3x 2 + x + 1 = 2x 2 – 3x 2 + 3x + x + 2 + 1 = x 2 + 4x + 3 EXAMPLE 8 Simplify (7a2 – 8ab – 6b2) – (11a2 – 5ab + 7b2). Solution: (7a2 – 8ab – 6b2) – (11a2 – 5ab + 7b2) = 7a2 – 8ab – 6b2 – (11a2) – (–5ab) – (+7b2) = 7a2 – 8ab – 6b2 – 11a2 + 5ab – 7b2 = 7a2 – 11a2 – 8ab + 5ab – 6b2 – 7b2 = – 4a2 – 3ab – 13b2 EXAMPLE 5 Simplify (6m + 12) + (4m2 – m – 5). Solution: (6m + 12) + (4m2 – m – 5) = 6m + 12 + 4m2 – m – 5 = 4m2 + 6m – m + 12 – 5 = 4m2 + 5m + 7 3 The remaining tiles is the simplest form of the sum. 2x2 + 3x – 1 Method 2: Add vertically. Align like terms, then add their coefficients. 3x 2 – x + 2 + –x 2 + 4x – 3 2x 2 + 3x – 1 Use the properties of integers Method 1: Use algebra tiles. Model each polynomial. From remove 2 Add zero pairs because there are not enough tiles to remove. Add 3 Remove 3x 2 – x – 1. –1 3x2 –x 4 The remaining tiles is the simplest form of the difference. –x 2 + 4x + 3
44 CHAPTER 2 Algebraic Expressions EXAMPLE 9 The diagram shows a rectangle. Write an algebraic expression, in simplest form, to represent the perimeter of the rectangle. Solution: Perimeter = (2x + 3) + (3x + 4) + (2x + 3) + (3x + 4) = 2x + 3 + 3x + 4 + 2x + 3 + 3x + 4 = 2x + 3x + 2x + 3x + 3 + 4 + 3 + 4 = 10x + 14 EXAMPLE 10 The diagram shows two rectangles, ABCD and AGFE. The area of the rectangle ABCD is 6x 2 + 17x + 12. The area of the rectangle AGFE is 6x 2 + 5x + 1. Write an algebraic expression, in simplest form, to represent the area of the shaded region. Solution: Area of the shaded region = (6x 2 + 17x + 12) – (6x 2 + 5x + 1) = 6x 2 + 17x + 12 – 6x 2 – 5x – 1 = 6x 2 – 6x 2 + 17x – 5x + 12 – 1 = 12x + 11 E Multiplying and dividing a polynomial by a number In multiplication or division of an algebraic expression by a number, every term in the expression is multiplied or divided by the number. EXAMPLE 11 Simplify each of the following. (a) 3(7a – 5) (b) –5(6k 2 – 2k + 9) (c) –4m2 + 6 2 (d) 16x 2 + 8x – 12 – 4 2x+ 3 3x+ 4 A B E F G D C
45 Algebraic Expressions CHAPTER 2 An equilateral triangle has three sides of equal lengths. Solution: (a) 3(7a – 5) = 3(7a) + 3(–5) = 21a – 15 (b) –5(6k 2 – 2k + 9) = – 5(6k 2 ) + (–5)(–2k) + (–5)(9) = –30k 2 + 10k – 45 (c) –4m2 + 6 2 = –4m2 2 + 6 2 = –2m2 + 3 (d) –16x 2 + 8x – 12 – 4 = 16x 2 – 4 + 8x – 4 + – 12 – 4 = –4x 2 – 2x + 3 EXAMPLE 12 Write an algebraic expression, in simplest form, to represent each of the following. (a) What is the length of side of an equilateral triangle with a perimeter of 6x + 9? (b) What is the volume of 3 cylindrical tanks where each of the cylindrical tanks has a volume of 6x 3 + 3x 2 ? Solution: (a) Length of side = 6x + 9 3 = 6x 3 + 9 3 = 2x + 3 (b) Volume = 3(6x 3 + 3x 2 ) = 3(6x 3 ) + 3(3x 2 ) = 18x 3 + 9x 2 Discuss with your friends, how do you simplify the following expressions using algebra tiles? (a) 3(7a – 5) (b) – 4m 2 + 6 2 INTERACTIVE ZONE
46 CHAPTER 2 Algebraic Expressions EXAMPLE 13 Simplify each of the following. (a) 2x(x – 3) (b) –3p(–7p 2 + 3p + 1) (c) –18n 2 + 8n 2n (d) –25h 3 – 15h 2 + 5h – 5h Solution: (a) 2x(x – 3) = 2x(x) + 2x(–3) = 2x 2 – 6x (b) –3p(–7p 2 + 3p + 1) = –3p(–7p 2) + (–3p)(3p) + (–3p)(1) = [(–3) × p × (–7) × p × p ] + [(–3) × p × 3 × p ] + [(–3) × p × 1] = [(–3) × (–7) × p × p × p ] + [(–3) × 3 × p × p ] + [(–3) × 1 × p ] = 21p 3 + (–9p 2 ) + (–3p ) = 21p 3 – 9p 2 – 3p (c) –18n 2 + 8n 2n = –18n 2 2n + 8n 2n = (–18) – 9 × n × n 1 1 2 × n 1 + 8 4 × n 1 1 2 × n 1 = –9n + 4 (d) –25h 3 – 15h 2 + 5h – 5h = –25h 3 – 5h + –15h 2 – 5h + 5h – 5h = (–25) 5 × h 1 × h × h ( 1 – 5) × h 1 + (–15) 3 × h 1 × h ( 1 – 5) × h 1 + 5 –1 × h 1 ( 1 – 5) × h 1 = 5h 2 + 3h – 1 Method 1: Use algebra tiles. Make a rectangle with a dimension of 2x and x – 3. Product The product is 2x 2 – 6x. F Multiplying and dividing a polynomial by a monomial In multiplication or division of an algebraic expression by a monomial, every term in the expression is multiplied or divided by the monomial.
47 Algebraic Expressions CHAPTER 2 team work Try the following activity in groups of 4. A student made these mistakes in a test. (a) The student simplified (b) The student simplified 2x + 3x as 5x 2 . 4 + 3x as 7x. Use algebra tiles to explain what the student did wrong. What are the correct answers? Present each group's answer in class. Practice 2.1 Basic Intermediate Advanced 1 Complete the following table with the algebraic expressions given. 5 d + e 6 pq y 2 + y + 1 7x + 3 9mn + m 7x 3 2a4 + a3 – 6a2 – a Monomials Polynomials B Identify coefficient, variable(s) and exponent of each variable for each of the following expressions. (a) 3a (b) 2b 2 (c) – 5hk (d) –m C State the degree of each of the following polynomials. (a) 5x 2 + 3xy – 2y (b) 3x 3 + 4xy – 2x 2 + y (c) 6x + 1 (d) 3xy + xy 2 + 11x 3 y D Model each of the following polynomials using algebra tiles. (a) –3x 2 – 4 (b) –2x + 5 (c) x 2 + 2x + 2 5 Simplify each of the following. (a) (2a – 3) + (5 – 4a) (b) (3b2 – 5b) + (b2 + 2b + 1) (c) (2hk + 3h2 – 3k 2 + 6h) + (7k 2 – 8k + 10) (d) (3m – 2) – (–m + 3) (e) (7n 2 – n + 1) – (3n2 – 2n – 4) (f) (5pq + 8p2 – 7q2 ) – (3q2 + 6pq – 4p2 ) 6 Simplify each of the following. (a) 3(2a + 9) (b) –3(2b2 + b – 4) (c) 6(5cd + 3d 2 – 4c 2 ) (d) 8h + 12 4 (e) 5m2 – 20 – 5 (f) 21p2 + 9p + 3 3 7 Simplify each of the following. (a) 2e(3e – 2) (b) –4h(5h2 – 3h + 1) (c) 3s 2 + 9s – 3s (d) 18t 3 – 12t 2 + 24t 6t 8 2n + 4 2n + 1 n + 4 The diagram shows a triangle. Write an algebraic expression, in simplest form, to represent the perimeter of the triangle. 9 P S Q R 3p + 5 The diagram shows a rectangle PQRS. The perimeter of the rectangle is 7p + 12. What is the length, in simplest form, of side PQ?
