Essential Mathematics Year 10 - Chapter 1 - Algebra, Equations and Linear Relationships PDF
Document Details
Uploaded by Deleted User
2024
Greenwood et al.
Tags
Summary
This textbook chapter covers foundational algebra concepts for Year 10 students in Australia. It includes key terms like expressions, equations, and coefficients. The material also reviews how to solve and manipulate algebraic expressions.
Full Transcript
1 Algebra, equations and linear relationships Maths in context: Computer software engineers and financial analysts Computer software engineers apply algebraic skills include renewable and traditional power generation, when coding. Th...
1 Algebra, equations and linear relationships Maths in context: Computer software engineers and financial analysts Computer software engineers apply algebraic skills include renewable and traditional power generation, when coding. There are many career opportunities for transmission lines, power storage, and staff wages. code development in the emerging technologies of A mathematical process called linear programming Artificial Intelligence, machine learning, cybersecurity, is used to determine the requirements for maximum cloud computing, and the automation of robots. profit. This involves graphing straight lines for each cost constraint, creating an area where all costs are Financial Analysts apply a knowledge of linear met. The profit equation is graphed over this feasible relations to investigate the conditions needed for a region to determine the maximum potential profit. company’s maximum potential profit. For example, Applying the maths of linear programming can save to make a profit, an energy company must reduce a large company many millions of dollars. its costs compared to its revenue. Typical costs Essential Mathematics for the Australian Curriculum ISBN 978-1-009-37313-5 © Greenwood et al. 2024 Cambridge University Press Year 10 Photocopying is restricted under law and this material must not be transferred to another party. Chapter contents 1A Review of algebra (CONSOLIDATING) 1B Solving linear equations 1C Linear inequalities 1D Linear equations involving more complex algebraic fractions (OPTIONAL) 1E Graphing straight lines (CONSOLIDATING) 1F Finding the equation of a line 1G Length and midpoint of a line segment 1H Parallel and perpendicular lines 1I Simultaneous equations using substitution 1J Simultaneous equations using elimination 1K Further applications of simultaneous equations 1L Regions on the Cartesian plane Australian Curriculum 9.0 ALGEBRA Solve linear inequalities and simultaneous linear equations in 2 variables; interpret solutions graphically and communicate solutions in terms of the situation (AC9M10A02) Use mathematical modelling to solve applied problems involving growth and decay, including financial contexts; formulate problems, choosing to apply linear, quadratic or exponential models; interpret solutions in terms of the situation; evaluate and modify models as necessary and report assumptions, methods and findings (AC9M10A04) Simplification of combinations of linear expressions with rational coefficients and the solution of related equations (Year 10 optional content) © ACARA Online resources A host of additional online resources are included as part of your Interactive Textbook, including HOTmaths content, video demonstrations of all worked examples, auto-marked quizzes and much more. Essential Mathematics for the Australian Curriculum ISBN 978-1-009-37313-5 © Greenwood et al. 2024 Cambridge University Press Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 4 Chapter 1 Algebra, equations and linear relationships 1A Review of algebra CONSOLIDATING LEARNING INTENTIONS To review the key words of algebra: term, coefficient, expression, equation To review how to combine like terms under addition and subtraction To review how to multiply and divide algebraic terms and apply the distributive law to expand brackets To review how to factorise an expression using the highest common factor To be able to substitute values for pronumerals and evaluate expressions Algebra involves the use of pronumerals (or variables), which are letters representing numbers. Combinations of numbers and pronumerals form terms (numbers and pronumerals connected by multiplication and division), expressions (a term or terms connected by addition and subtraction) and equations (mathematical statements that include an equals sign). Skills in algebra are important when dealing with the precise and concise nature of mathematics. The complex nature of many problems in finance and engineering usually result in algebraic Stockmarket traders rely on financial modelling based on expressions and equations that need to be simplified complex algebraic expressions. Financial market analysts and computer systems analysts require advanced and solved. algebraic skills. Lesson starter: Mystery problem Between one school day and the next, the following problem appeared on a student noticeboard. 3x − 9 + 5(x − 1) −x(6 − x) = 0. Prove that 8 − x 2 + _ 3 By working with the left-hand side of the equation, show that this equation is true for any value of x. At each step of your working, discuss what algebraic processes you have used. KEY IDEAS Key words in algebra: 2a , 7 (a constant term) term: 5x, 7x 2y, _ 3 coefficient: −3 is the coefficient of x 2 in 7 − 3x 2; 1 is the coefficient of y in y + 7x. _ expression: 7x, 3x + 2xy, _ x + 3 , √ 2a 2 − b 2 equation: x = 5, 7x − 1 = 2, x 2 + 2x = −4 Expressions can be evaluated by substituting a value for each pronumeral (variable). Order of operations are followed: First brackets, then indices, then multiplication and division, then addition and subtraction, working then from left to right. Essential Mathematics for the Australian Curriculum ISBN 978-1-009-37313-5 © Greenwood et al. 2024 Cambridge University Press Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 1A Review of algebra 5 Like terms have the same pronumeral part and, using addition and subtraction, can be collected to form a single term. For example, 3x − 7x + x = − 3x 2b − b a 6 a 2= 5 a 2b b = b a Note that a2 2 The symbols for multiplication ( ×)and division (÷)are usually not shown. 7x 7 × x ÷ y = _ y 2b ÷ (ab) = _ − 6 a − 6 a 2b ab = − 6a The distributive law is used to expand brackets. a(b + c) = ab + ac 2(x + 7) = 2x + 14 a(b − c) = ab − ac − x(3 − x) = − 3x + x 2 Factorisation involves writing expressions as a product of factors. Many expressions can be factorised by taking out the highest common factor (HCF). 15 = 3 × 5 3x − 12 = 3(x − 4) 2y − 6xy + 3x = 3x(3xy − 2y + 1) 9 x Other general properties are: associative a × (b × c) = (a × b) × c a + (b + c) = (a + b) + c or ( ) commutative ab = ba or a + b = b + a Note : _a ≠ _ b and a − b ≠ b − a b a a × 1 = a or a + 0 = a identity inverse a×_ 1 = 1 or a + (− a) = 0 a BUILDING UNDERSTANDING 1 Which of the following is an equation? A 3x − 1 B _ x + 1 C 7x + 2 = 5 D 3 x 2y 4 2 Which expression contains a term with a coefficient of − 9? A 8 + 9x B 2x + 9 x 2y C 9a − 2ab D b − 9 a 2 3 State the coefficient of a 2in these expressions. 