48 CHAPTER 2 Algebraic Expressions Write an algebraic expression, in simplest form, for each of the following. (a) What is the perimeter of a square with a side length of 7r + 2? (b) A cube has 6 surfaces. What is the area of one surface of a cube with a total surface area of 54x 2 ? Square P has a side length of 7x + 6. Square Q has a side length that is 3 times the side length of square P. (a) Determine the perimeter, in simplest form, of (i) square P, (ii) square Q. (b) Write an algebraic expression, in simplest form, to represent the difference in the perimeter of squares P and Q. The area of circle A is given by πx 2 . The area of circle B is given by 9πx 2 . Find the difference in the area of circles A and B. Give your answer in the simplest form. The diagram shows a shape. Write an algebraic expression, in simplest form, to represent the area of the shape. A student multiplied (–x 2 – 7x + 3) by (–3x) as follows. (–3x)(–x2 – 7x + 3) = (–3x)(–x2 ) – (–3x)(–7x) + (–3x)(3) = –3x3 – (21x2 ) + (9x) = –3x3 – 21x2 + 9x Is the student’s work correct? If not, explain where the student made any errors and write the correct solution. Given that the perimeter of an isosceles triangle is 7y + 5x + 9. Write three algebraic expressions that could represent the side lengths of the triangle. 2.2 Determining the Values of Algebraic Expressions The value of an algebraic expression can be determined by substituting the variables with the given value. EXAMPLE 14 If x = –5 and y = 4, determine the value of 4(2x – y) – (6x – 7y). Solution: 4(2x – y) – (6x – 7y) = 4(2x) + 4(–y) – (6x) – (–7y) = 8x – 4y – 6x + 7y = 8x – 6x – 4y + 7y = 2x + 3y = 2(–5) + 3(4) = –10 + 12 = 2 Simplify first. 15x 4x 6x 9x
49 Algebraic Expressions CHAPTER 2 EXAMPLE 15 The perimeter of a triangle is given by 12x + 5. The lengths of two sides of the triangle are 5x – 2y and x + y. If x = 3 cm and y = 2 cm, find the length of the third side. Solution: Length of the third side = 12x + 5 – (5x – 2y) – (x + y) = 12x + 5 – 5x + 2y – x – y = 12x – 5x – x + 5 + 2y – y = 6x + 5 + y = 6(3) + 5 + (2) = 18 + 5 + 2 = 25 cm Practice 2.2 Basic Intermediate Advanced 1 If x = 5 and y = –3, find the value of each of the following. (a) (–2x + y) + 3(x + 2y) (b) 4(x – 2y) – 2(y – 3x) (c) –18x 2y – 6x B The perimeter of an equilateral triangle is represented by 27x 2 + 21x + 18. (a) What is the length, in simplest form, of one side of the triangle? (b) If x = 2 cm, find the length of one side of the triangle. C The cost, in dollars, to produce x magazines is 2x + 2000. The cost, in dollars, to produce y books is 4y + 2400. (a) Write an algebraic expression, in simplest form, to represent the total cost of the magazines and books. (b) If the company wants to produce 2000 magazines and 1000 books, what is the total cost, in dollars, of the magazines and books? D The surface area of cylinder P is 2πr (r + h) where r is its radius and h is its height. Cylinder Q has the same height as cylinder P, but its radius is twice the radius of cylinder P. (a) Write an algebraic expression, in simplest form, to represent the surface area of cylinder Q. (b) If the radius and height of cylinder P are 2 cm and 5 cm respectively, find the surface area of cylinder Q. 5 3x + 2 2x x + 1 2x Q B A D P C R S The diagram shows two rectangles, ABCD and PQRS. (a) Write an algebraic expression, in simplest form, to represent (i) the difference in the perimeter of rectangles ABCD and PQRS, (ii) the area of the shaded region. (b) Given that x = 1.5 cm, find (i) the difference in the perimeter of rectangles ABCD and PQRS, (ii) the area of the shaded region.
50 CHAPTER 2 Algebraic Expressions 2.3 Applying Algebraic Expressions A Making generalisation about patterns Objective: To generalise the patterns of sequences. Instruction: Do this activity in groups of four. Each group moves between stations. Material: Manila cards, matchsticks 1. At Station 1, a few fraction cards are arranged in a row as follows. 1 11 2 11 3 11 4 11 5 11 Card 1 Card 2 Card 3 Card 4 Card 5 Write an algebraic expression for the pattern. Card 1 2 3 4 5 ... n Fraction 1 11 2 11 3 11 4 11 5 11 ... 2. At Station 2, a few number cards are arranged in a row as follows. Card 1 Card 2 Card 3 Card 4 Card 5 2 4 6 8 10 Write an algebraic expression for the pattern. Card 1 2 3 4 5 ... n Number 2 4 6 8 10 ... 3. At Station 3, a few matchsticks are given. Arrange the matchsticks to form 1 square, 2 squares, 3 squares and so on. Diagram 1 Diagram 2 Diagram 3 Write an algebraic expression for the pattern. Diagram 1 2 3 ... n Number of matchsticks 4 7 10 ... 4. Present your findings in class. 1
51 Algebraic Expressions CHAPTER 2 EXAMPLE 16 Make a generalisation for the pattern of each of the following sequences using algebraic expression. (a) 3, 5, 7, 9, 11, … (b) 2, 8, 14, 20, 26, … (c) 110, 90, 70, 50, 30, … Solution: (a) 3, 5, 7, 9, 11, ... +2 +2 +2 +2 3 = 2(1) + 1 5 = 2(2) + 1 7 = 2(3) + 1 9 = 2(4) + 1 11 = 2(5) + 1 Thus, the pattern is 2n + 1 where n = 1, 2, 3, 4, 5, … (b) 2, 8, 14, 20, 26, ... +6 +6 +6 +6 2 = 6(1) – 4 8 = 6(2) – 4 14 = 6(3) – 4 20 = 6(4) – 4 26 = 6(5) – 4 Thus, the pattern is 6n – 4 where n = 1, 2, 3, 4, 5, … (c) 110, 90, 70, 50, 30, ... –20 –20 –20 –20 110 = 130 – 20(1) 90 = 130 – 20(2) 70 = 130 – 20(3) 50 = 130 – 20(4) 30 = 130 – 20(5) Thus, the pattern is 130 – 20n where n = 1, 2, 3, 4, 5, … EXAMPLE 17 The diagram shows an arrangement of canned drinks in a supermarket. The first row has 3 cans. The second row has 4 cans, the third row has 5 cans and so on. Make a generalisation for the pattern of the arrangement of canned drinks using numbers, words and algebraic expressions. First row Second row Third row To determine the algebraic expression of the pattern of a sequence where the order of numbers is decreasing, we subtract a multiple of n from a fixed number. For instance, the pattern of 15, 13, 11, 9, … is 17 – 2n. 2 × 1 + 1 Difference × Position + or – Number
52 CHAPTER 2 Algebraic Expressions Solution: Using numbers: 3, 4, 5, 6, ... Using words: The first row has 3 cans and each row below it has one can more than the row before it. Using algebraic expressions: Row Sequence 1 1 + 2 = 3 2 2 + 2 = 4 3 3 + 2 = 5 4 4 + 2 = 6 5 5 + 2 = 7 ... ... n n + 2 n + 2, where n = 1, 2, 3, ... represents the number of canned drinks in the n row (from the top). EXAMPLE 18 1 table 8 seats 2 tables 12 seats The tables in Restaurant Q can be arranged in the above pattern. (a) Write an algebraic expression to represent the number of seats if n tables are arranged in a row. (b) How many seats can be arranged if seven tables are to be arranged in a row? Solution: (a) Number of tables Number of seats 1 8 = 4 × 1 + 4 2 12 = 4 × 2 + 4 3 16 = 4 × 3 + 4 ... ... n 4 × n + 4 Hence, 4n + 4 where n = 1, 2, 3, … = 12 = 16 = 10 = 8 = What is the number that should be written in the box?