3 − 4 a 2 2 a a + a 2 b _ 2 c 1 − _a d − _ − 1 7 a 2 5 3 4 Decide whether the following pairs of terms are like terms. a xyand 2yx b 7 a 2band − 7b a 2 c − 4ab c 2and 8a b 2c 5 Evaluate: a (− 3)2 2)3 b (− 3 c −2 d −3 2 Essential Mathematics for the Australian Curriculum ISBN 978-1-009-37313-5 © Greenwood et al. 2024 Cambridge University Press Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 6 Chapter 1 Algebra, equations and linear relationships Example 1 Collecting like terms Simplify by collecting like terms. a 7a + 3a 2b − 2 a b 3 a 2b c 5xy + 2x y 2− 2xy + 3 y 2x SOLUTION EXPLANATION a 7a + 3a = 10a Keep the pronumeral and add the coefficients. 2b − 2 a b 3 a 2b = a 2b 2band 2 a 3 a 2bhave the same pronumeral part, so they are like terms. Subtract coefficients and recall that 1 a 2b = a 2b. 2− 2xy + 3 y c 5xy + 2x y 2x = 3xy + 5x y 2 2x = 3x y Collect like terms, noting that 3 y 2. The +or −sign belongs to the term that directly follows it. Now you try Simplify by collecting like terms. a 4a + 13a 2− 2a b b 5a b 2 c 3xy + 4 x 2y − xy + 2y x 2 Example 2 Multiplying and dividing expressions Simplify the following. 7xy a 2h × 7l b − 3pr × 2p c − _ 14y SOLUTION EXPLANATION a 2h × 7l = 14hl Multiply the numbers and remove the × sign. b − 3pr × 2p = − 6 p 2r Multiply the numbers and recall that p × p is 2. written as p 7xy c − _ x = − _ Cancel the highest common factor of 7and 14y 2 14and cancel the y. Now you try Simplify the following. a 3a × 6b b − 2xy × 5x 4ab c − _ 8a Essential Mathematics for the Australian Curriculum ISBN 978-1-009-37313-5 © Greenwood et al. 2024 Cambridge University Press Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 1A Review of algebra 7 Example 3 Expanding brackets Expand the following using the distributive law. Simplify where possible. a 2(x + 4) b − 3x(x − y) c 3(x + 2) − 4(2x − 4) SOLUTION EXPLANATION a 2(x + 4) = 2x + 8 2(x + 4) = 2 × x + 2 × 4 b − 3x(x − y) = − 3 x 2+ 3xy Note that x × x = x 2and − 3 × (− 1) = 3. c 3(x + 2) − 4(2x − 4) = 3x + 6 − 8x + 16 Expand each pair of brackets and simplify = − 5x + 22 by collecting like terms. Now you try Expand the following using the distributive law. Simplify where possible. a 3(x + 2) b − 2x(x − y) c 2(x + 3) − 3(2x − 1) Example 4 Factorising simple algebraic expressions Factorise: a 3x − 9 2+ 4x b 2 x SOLUTION EXPLANATION a 3x − 9 = 3(x − 3) HCF of 3 xand 9 is 3. Check that 3(x − 3) = 3x − 9. 2+ 4x = 2x(x + 2) b 2 x 2and 4xis 2x. HCF of 2 x Check that 2x(x + 2) = 2x2 + 4x. Now you try Factorise: a 2x − 10 2+ 9x b 3 x Essential Mathematics for the Australian Curriculum ISBN 978-1-009-37313-5 © Greenwood et al. 2024 Cambridge University Press Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 8 Chapter 1 Algebra, equations and linear relationships Example 5 Evaluating expressions 2− 2bcif a = − 3, b = 5and c = − 1. Evaluate a SOLUTION EXPLANATION 2− 2bc = (− a 3)2 − 2(5) (− 1) Substitute for each pronumeral: = 9 − (− 10) (−3)2= (−3)× (−3)and 2 × 5 × (− 1) = − 10 = 19 To subtract a negative number, add its opposite. Now you try 2− 3acif a = 1, b = − 2and c = − 3. Evaluate b Exercise 1A FLUENCY 1−6(1/2) 1−6( 1/2) 1−6( 1/3) 1 Simplify by collecting like terms. Example 1a a 6a + 4a b 8d + 7d c 5y − 5y d 2xy + 3xy e 9 ab − 5ab f 4 t + 3t + 2t Example 1b 2b − 2 a g 4 a 2b h 2y − 4 x 5 x 2y i 3s t2− 4s t2 n − 7n m j 4 m 2 2 k 0.3 a b − b a 2 2 l 0.2a b 2− 2 b 2a Example 1c m 4gh + 5 − 2gh n 7xy + 5xy − 3y o 4a + 5b − a + 2b p 3jk − 4j + 5jk − 3j 2+ 5 a q 2a b 2b − a b 2+ 5b a 2 r 3mn − 7 m 2n + 6n m 2− mn 2 Simplify the following. Example 2a, b a 4a × 3b b 5 a × 5b c − 2a × 3d d 5 h × (− 2m) e − 6h × (− 5t) f − 5b × (− 6l) 2× 6t g 2 s h − 3 b 5 2× 7 d i 4ab × 2a j − 6p × (− 4pq) k 3ab × (− 5b) l 7mp × 9mr Example 2c m _7x 6ab n _ o − _3a 2ab p − _ 7 2 9 8 q _4ab 15xy 4xy _ t − 28ab 2a r − _ s − _ 56b 5y 8x Example 3a, b 3 Expand the following, using the distributive law. a 5(x + 1) b 2(x + 4) c 3 (x − 5) d − 5(4 + b) e − 2(y − 3) f − 7(a + c) g − 6(− m − 3) h 4(m − 3n + 5) i − 2(p − 3q − 2) j 2x(x + 5) k 6a(a − 4) l − 4x(3x − 4y) m 3y(5y + z − 8) n 9g(4 − 2g − 5h) o − 2a(4b − 7a + 10) p 7y(2y − 2 y − 4) 2 q − 3a(2 a 2− a − 1) r − t(5 t3+ 6 t2+ 2) s 2m(3 m3 − m + 5m) 2 t − x(1 − x 3) u − 3s(2t − s 3) Essential Mathematics for the Australian Curriculum ISBN 978-1-009-37313-5 © Greenwood et al. 2024 Cambridge University Press Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 1A Review of algebra 9 Example 3c 4 Expand and simplify the following, using the distributive law. a 2(x + 4) + 3(x + 5) b 4(a + 2) + 6(a + 3) c 6(3y + 2) + 3(y − 3) d 3(2m + 3) + 3(3m − 1) e 2(2 + 6b) − 3(4b − 2) f 3(2t + 3) − 5(2 − t) g 2x(x + 4) + x(x + 7) h 4(6z − 4) − 3(3z − 3) Example 4 5 Factorise: a 3x − 9 b 4 x − 8 c 1 0y + 20 d 6y + 30 e 2+ 7x x f 2+ 8a 2 a 2− 5x g 5 x h 2− 63y 9 y i xy − x y 2 2y − 4 x j x 2y 2 k 8 a 2b + 40 a 2 l 2b + ab 7 a m − 5 t − 5t 2 n − 6mn − 18m n 2 o −y 2− 8yz Example 5 6 Evaluate these expressions if a = − 4, b = 3and c = − 5. a − 2 a 2 b b − a c abc + 1 d − ab _ √a 2 2 + b a + b 3b − a − b 2 2 _ e _ f _ g a _ h _ 2 5 c √c 2 PROBLEM–SOLVING 7 7, 8 8 7 Find an expression for the area of a floor of a rectangular room with the following side lengths. Expand and simplify your answer. a x + 3and 2x b xand x − 5 8 Find expressions in simplest form for the perimeter (P)and area (A)of these shapes. (Note: All angles are right angles.) a x b c 3 1 x−3 x−1 x−2 x x x−4 2 4 Essential Mathematics for the Australian Curriculum ISBN 978-1-009-37313-5 © Greenwood et al. 2024 Cambridge University Press Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 10 Chapter 1 Algebra, equations and linear relationships REASONING 9 9, 10 10, 11 9 When a = −2 give reasons why: a a2 > 0 b −a 2 < 0 c a3 < 0 10 Decide whether the following are true or false for all values of a and b. If false, give an example to show that it is false. a a+b=b+a b a−b=b−a c ab = ba d _a=_ b b a e a + (b + c) = (a + b) + c f a − (b − c) = (a − b) − c g a × (b × c) = (a × b) × c h a ÷ (b ÷ c) = (a ÷ b) ÷ c 11 a Write an expression for the statement ‘the sum of x and y divided by 2’. b Explain why the statement above is ambiguous. c Write an unambiguous statement describing _ a + b. 2 ENRICHMENT: Algebraic circular spaces − − 12 12 Find expressions in simplest form for the perimeter (P) and area (A) of these shapes. Your answers may contain π, for example 4π. Do not use decimals. a x+1 b c x x x 2x − 3 Architects, builders, carpenters and landscapers are among the many occupations that use algebraic formulas to calculate areas and perimeters in daily work. Essential Mathematics for the Australian Curriculum ISBN 978-1-009-37313-5 © Greenwood et al. 2024 Cambridge University Press Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 1B Solving linear equations 11 1B Solving linear equations LEARNING INTENTIONS To know the form of a linear equation To understand that an equivalent equation can be generated by applying the same operation to each side of the equation To be able to solve a linear equation involving two or more steps, including brackets and variables on both sides To be able to solve linear equations involving algebraic fractions To be able to combine simple algebraic fractions under addition or subtraction To understand that solutions can be checked by substituting into both sides of an equation A linear