53 Algebraic Expressions CHAPTER 2 (b) When n = 7, 4n + 4 = 4(7) + 4 = 28 + 4 = 32 Therefore, 32 seats can be arranged if 7 tables are to be arranged in a row. B Forming algebraic formulae An expression is a mathematical statement with no equal symbol. An equation links two expressions with an equal symbol. A formula is an equation that shows relationship between a few variables. For instance, the formula for calculating the area of a triangle, A = 1 2 bh where b is the base of the triangle and h is the height of the triangle. b and h are variables because they do not have a fixed value. The subject of a formula is the variable that is being calculated. For instance, A is the subject of the formula A = 1 2 bh. The subject of the formula is usually written on the left of the equation and the expression is on the right. Sometimes we need to rearrange the formula to find the value we are looking for. For instance, we can rearrange the formula A = 1 2 bh to find the height of the triangle, h. Then, we have h = 2A b with h becomes the subject of the formula. The polynomials 4x – 3y and 2x + y represent the lengths of two sides of a triangle. The perimeter of the triangle is 9x + 2. Determine the length of the third side. Critical Thinking EXAMPLE 19 Write a formula for area, A, of the rectangle shown in the diagram. Solution: It is known that the area of a rectangle is length × width. Thus, A = 2p × p A = 2p2 2p p In the formula p = pq + r 2 , p is not the subject of the formula because the term p is on both sides of the equation.
54 CHAPTER 2 Algebraic Expressions EXAMPLE 20 Express the variables in the bracket as the subject of the formula. (a) a = b + c [b] (b) e = f – 3g [f ] (c) h = i 2j [i ] (d) k = ml [m] Solution: (a) a = b + c a – c = b + c – c a – c = b Thus, b = a – c. (b) e = f – 3g e + 3g = f – 3g + 3g e + 3g = f Thus, f = e + 3g. (c) h = i 2j h × 2j = i 2j × 2j 2hj = i Thus, i = 2hj. (d) k = ml k l = ml l k l = m Thus, m = k l . Subtract both sides of the equation by c. Add both sides of the equation by 3g. Multiply both sides of the equation by 2j. Divide both sides of the equation by l. EXAMPLE 21 The price, in $, of a customer who stays in a hotel for t days follows the formula of P = 215 + 75t. (a) Find the value of P when t = 3. (b) Calculate the number of days the customer stays in the hotel if he pays $815. The total age for Suryadi and his brother is Q. Given that Suryadi's age is k years old and his brother's age is 5 years younger than him. (a) Construct a formula for their total ages, Q. (b) Find the value of k if Q is 27.
55 Algebraic Expressions CHAPTER 2 Solution: (a) P = 215 + 75t = 215 + 75(3) = 215 + 225 = 440 (b) P = 215 + 75t P – 215 = 75t 75t = P – 215 t = P – 215 75 When P = 815, t = 815 – 215 75 = 600 75 = 8 The number of days the customer stays in the hotel is 8. Practice 2.3 Basic Intermediate Advanced 1 Make a generalisation for the pattern of each of the following sequences using algebraic expression. (a) 1, 5, 9, 13, … (b) 6, 11, 16, 21, … (c) 35, 32, 29, 26, … (d) 40, 35, 30, 25, … B Write a formula for each of the following based on the given diagrams. (a) Volume, V p q r (b) Area, A t s (c) Perimeter, P a C Write a formula to show the relationship between the variables in each of the following statements. (a) The sum of nine and square of a number, x, is y. (b) Power, P, is the measure of work done, W, divided by the time taken, t. (c) Einstein’s equation states that the energy, E, is the product of mass, m, and the square of speed of light, c. D Express the variables in brackets as the subject of the formula. (a) h = e + 4 [e] (b) c + 2d = e [c] (c) l = b – d [b] (d) –p + r 2 = h [p] (e) m n = 4r [m] (f) V I = R [I ] (g) 2rt = s [r] (h) 3v = nq [n] (i) y = 7(2 – w) [w]
CHAPTER 2 Algebraic Expressions 56 5 A stone is thrown vertically upwards with initial velocity, u m/s. The final velocity, v m/s, of the stone at t, in seconds, after throwing is v = u + gt. Express t in terms of g, u and v. 6 Given that the formula of the volume of a cylinder is V = πr 2 h, where r and h are the radius and height of cylinder respectively. Calculate the volume, in cm3 , for the cylinder with a radius of 5 cm and a height of 2 cm. Use π = 22 7 7 Diagram 1 Diagram 2 Diagram 3 Diagram 4 Diagram 5 The diagram above shows the pattern of straight lines drawn passing through a number of points. Make a generalisation of pattern of the number of points passed through by each of the straight lines using numbers, words and algebraic expressions. 8 A grocery store exhibits cereal boxes in an attractive arrangement to attract customers. The diagram above shows the arrangement for the first six top rows. Make a generalisation of pattern of the number of cereal boxes by using numbers, words and algebraic expressions. 9 1 triangle 2 triangles 3 triangles The toothpicks are arranged to form a few triangles as shown in the diagram above. (a) How many toothpicks are used to arrange 5 triangles? (b) Write an algebraic expressions to represent the number of toothpick, p, if n triangles are formed. n = 1 n = 2 n = 3 The diagram shows a sequence made from circles and dots. (a) Complete the following table. Diagram (n) 1 2 3 Number of circles 1 2 Number of dots (b) Write an algebraic expression to represent the number of dots in a diagram with n circles. (c) How many dots would there be in a diagram with 16 circles? The diagram shows a cuboid with square base of y cm and height of h cm. Given the total surface area of the cuboid is A. (a) Express h in terms of y and A. (b) Hence, calculate the value of h when y = 2 and A = 48. This year, Phang is k years old and his mother is three times of his age. The sum of their ages five years ago is q years. Express k in terms of q. Write an algebraic expression, in simplest form, to represent the sum of three consecutive whole numbers. 1 + 2 + 3 = 6 + 7 + 8 = 11 + 12 + 13 = h cm y cm y cm