equation is a statement that contains an equals sign and includes constants and pronumerals with a power of 1 only. Here are some examples: 2x − 5 = 7 _x+1 = x+4 3 −3(x + 2) = _1 2 We solve linear equations by operating on both sides of the equation until a solution is found. A small business, such as a garden nursery, generates revenue from its sales. To calculate the number of employees (x) a business can afford, a linear revenue equation is solved for x: Revenue (y) = pay (m) × employees (x) Lesson starter: What’s the best method? + costs (c) Here are four linear equations. Discuss what you think is the best method to solve them using ‘by hand’ techniques. Discuss how it might be possible to check that a solution is correct. a _7x − 2 = 4 b 3(x − 1) = 6 c 4x + 1 = x − 2 3 KEY IDEAS An equation is true for the given values of the pronumerals when the left-hand side equals the right-hand side. 2x − 4 = 6 is true when x = 5 but false when x ≠ 5. A linear equation contains pronumerals with a highest power of 1. Useful steps in solving linear equations are: using inverse operations (backtracking) collecting like terms expanding brackets multiplying by the denominator. Algebraic fractions can be added or subtracted by first finding the lowest common denominator (LCD) and then combining the numerators. Essential Mathematics for the Australian Curriculum ISBN 978-1-009-37313-5 © Greenwood et al. 2024 Cambridge University Press Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 12 Chapter 1 Algebra, equations and linear relationships BUILDING UNDERSTANDING 1 Decide whether the following are linear equations. _ a x 2− 1 = 0 b √ x + x = 3 x − 1 = 5 c _ 3x = 2x − 1 d _ 2 4 2 Decide whether these equations are true when x = 2. a 3x − 1 = 5 b 4 − x = 1 2x + 1 = x + 4 c _ 5 3 Decide whether these equations are true when x = − 6. a − 3x + 17 = x b 2(4 − x) = 20 2 − 3x = _ c _ − 12 10 x 4 Solve the following equations and check your solution using substitution. a x + 8 = 13 b x − 5 = 3 c − x + 4 = 7 d − x − 5 = − 9 5 Simplify the following by firstly finding the lowest common denominator (LCD). 1 + _ a _ 1 4 − _ b _ 1 3 − _ c _ 1 5 + _ d _ 7 2 3 3 5 7 14 3 6 Example 6 Solving linear equations Solve the following equations and check your solution using substitution. a 4x + 5 = 17 b 3(2x + 5) = 4x SOLUTION EXPLANATION a 4x + 5 = 17 Subtract 5from both sides and then divide both 4x = 12 sides by 4. x = 3 Check: LHS = 4 × 3 + 5 = 17, RHS = 17 Check by seeing if x = 3makes the equation true. b 3(2x + 5) = 4x Expand the brackets. 6x + 15 = 4x Gather like terms by subtracting 4xfrom both 2x + 15 = 0 sides. 2x = − 15 Subtract 15from both sides and then divide x = −_ 15 both sides by 2. 2 Check: 15 makes the Check by seeing if x = − _ 2 ( ( 2 ) ) LHS = 3 2 × − _15 + 5 equation true by substituting into the equation’s left-hand side (LHS) and right-hand = − 30 side (RHS) and confirming they are equal. ( 2 ) 15 RHS = 4 × − _ = − 30 Now you try Solve the following equations and check your solution using substitution. a 2x + 7 = 13 b 4(2x + 1) = 2x Essential Mathematics for the Australian Curriculum ISBN 978-1-009-37313-5 © Greenwood et al. 2024 Cambridge University Press Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 1B Solving linear equations 13 Example 7 Solving equations involving fractions Solve the following equations and check your solution using substitution. x + 3 = 2 a _ b 4 − _x = 7 5 − x = x − 3 c _ 4 2 4 SOLUTION EXPLANATION x + 3 = 2 a _ Multiply both sides by 4, since all of x + 3is divided 4 by 4. x+3=8 Subtract 3from both sides. x=5 Check: LHS = _5 + 3 = 2, RHS = 2 Check by seeing if x = 5makes the equation true. 4 b 4 − _x = 7 Subtract 4from both sides of the equation. The 2 minus sign stays with _x . Multiply both sides of the − x = 3 _ 2 2 x = − 6 equation by − 2. (− 6) Check by substituting x = − 6into the left-hand side. Check: LHS = 4 − _= 4 + 3 = 7 2 RHS = 7 _5−x Multiply both sides of the equation by 4 and include c 4 = x − 3 brackets. 