57 Algebraic Expressions CHAPTER 2 Summary Summary Summary Algebraic Expressions Monomial Applying Algebraic Expressions Consists of one algebraic term. For instance, 3xy 2 , 5x, 29 Polynomial Consists of two or more algebraic terms. 5x 2 + 2y – 7 Coefficient Exponent Variable Constant Operator Pattern • An arrangement of numbers, shapes, colours, letters and so on that follows a particular rule. Formula • An equation that shows relationship between a few variables. Adding and subtracting polynomials Add or subtract the coefficient of the same variable. Multiplying and dividing a polynomial by a number or a monomial • Multiply or divide every term by the number. • Multiply or divide every term by the monomial. Section A 1. Which of the following polynomials has a degree of 1? A x 2 – 2x + 2 C 5x + 3 – xy B 5 – 3x D 4xy + 9 2. Which expression does not have zero as a constant term? A –2m 2 C 9t – 2t 2 B pq + 4q D 2n3 + n2 – 13 3. Which of the following set of algebra tiles represents –x + 2x 2 + 1? A B C D 4. Simplify (4y 2 – 5x + xy) – (7y 2 – 4xy – 8x). A –3y 2 – x + 9xy B – 3y 2 + 3x + 5xy C –3y 2 – 9x – 7xy D –3y 2 + 13x – 3xy 5. 5x(2x 2 – 8x + 1) = A 10x 2 – 8x + 1 B 10x 2 – 40x + 5 C 10x 3 – 8x 2 + x D 10x 3 – 40x 2 + 5x 2
CHAPTER 2 Algebraic Expressions 58 6. 2w + 1 4w + 3 w The diagram shows a triangle. The expression, in simplest form, for the perimeter of the triangle is A 7w + 4 C 5w + 3 B 6w + 4 D 3w + 3 7. Given that the nth number of a number sequence is 4n – 11. Find the 7th number of the sequence. A 17 C 19 B 18 D 20 8. Given that k = l + 3 2l , express l in terms of k. A l = 2k – 1 3 C l = 2(3k – 1) B l = 3 2k – 1 D l = 2k – 3 9. Chong has $k. After spending $(h – 3) a day for 5 days, he still has $10. Express h in terms of k. A h = 5k – 5 C h = k + 5 5 B h = 5k + 25 D h = k – 25 5 10. The distance, s, in m, travelled by an accelerating rocket is given by s = ht + 1 2 at 2 . Find the value of s when h = 3 m/s, t = 100 s and a = 0.5 m/s2 . A 300 C 1400 B 700 D 2800 Section B 1. Identify the variables, coefficients and constant terms in each of the following expressions. (a) – 3a2 + 7 (b) 3m – 11 (c) –2 – p + 5p 2 2. State the degree of each of the following polynomials. (a) 3b + 2 (b) –k 2 – 2k + 6 (c) 2n2 + 9 (d) 3p3 – 11pq – 3 (e) x – 2xy + 7 + 5y 3. Complete the following table using the polynomials given. 4x + 3x 2 5x 3 + 4x –3 + 5x 5x 2 – 6 – x 4 Types Polynomial Cubic polynomial Linear polynomial Quartic polynomial Quadratic polynomial 4. Model the polynomial 3h + 5 – h 2 using algebra tiles. 5. Simplify each of the following. (a) (3y 2 + 4y – 2) + (3y 2 – 8y – 3) (b) (4x 2 – 2x + 3) – (–11x 2 + 6x – 3) (c) (2w + 7 – 3w2 ) + (–2 – 4w + 6w 2 ) (d) (3 + 2s 2 + 3s ) – (–s 2 – 4 – s ) (e) (4p 2 – 5pq – 3q 2 ) + (9q 2 – 3pq – 4p 2) (f) (10mn – 3n 2 + 2m) – (5n – 4m 2 + mn) 6. x + x = x2 (x)(x) = 2x The mathematical statements shown in the diagram are not correct. Write the correct mathematical statements and explain it using algebra tiles. 7. (a) The sum of two polynomials is 10x + 7. One polynomial is 3x – 5. What is the other polynomial? (b) The difference of two polynomials is x 2 + 2x + 5. One polynomial is 2x 2 + x + 2. What is the other polynomial? List down two possible polynomials. 8. Write an algebraic expression with 6 terms that has only 2 terms after simplified.
Algebraic Expressions CHAPTER 2 59 9. Simplify each of the following. (a) 2(–7 + 2h + 3h2 ) (b) (10k 2 – 2k + 5)(– 3) (c) – 4(ef – e2 + f 2) (d) 2c (–3c + 4) (e) 12 – 16y 4 (f) 25 – 35x + 10x 2 – 5 (g) (–6w 2 – 9w) ÷ 3x (h) 18s 2 – 27s – 9s 10. Write an algebraic expression, in simplest form, for the perimeter of each of the following shapes. Hence, determine the perimeter when x = 5 cm. (a) 7x + 2 6x (b) 8x + 3 11. The area of a rectangle is represented by expression 15x 2 + 10x . The length of the rectangle is 5x . (a) Write an algebraic expression, in simplest form, to represent the width of the rectangle. (b) What are the dimensions and area of the rectangle when x = 6 cm? 12. Given that x 4 – 2y 2 = z 3 . (a) Express z in terms of x and y. (b) Calculate the value of z when x = –8 and y = –2. 13. The formula of the area of a trapezium is L = 1 2 (a + b)h. Find the value of a when L = 128, b = 14 and h = 16. 14. n = 1 n = 2 n = 3 The diagram shows a sequence of diagrams that are formed by squares and dots. (a) Complete the following table. Number of squares 1 2 3 Number of dots (b) Write down the number of dots in terms of n in its simplest form for diagram n. 15. n = 1 n = 2 n = 3 Each diagram in the sequence above consists of a number of toothpicks. (a) Draw diagram 4 of the sequence. (b) Write down the number of toothpicks in terms of n in its simplest form for diagram n. (c) Using the expression formed in (b), find the number of toothpicks in (i) diagram 10, (ii) diagram 55. 16. (a) Write an algebraic expression, in simplest form, for each of the following. (i) Even numbers (ii) Odd numbers (b) The sum of an even number and an odd number is an even number or an odd number? Justify your answer using the concept of algebra. 17. Nurul Huda is 10 years older than Najwa. After eight years, Aisyah’s age is twice Najwa’s age. (a) If p represents Najwa’s present age and J represents the total ages of Nurul Huda and Aisyah after eight years, express J in terms of p. (b) Find Aisyah’s age after eight years if J = 46.