5 − x = 4(x − 3) Expand the brackets and gather like terms by adding − x = 4x − 12 5 xto both sides. Add 12to both sides and then divide 5 = 5x − 12 ∴ 5x = 17 both sides by 5. x =_ 17 5 5−_ 17 5 Check: LHS = _ 8 ÷ 4 = _ = _ 2 Check that LHS = RHS using substitution. 4 5 5 17 − 3 = _2 HS = _ R 5 5 Now you try Solve the following equations and check your solution using substitution. x + 1 = 4 a _ 3 b 2 − _x = 6 5 c _ 4 − x = x + 2 3 Essential Mathematics for the Australian Curriculum ISBN 978-1-009-37313-5 © Greenwood et al. 2024 Cambridge University Press Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 14 Chapter 1 Algebra, equations and linear relationships Example 8 Solving equations by combining simple algebraic fractions Consider these algebraic fractions. a Simplify _ 2a a+_ 2a = 2 a+_ 6 9 b Hence, solve _ 6 9 SOLUTION EXPLANATION a+_ _ 3a + _ 2a = _ 4a The LCD of 6 and 9 is 18. Express each a 6 9 18 18 fraction as an equivalent fraction with a 7a = _ denominator of 18. Then add the numerators. 18 a+_ b From part a, _ 2a = _ 7a Combine the fractions on the left-hand side 6 9 18 using the result from part a. Then multiply Solve 7a = 2 _ both sides by 18 and divide both sides by 7. 18 7a = 36 ∴a = _36 7 Now you try Consider these algebraic fractions. a Simplify _3a + _ a 3a + _ a=2 8 6 b Hence, solve _ 8 6 Using calculators to solve equations 5x − 4 = 12. Solve the equation _ 3 Using the TI-Nspire: Using the ClassPad: In a Calculator page use menu >Algebra>Solve and Tap solve(, then and type the equation as shown. type the equation as shown. Hint: use the fraction template ( ctrl ÷ ) Essential Mathematics for the Australian Curriculum ISBN 978-1-009-37313-5 © Greenwood et al. 2024 Cambridge University Press Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 1B Solving linear equations 15 Exercise 1B FLUENCY 1−3( 1/2) , 5(1/2) 1−5(1/3) 1−5( 1/3) Example 6a 1 Solve the following equations and check your solution using substitution. a 2x + 9 = 15 b 4x + 3 = 15 c 3 x − 3 = − 4 d 6x + 5 = − 6 e − 3x + 5 = 17 f − 2x + 7 = 4 g − 4x − 9 = 9 h − 3x − 7 = − 3 i 8 − x = 10 j 5 − x = − 2 k 6 − 5x = 16 l 4 − 9x = − 7 Example 6b 2 Solve the following equations and check your solution using substitution. a 4(x + 3) = 16 b 2(x − 3) = 12 c 2 (x − 4) = 15 d 3(1 − 2x) = 8 e 3(2x + 3) = − 5x f 2(4x − 5) = − 7x g 3(2x + 3) + 2(x + 4) = 25 h 2(2x − 3) + 3(4x − 1) = 23 i 2(3x − 2) − 3(x + 1) = 5 j 5(2x + 1) − 3(x − 3) = 63 k 5(x − 3) = 4(x − 6) l 4(2x + 5) = 3(x + 15) m 5(x + 2) = 3(2x − 3) n 3(4x − 1) = 7(2x − 7) o 7(2 − x) = 8 − x Example 7a, b 3 Solve the following equations and check your solution using substitution. x − 4 = 3 a _ x + 2 = 5 b _ x + 4 = − 6 c _ 2 3 3 d _2x + 7 = 5 2x + 1 = − 3 e _ f 3x − 2 = 4 _ 3 3 4 _x g − 5 = 3 3x 2x − 2 = − 8 2 _ h + 2 = 8 i _ 2 3 j 5 − _x = 1 2x = 0 k 4 − _ l 4x = 9 5 − _ 2 3 7 Example 7c 4 Solve the following equations. 