LINEAR EQUATIONS IN 3TWO VARIABLES Applications of this chapter Many people use linear equations every day, even if they do the calculations in their heads without drawing a line graph. Any linear system can be described with a linear equation. We can apply linear equations to various real life situations, such as figuring out income over time, estimating recipe ingredients, calculating mileage rates, figuring out budgets, predicting profit or weather and others. What do you need to know to organise a trip? 60
61 Learning Outcomes • Understand and use the concept of linear equations in two variables. • Determine the possible solutions for linear equations in two variables. • Understand the simultaneous linear equations in two variables. • Solve the simultaneous linear equations in two variables. • Solve problems involving simultaneous linear equations in two variables. Maths History Diophantus of Alexandria was a Greek mathematician who sometimes known as ‘the father of algebra’. He was the author of a series of books called Arithmetica. Arithmetica is a work on the solution of algebraic equations. Linear Equations in Two Variables Concept of Linear Equations in Two Variables Representation of Linear Equations in Two Variables Possible Solutions for Linear Equations in Two Variables Graphical Representation of Simultaneous Linear Equations in Two Variables Solution of Simultaneous Linear Equations in Two Variables Simultaneous Linear Equations in Two Variables • Linear equation • Variable • Solution • Simultaneous linear equations • Graphical method • Substitution method • Elimination method Key Terms Concept Map Linear Equations
62 CHAPTER 3 Linear Equations in Two Variables Flashback 1. Write the relationship between the two quantities below using the symbol ‘=’ or ‘≠’. (a) 4 × 3 1 4 4 (b) 3 – (–11) 16 (c) 1 tonne 100 kg Ticket rates Price per day ($) Price per half day ($) Child (6 to 12 years) 17 14 Youth (13 to 17 years) 37 29 Adult (18 to 64 years) 53 42 Senior (65 years and older) 42 34 Equipment rental Price per 1 day ($) Price for 3 or more days ($) Complete package 30 32 Skis or boards 22 23 Boots 8 9 Poles 7 7 Jason is planning a ski trip for his class. Based on the above information, he wants to calculate the total cost for all 30 students. They are Grade 8 students. (a) What assumptions does Jason have to make during calculating cost? (b) What other things does he need to consider? 1 It is given that the number of boys in the class is 6 more than the number of girls. How do you determine the number of groups that can be formed such that each group has 2 girls? Critical Thinking 2. In the diagram, PQR is a straight line. Find the value of x. (9x – 2)º 2xº (4x + 7)º 40º P Q R
63 Linear Equations in Two Variables CHAPTER 3 3.1 Linear Equations in Two Variables A Recognising linear equations in two variables Objective: To identify linear equations of one variable and two variables. Instruction: Do this activity in pairs. 1. Consider the following situation. Ben has 5 marbles. After John gives him some marbles, he has 9 marbles now. Let the number of marbles given by John is x, the linear equation used to represent the situation is 5 + x = 9. This equation is a linear equation in one variable. 2. Observe the equation formed in Step 1. (a) How many variables are there in the equation? (b) What is the highest power of its variable? 3. Observe the linear equations written on the cards. 7h + 2 = 16 k − 3k = 6 6m + 4n = 9 p + 6 = 10 3x − 1 = 5y Which of the equations are linear equation in one variable? 4. Why are some other linear equations not classified as linear equations of one variable? What type of linear equations are these? 5. Present your findings in class. 1 A linear equation in two variables is an equation which has two variables and the power of each variable is one. For example, u = v, 3x + 4y = 6, p – 2q = 3 – q. Linear equation in two variables can be expressed in general form, ax + by = c where a and b are non-zero.
64 CHAPTER 3 Linear Equations in Two Variables EXAMPLE 1 Determine whether each of the following equations is a linear equation in two variables. (a) 6x – 2y = 7 (b) x – y 2 = 4 (c) p + 4q = 5r (d) v = 1 v + 3 Solution: (a) Yes, because there are two variables, x and y, with the power of 1. (b) No, because the power of variable y is 2. (c) No, because there are three variables. (d) No, because the power of variable v is not 1. EXAMPLE 2 Write an equation in two variables for each of the following. (a) The total mass of an iron rod and 2 steel bars is 40 kg. (b) The selling price of x pens costing $3 each is $10 more than the selling price of y pencils costing 90 cents each. Solution: (a) Let p = mass of an iron rod, in kg q = mass of a steel bar, in kg Therefore, p + 2q = 40. (b) Selling price of x pens = 300x cent Selling price of y pencils = 90y cent Therefore, 300x – 90y = 1000 30x – 9y = 100. Unit for each item is in cent. Variables for linear equations in two variables can be represented by any lowercase letters. B Writing linear equations in two variables from given information When writing a linear equation in two variables, we represent the two variables with two suitable letters (usually small letters), then form the equation according to the given information. Unit for every item in the equation must be the same. Discuss why xy + 2y = 3 and 1 x + 1 y = 5 are not linear equations in two variables. Critical Thinking
65 Linear Equations in Two Variables CHAPTER 3 C Determining the possible solutions for linear equations in two variables A linear equation in two variables may have many possible pairs of solutions. For example, x = –2, y = 10; x = –1, y = 8; x = 0, y = 6 and x = 1, y = 4 are the possible pairs of solutions for 2x + y = 6. EXAMPLE 3 Given that 4x + 3y = 14, (a) find the value of y when x = –1. (b) find the value of x when y = 2. Solution: (a) When x = –1, 4(–1) + 3y = 14 – 4 + 3y = 14 3y = 18 y = 6 (b) When y = 2, 4x + 3(2) = 14 4x + 6 = 14 4x = 8 x = 2 EXAMPLE 4 Determine whether each of the following is the possible pairs of solutions for 3s + 2t = –7. (a) s = – 3, t = 1 (b) s = 2, t = –2 Solution: (a) Substitute s = –3 and t = 1 into 3s + 2t (left hand side). 3(–3) + 2(1) = –9 + 2 = –7 = right hand side Thus, s = –3 and t = 1 is one of the possible pairs of solutions. (b) Substitute s = 2 and t = –2 into 3s + 2t. 3(2) + 2(–2) = 6 – 4 = 2 ≠ right hand side Thus, s = 2 and t = –2 is not one of the possible pairs of solutions. Note that x = –1, y = 6 and x = 2, y = 2 are the two possible pairs of solutions for 4x + 3y = 14.
66 CHAPTER 3 Linear Equations in Two Variables EXAMPLE 5 Determine three pairs of possible solutions for 3x + 2y = 7. Solution: 3x + 2y = 7 y = 7 – 3x 2 When x = –3, y = 7 – 3(–3) 2 = 7 + 9 2 = 8 When x = 0, y = 7 – 3(0) 2 = 7 2 = 3 1 2 When x = 1, y = 7 – 3(1) 2 = 4 2 = 2 Therefore, three pairs of the possible solutions are x = –3 and y = 8, x = 0 and y = 3 1 2 , x = 1 and y = 2. Express y as the subject of the equation. Practice 3.1 Basic Intermediate Advanced 1 Determine whether each of the following equations is a linear equation in two variables. (a) 7p – 2q = 4 (b) 2xy − y = 6 (c) m = 5n 2 − 8 (d) h = 4g − 2 Write a linear equation in two variables for each of the following. (a) The difference in age between Mr Khoo and his youngest sister is 11 years. (b) The price of a pair of trousers is three times the price of a pair of shorts. (c) The average score of two students in a bowling competition is 781. Mr Albert paid $4.80 for p pieces of pizza at a price of 30 cents each and q bottles of water at a price of $1.20 each. Write an equation from the given information. Mrs Kumar employs a driver with a daily wage of $20 to help her deliver packed lunch to some office workers. The cost of each packet of lunch is $1.40 and Mrs Kumar charges the office workers $3 for each packet. Write an equation relating Mrs Kumar’s daily profit, $p, and the number of packets, n, of lunch delivered.
Linear Equations in Two Variables CHAPTER 3 67 Complete each of the following tables. (a) 3x – y = 1 (b) 1 2 m + 3n – 2 = 0 x y 5 4 –10 m n 4 –2 5 Given that the value of a variable in each of the following linear equations, determine the value of the other variable. (a) b = 4a + 7 [a = 2] (b) 2 3 m – 4n = 8 [n = 5] (c) 6p + 5q – 7 = 0 [p = 7] (d) 2.7u + 1.7t = 2 [u = 2] (e) 4 5 x – 3 4 y = 20 [y = –8] Determine whether the values in the brackets is a possible pair of solutions for the given equations. (a) 4x + y = 2 (x = 1, y = –2) (b) 2x + 5y = 10 (x = −2, y = 3) Determine three pairs of possible solutions for each of the following linear equations in two variables. (a) y = 7x + 6 (b) 6t – 3u = 10 (c) 8k – 3h + 5 = 0 (d) 3a + b 4 = 6 Susan’s purse contained x coins and y $5 notes. Given that the total amount of notes in her purse is equal to the total amount of the coins and this situation can be represented by a linear equation in two variables, 5y = 0.50x. (a) What is meant by 0.50 in the equation? (b) If Susan had 8 pieces of $5 notes, calculate the number of coins in her purse. Jane brought a pouch containing 20 pieces of 20-cent coins and 12 pieces of 50-cent coins to buy stationery. She spent $5.20 altogether. (a) Write a linear equation in two variables for the given information. (b) List all the possible ways that Jane could pay with the coins in her pouch. The perimeter of a rectangle with length of p cm and width of q cm is 10 cm. Find the possible values of p and q. Wani and her brother studied at St. Andrew Primary School. Their total age is 15 years. State the possible age of Wani. 3.2 Simultaneous Linear Equations in Two Variables A Understanding simultaneous linear equations in two variables graphically Simultaneous linear equations in two variables are two linear equations in two variables such that the two variables in both equations are the same and the power of each variable is 1. Simultaneous linear equations in two variables can be written as follows, ax + by = e cx + dy = f where x and y are variables, a, b, c and d are coefficients, e and f are constants. Simultaneous linear equations in two variables also known as a system of linear equations in two variables.
68 CHAPTER 3 Linear Equations in Two Variables Objective: To represent the linear equations in two variables graphically. Instruction: Do this activity in groups of four. Material: Grid papers, pencil, ruler 1. Construct tables of values for the following linear equations in two variables. (a) −x + y = 2 (b) x + y = 10 x y (x, y) 0 … (0, …) 2 … (2, …) 4 … (4, …) 6 … (6, …) 8 … (8, …) x y (x, y) 0 … (0, …) 2 … (2, …) 4 … (4, …) 6 … (6, …) 8 … (8, …) 2. Draw graphs on a grid paper based on the tables of values above. 3. What is the shape of the graph of each linear equation in two variables? 4. Are the points on the graph of the linear equation the solutions to the linear equation? 5. What can you conclude on the point of intersection of the graphs of the two linear equations? 6. Present your findings in class. 2 From the findings of Activity 2, we found that each linear equation in two variables is represented by a straight line. All points on the straight line are solutions to the linear equation. So, the point of intersection is a solution for both the linear equations. x = 4 and y = 6 is the solution for the linear equations –x + y = 2 and x + y = 10. Thus, simultaneous linear equations in two variables are represented by two straight lines separately. x-coordinates and y-coordinates of the intersection of the two straight lines are the solution of the simultaneous linear equations. Given that simultaneous linear equations ax + by = e and cx + dy = f. • If a c ≠ b d , there is one unique solution for the simultaneous linear equations. • If a c = b d ≠ e f , there is no solution for the simultaneous linear equations. • If a c = b d = e f , there is infinite many solutions for the simultaneous linear equations. 2 4 6 8 10 –x + y = 2 x + y = 10 (4, 6) 0 x y 2 4 6 8 10
69 Linear Equations in Two Variables CHAPTER 3 Objective: To explore the simultaneous linear equations in two variables graphically. Instruction: Do this activity in pairs. 1. Open the file of Simultaneous Linear Equations using GeoGebra. 2. The following are simultaneous linear equations in two variables. Input each of the linear equations. x + y = 4 2x + y = 5 3. Observe the straight lines displayed. Determine whether the straight lines are intersecting, parallel or overlapping. 4. How many points of intersection are there? 5. What can you conclude on the type of solution of the simultaneous linear equations? 6. Repeat Steps 1 to 5 for each of the following simultaneous linear equations in two variables. (a) 2x − 3y = –3 (b) x + y = 3 4x − 6y = 12 5x + 5y = 15 7. Present your findings in class. 3 Scan or click the above QR code to download this activity file. GeoGebra From the findings of Activity 3, we found that there are three cases involving solution of simultaneous linear equations as shown in the table below. x + y = 4 2x + y = 5 y x –1 1 2 3 4 0 1 2 3 4 2x – 3y = –3 4x – 6y = 12 y x –3 –2 –1 1 2 0 –2 –1 1 2 x + y = 3 5x + 5y = 15 y x –1 1 2 3 4 0 1 2 3 4
70 CHAPTER 3 Linear Equations in Two Variables team work When given a pair of simultaneous equations, how can we determine, without solving, that the pair of simultaneous equations has (a) only one solution? (b) infinitely many solutions? (c) no solution? Simultaneous linear equations in two variables Condition of both straight lines Point of intersection Type of solution x + y = 4 2x + y = 5 Intersecting One Unique solution 2x – 3y = – 9 4x – 6y = 12 Parallel Zero No solution x + y = 3 5x + 5y = 15 Overlapping Infinite number Infinite solutions B Solving simultaneous linear equations in two variables Solving simultaneous linear equations in two variables is to find a pair of values that satisfies both equations. We can use graphical method, substitution method or elimination method as follows. (I) Graphical method 1 Construct a table of values for each equation separately. Draw the two straight lines based on the constructed table of values. Find the x-coordinates and y-coordinates for the point of intersection of the two straight lines. EXAMPLE 6 Solve the following simultaneous linear equations by using graphical method. x + y = 6 and 3y = 2x + 3 Solution: For x + y = 6: For 3y = 2x + 3: x 0 6 x 0 6 y 6 0 y 1 5 From the graph, the point of intersection is (3, 3). Thus, the solution is x = 3 and y = 3. 2 4 6 3y = 2x + 3 x + y = 6 0 x y 2 4 6 8 Two points are needed to draw a straight line graph.