2 − x = x + 1 a _ 3 − x = x − 1 b _ x + 2 = 2 − x c _ 3 4 5 d _x − 3 − 2 = − 6 e _x + 1 + 2 = 9 x f _ − 3 − 4 = 2 5 3 2 g 4 + _x − 5 = − 3 2 − x = 2 h 1 − _ 1 − x = − 1 i 5 − _ 2 3 2 5 For each of the following statements, write an equation and solve it to find x. a When 3is added to x , the result is 7. b When xis added to 8, the result is 5. c When 4is subtracted from x , the result is 5. d When xis subtracted from 1 5, the result is 22. e Twice the value of xis added to 5and the result is 13. f 5less than xwhen doubled is − 15. g When 8is added to 3 times x, the result is 23. h 5less than twice x is 3less than x. Essential Mathematics for the Australian Curriculum ISBN 978-1-009-37313-5 © Greenwood et al. 2024 Cambridge University Press Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 16 Chapter 1 Algebra, equations and linear relationships PROBLEM–SOLVING 6−9 6( 1/2), 7, 9, 10 6−7( 1/2), 10−12 Example 8 6 Consider the following algebraic fractions. a i Simplify _ x + _x x + _ ii Hence, solve _ x = 7 4 3 4 3 b i Simplify x + _ _ 2x 2x = 2 x + _ ii Hence, solve _ 6 5 6 5 c i Simplify _− _3x x ii Hence, solve _3x − _ x = 5 8 4 8 4 d Solve: 2x = 7 i _x + _ 2 3 3x − _ ii _ 2x = 1 5 3 2x _ _x iii − = 3 5 4 7 Substitute the given values and then solve for the unknown in each of the following common formulas. a v = u + at Solve for agiven v = 6, u = 2and t = 4. 1 b s = ut + _a t2 Solve for u given s = 20, t = 2and a = 4. 2 ( 2 ) c A = h + b Solve for bgiven A = 10, h = 4and a = 3. _a ( 100 ) d A = P 1 + _ r Solve for r given A = 1000and P = 800. 8 A service technician charges $30up front and $ 46 for each hour that she works. a What will a 4-hour job cost? b If the technician works on a job for 2 days and averages 6hours per day, what will be the overall cost? c Find how many hours the technician worked if the cost is: i $76 ii $513 iii $1000(round to the nearest half hour). 9 The perimeter of a square is 68 cm. Use an equation to determine its side length. 10 The sum of two consecutive numbers is 35. What are the numbers? 11 I ride four times faster than I jog. If a trip took me 45minutes and I spent 15of these minutes jogging 3 km, use an equation to determine how far I rode. Essential Mathematics for the Australian Curriculum ISBN 978-1-009-37313-5 © Greenwood et al. 2024 Cambridge University Press Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 1B Solving linear equations 17 12 The capacity of a petrol tank is 80 litres. If it initially contains 5litres and a petrol pump fills it at 3 litres per 1 0 seconds, find: a the amount of fuel in the tank after 2 minutes b how long it will take to fill the tank to 32 litres c how long it will take to fill the tank. REASONING 13 13 14 13 Solve 2(x − 5) = 8using the following two methods and then decide which method you prefer. Give a reason. a Method 1: First expand the brackets. b Method 2: First divide both sides by 2. 14 A family of equations can be represented using other pronumerals (sometimes called parameters). For 4 + a . example, the solution to the family of equations 2x − a = 4is x = _ 2 Find the solution for x in these equation families. a x + a = 5 b 6x + 2a = 3a c ax + 2 = 7 d ax − 1 = 2a e _ax − 1 = a f ax + b = c 3 ENRICHMENT: Equations with more than one pronumeral − − 15, 16 15 Make athe subject in these equations. a a(b + 1) = c b ab + a = b 1 + b = c c _ a a = 1 d a − _ 1 + _ e _ 1 = 0 1 + _ f _ 1 = _ 1 b a b a b c 16 Solve for x in terms of the other pronumerals. a _x − _x = a b _x + _x = 1 c _x − _ x = c 2 3 a b a b Essential Mathematics for the Australian Curriculum ISBN 978-1-009-37313-5 © Greenwood et al. 2024 Cambridge University Press Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 18 Chapter 1 Algebra, equations and linear relationships 1C Linear inequalities LEARNING INTENTIONS To know the meaning of the term inequality To be able to use and interpret the symbols >, ⩾, ⩽, < To know how to interpret and represent an inequality on a number line To