71 Linear Equations in Two Variables CHAPTER 3 EXAMPLE 7 Solve the following simultaneous equations by substitution method. 2x – 3y = 5 2x + 3y = –6 Solution: 2x – 3y = 5 ........................ 1 2x + 3y = –6 ...................... 2 From 1, x = 5 + 3y 2 .............. 3 Substitute 3 into 2, 2 5 + 3y 2 + 3y = –6 5 + 3y + 3y = –6 6y = –6 – 5 y = – 11 6 Substitute y = – 11 6 into 3, x = 5 + 1 3 – 11 6 2 2 = 5 – 11 2 ÷ 2 = 10 – 11 2 × 1 2 = – 1 4 Therefore, the solution is x = – 1 4 and y = – 11 6 . Label the equations. Express one variable as the subject of the equation. (II) Substitution method 1 Choose one of the equations and express one variable as the subject of the equation. 2 Substitute the expression for the variable (the subject) into the other equation. 3 Solve the linear equation in one variable to obtain the value of the variable. 4 Substitute the value obtained into the equation in 1 to obtain the value of the other variable. (III) Elimination method 1 Choose a variable to be eliminated. Ensure that its coefficient in both equations are numerically equal. 2 Eliminate the variable by subtracting or adding the two equations to obtain a linear equation in one variable. 3 Solve the linear equation in one variable. Then, substitute the value into one of the original equations to obtain the value of the other variable.
72 CHAPTER 3 Linear Equations in Two Variables EXAMPLE 8 Solve the following simultaneous equations by elimination method. 2x + 3y = 1 3x + 2y = 4 Solution: 2x + 3y = 1 ........... 1 3x + 2y = 4 ........... 2 1 × 2: 4x + 6y = 2 ........... 3 2 × 3: 9x + 6y = 12 .......... 4 Eliminate the variable y by subtracting equation 3 from 4. 4 – 3: 5x = 10 x = 10 5 = 2 Substitute x = 2 into 1, 2(2) + 3y = 1 3y = 1 – 4 y = –3 3 = –1 Therefore, the solution is x = 2 and y = –1. Label the equations. If the coefficients of the variable to be eliminated have • the same sign, subtract the two equations. • different signs, add the two equations. C Solving problems We can use the following steps to solve problems involving simultaneous linear equations in two variables. 1 Identify the variable quantities and represent them with any suitable letters. 2 Based on the given information, form the two equations in terms of the letters chosen. 3 Solve the simultaneous equations. 4 Interpret the result by relating the answer back to the problem. EXAMPLE 9 6 adults and 4 senior citizens have to pay $228 while 13 adults and 7 senior citizens have to pay $459 to visit an exhibition at the Art Science Museum. Find the total amount 2 adults and a senior citizen have to pay to visit the exhibition. • Multiply 1 by 2 and 2 by 3 to make the coefficients of y numerically equal. • 6 is the LCM for 3 and 2.
73 Linear Equations in Two Variables CHAPTER 3 Solution: Stage 1: Understand the problem List the facts and the question. Facts: Total cost of tickets for 6 adults and 4 senior citizens = $228 Total cost of tickets for 13 adults and 7 senior citizens = $459 Question: Find the cost of tickets for 2 adults and a senior citizen. Stage 2: Think of a plan • Let the cost of the ticket for an adult = $x • Let the cost of the ticket for a senior citizen = $y • Form two simultaneous linear equations and solve them. Stage 3: Carry out the plan 6x + 4y = 228......... 1 13x + 7y = 459......... 2 1 × 7: 42x + 28y = 1596......... 3 2 × 4: 52x + 28y = 1836......... 4 4 – 3: 10x = 240 x = 24 Substitute x = 24 into the equation 1, 6(24) + 4y = 228 4y = 84 y = 21 Thus, the cost of the ticket for an adult is $24 and the cost of the ticket for a senior citizen is $21. The cost of the tickets for 2 adults and a senior citizen = $2x + $y = $2(24) + $21 = $69 Stage 4: Look back Work backwards to check. 69 = 48 + y y = 21
74 CHAPTER 3 Linear Equations in Two Variables EXAMPLE 10 The difference between two numbers is 31. If 5 is added to the smaller number, the result is equal to half of the larger number. What is the sum of the two numbers? Solution: Stage 1: Understand the problem List the facts and the question. Facts: The difference between two numbers is 31. 5 is added to the smaller number, the result is equal to half of the larger number. Question: What is the sum of the two numbers? Stage 2: Think of a plan • Let the larger number = x • Let the smaller number = y • The difference between the two numbers is 31. Stage 3: Carry out the plan The difference between the two numbers is 31. x – y = 31 ......... 1 5 added to the smaller number is equal to 1 2 x. y + 5 = 1 2 x ......... 2 2 × 2: 2y + 10 = x x – 2y = 10 ......... 3 1 – 3: –y – (–2y) = 31 – 10 y = 21 Substitute y = 21 into 1, x – 21 = 31 x = 31 + 21 = 52 Sum of the two numbers = x + y = 52 + 21 = 73
75 Linear Equations in Two Variables CHAPTER 3 A pet shop has a total of 37 cats and birds. It is given that the total number of legs of the cats and birds is 92. What is the number of birds in the shop? EXAMPLE 11 2x m (15 – y) m (x + 3) m 2y m P S Q R The diagram shows a rectangular land. Calculate the area, in m2 , of the land. Solution: Stage 1: Understand the problem List the facts and the question. Facts: Length of PS = 2x m Length of QR = (15 – y) m Breadth of PQ = (x + 3) m Breadth of SR = 2y m Question: Find the area of a rectangular land PQRS. Stage 2: Think of a plan • Area of rectangle = Length × Breadth 2y = x + 3 …… 1 15 – y = 2x y = 15 – 2x …… 2 SR = PQ QR = PS Stage 4: Look back Work backwards to check. Find the value of y. 73 = 52 + y y = 73 – 52 = 21
76 CHAPTER 3 Linear Equations in Two Variables Stage 3: Carry out the plan 2y = x + 3 …… 1 15 – y = 2x y = 15 – 2x …… 2 Substitute 2 into 1: 2(15 – 2x) = x + 3 30 – 4x = x + 3 27 = 5x x = 5.4 Substitute x = 5.4 into 2: y = 15 – 2(5.4) = 15 – 10.8 = 4.2 m Length of the land, PS = 2 × 5.4 = 10.8 m Width of the land, RS = 2 × 4.2 = 8.4 m Thus, the area of the land = 10.8 × 8.4 = 90.72 m2 Stage 4: Look back Work backwards to check. 90.72 m2 ÷ 10.8 m = 8.4 m SR = PQ QR = PS When Hazura added three consecutive odd numbers, the result was 87. What is the first odd number? Practice 3.2 Basic Intermediate Advanced 1 Solve each of the following simultaneous linear equations by using graphical method. (a) y = 8 − x and y = 2x − 1 (b) y = x + 3 and y = –2x + 9 (c) y = 2x − 6 and y = −x + 6 (d) y = 9 − 2x and y − x = 6 Solve each of the following simultaneous linear equations by using substitution method. (a) −p + 3q = 5 and 6p + 4q = 14 (b) 4m − 7n = 26 and 5m + n = 13 (c) 3u + 7v = 13 and 5u + 2v = 12 (d) 5x − 9y = 17 and 3x − 8y = 5 Solve each of the following simultaneous linear equations by using elimination method. (a) 6a − 2b = 26 and 5a + 2b = 18 (b) 7p + 3q = 1 and 4p − 3q = −23 (c) 3h + 5k = 1 and 4h + 7k = 2 (d) 3x − 8y = 27 and 5x − 9y = 32 Ai Ling’s house number is a two-digit number. The sum of the digits is 11 and its difference is 3. What is Ai Ling’s house number if the number is greater than 50?
Linear Equations in Two Variables CHAPTER 3 77 Section A 1. Which of the following equations is not a linear equation in two variables? A 3x – 2y = 1 B x + 3y – 5 = 0 C y = 1 x + 4 D 2x – y = 3y + 1 2. The prices of a pear and an orange are x cent and y cent respectively. Mr Zainal paid $29.85 for 9 pears and 20 oranges. The equation that relates x and y is A x + y = 29.85 B 9x + 20y = 29.85 C 9x + 20y = 2985 D x + y = 2985 3 56 marbles are divided between two boys such that one-third of one boy’s marbles is one-quarter of the other. How many marbles will each boy receive? 8 buns and 9 muffins cost $9 whereas 6 buns and 11 muffins cost $9.30. What is the cost of each item? Mr Smith bought two adult tickets and two child tickets for a circus show with $50. Mr Steve paid $95 for four adult tickets and three child tickets for the same circus show. How much does Mr Chandra have to pay for five adults and one child? Given that the total age of Anita and her younger sister is 10 years and the difference of their age is 4 years. Find the age of Anita’s sister. (3x – y) cm (2y + x) cm (2x + 1) cm The diagram shows an equilateral triangle. Calculate the perimeter, in cm, of the triangle. (x + 7) cm (4x – y) cm 20y – 6x 10x The diagram shows a rhombus. Calculate (a) the length of one side of the rhombus, (b) the size of the angles. Summary Summary Summary Linear Equations Linear equations in two variables A linear equation in two variables is an equation consisting of only a numeral and two linear algebraic terms of different variables. For example: 3x + 2y = 5 y = 3x – 7 Simultaneous linear equations in two variables Simultaneous linear equations in two variables consist of two linear equations in two variables with a common solution that satisfies both equations. For example, x + 3y = 5 and 3x – 2y = 4 are simultaneous linear equations with a common solution x = 2 and y = 1.
CHAPTER 3 Linear Equations in Two Variables 78 3. Given that 4x + 7y = 9, find the value of x when y = 3. A – 3 B 3 C 7 D 7.5 4. Given that 2x + 5y – 4 = 0, if x = 7, y = A –3.6 B –2 C 2 D 3.6 5. Which of the following is a possible solution for the equation 10x + 3y = 15? A x = –3, y = –5 B x = –3, y = 5 C x = 3, y = –5 D x = 3, y = 5 6. Given that 7x – 2y – 8 = 0, find the value of x when y = 0.2. A 8.4 B 4.2 C 4.1 D 1.2 7. Given that x + y = 1 and x – y = 9, then y = A – 5 C 4 B – 4 D 5 8. Given that 5m – 2n = 9 and 4m = 5n + 5.5, find the value of n. A – 2 B – 0.5 C 0.5 D 2 9. Given that x + 3y = 4 and 4x – 5y = 9. The following shows part of the steps in solving the above simultaneous equations. 4x + 12y = 4 4x – 5y = 9 Which term is wrongly written? A 9 B 4 C 4x D 12y 10. The ratio of the number of sides of two regular polygons is 1 : 3 and the sum of all the interior angles of the two polygons is 1440°. Name the polygon with more sides. A Decagon B Nonagon C Octagon D Hexagon Section B 1. (a) Given that x – 4y = –11 and y = –1, calculate the value of x. (b) The perimeter of a rectangle with length 3a cm and breadth 4b cm is 54 cm. Write an equation from the given information. 2. (a) Given that 3m – 5n = 21, calculate the value of n where m = 2. (b) 6 cm (2g – h) cm The area of the triangle is 42 cm2 . Write an equation from the given information. 3. (a) A pair of shoes and a blouse cost $100 and $50 respectively. Sheila bought x pairs of shoes and y blouses. The total amount that she spent was $350. Write an equation from the given information. (b) 14 cm 16 cm (4x – 5y) cm (3x – y) cm P Q S R The diagram shows a rectangle PQRS. Write two linear equations based on the given information.
Linear Equations in Two Variables CHAPTER 3 79 4. Solve the simultaneous equations. 2x – 3y = 9 4x + 3y = 9 Find the values of x and y. 5. Solve the simultaneous equations. 3p + 5q = 3 3q – 5p = 29 Find the values of p and q. 6. Solve the simultaneous equations. 5x – y = 12 y = 2x 7. Solve the simultaneous equations. p – 2q = 4 3p – 4q = 10 8. Solve the simultaneous equations. 3x + 2y = 10 2x + 3y = 5 9. Solve the simultaneous equations. 4p + 3q = 6 3p – 2q = 13 10. (a)Solve the simultaneous equations. 3x – 4y = 18 4x + 3y = –1 (b) What is the value of x – y? 11. The sum of two numbers is 4 and their difference is 10. Find the two numbers. 12. The difference between the length and the breadth of a rectangle is 14 cm. If 5 cm is added to its length, then it becomes twice as long as the breadth. Find the original length and breadth of the rectangle. 13. 4 cm (a + b) cm (a – b) cm The diagram shows a parallelogram.The area of the parallelogram is 76 cm2 and its perimeter is 48 cm. Find the values of a and b. 14. The average age of Mrs Heng and her daughter is 30 years. The age of Mrs Heng is three times the age of her daughter. In how many years’ time will she be twice as old as her daughter? 15. 3(x – 2) 3y – 10 2x + y Fiction Children’s books Non-fiction The pie chart shows the types of books borrowed by the members of a library on a certain day. The total number of books borrowed on that day was 1080. If 354 of the books borrowed are non-fiction books, find the values of x and y. 16. The sum of two prime numbers is 64. If 5 is subtracted from each numbers, the bigger number becomes twice the other number. What is the difference between these two prime numbers?
JBRB221242 ISBN 978-981-17293-2-4 FOCUS-ON TEXTBOOK MATHS 8 FOCUS-ON MATHS is a complete mathematics programme specially written in line with the latest Indonesian Mathematics syllabus (Phase D) for Grade 7 to Grade 9 students. The topic coverage in each grade is arranged to address all the learning achievements (Capaian Pembelajaran) as prescribed by the Indonesian Ministry of Education. The series adopts the Singapore Maths method which is a world-class maths teaching approach. This comprehensive series builds on the foundations laid in primary mathematics and prepares learners for embarking on higher-level mathematics. With 21st Century Skills and Higher Order Thinking Skills infused in the contents; this series challenges students with engaging problem-solving tasks in real-word contexts, enabling them to become independent maths learners and build foundations for future success. Focus-on Maths comprises: • Textbook • Workbook • Teacher’s Guide • Teaching Aids