understand when to reverse the direction in an inequality To be able to solve a linear inequality There are many situations in which a solution to the problem is best described using one of the symbols or ⩾. For example, a pharmaceutical company might need to determine the possible number of packets of a particular drug that need to be sold so that the product is financially viable. This range of values may be expressed using inequality symbols. An inequality is a mathematical statement that uses an is less than () or an is greater than or equal express the dosage range of a medication from the lowest to (⩾) symbol. Inequalities may result in an infinite effective level to the highest safe level. number of solutions and these can be illustrated using a number line. Lesson starter: What does it mean for x? The following inequalities provide some information about the value of x. a 2⩾x b −2x < 4 c 3 − x ⩽ −1 Can you describe the possible values for x that satisfy each inequality? Test some values to check. How would you write the solution for x? Illustrate this on a number line. KEY IDEAS The four inequality symbols are and ⩾. x > a means x is greater than a. x a x ⩾ a means x is greater than or equal to a. x a x < a means x is less than a. x a x ⩽ a means x is less than or equal to a. x a Also a < x ⩽ b could be illustrated as shown. x a b Essential Mathematics for the Australian Curriculum ISBN 978-1-009-37313-5 © Greenwood et al. 2024 Cambridge University Press Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 1C Linear inequalities 19 An open circle is used for < or >where the endpoint is not included. A closed circle is used for ⩽ or ⩾where the endpoint is included. Solving linear inequalities follows the same rules as solving linear equations, except: We reverse the inequality sign if we multiply or divide by a negative number. For example, − 5 < − 3is equivalent to 5 > 3and if − 2x < 4, then x > − 2. We reverse the inequality symbol if the sides are switched. For example, if 2 ⩾ x, then x ⩽ 2. BUILDING UNDERSTANDING 1 State three numbers that satisfy each of these inequalities. a x ⩾ 3 b x < − 1.5 c 0 < x ⩽ 7 d − 8.7 ⩽ x < − 8.1 2 Match the graph a–c with the inequality A–C. A x > 2 B x ⩽ − 2 C 0 ⩽ x < 4 a x b x c x −3 −2 −1 0 1 2 −1 0 1 2 3 4 5 0 1 2 3 3 Phil houses xrabbits. If 1 0 < x ⩽ 13, how many rabbits could Phil have? 4 Insert the correct inequality symbol. a If − x < 2then x — − 2 b If − a > − 4then a — 4 Example 9 Writing inequalities from number lines Write as an inequality. a x b x 1 2 3 −3 −2 −1 0 SOLUTION EXPLANATION a x > 2 An open circle means 2is not included and x is greater than 2. b − 3 ⩽ x < 0 − 3is included but 0is not, and x lies between −3 and 0. Now you try Write as an inequality. a x b x 0 1 2 –4 –3 –2 –1 Essential Mathematics for the Australian Curriculum ISBN 978-1-009-37313-5 © Greenwood et al. 2024 Cambridge University Press Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 20 Chapter 1 Algebra, equations and linear relationships Example 10 Solving linear inequalities Solve the following inequalities and graph their solutions on a number line. a 3x + 4 > 13 x ⩽ 6 b 4 − _ c 3x + 2 > 6x − 4 3 SOLUTION EXPLANATION a 3x + 4 > 13 Subtract 4from both sides and then divide both sides by 3. 3x > 9 ∴x > 3 x Use an open circle since xdoes not include 3. 1 2 3 4 5 b 4−_x ⩽ 6 Subtract 4from both sides. 3 x ⩽ 2 −_ Multiply both sides by − 3and reverse the inequality symbol. 3 ∴ x ⩾ −6 x Use a closed circle since xincludes the number − 6. −7 −6 −5 −4 −3 c 3x + 2 > 6x − 4 Subtract 3xfrom both sides to gather the terms containing x. 2 > 3x − 4 Add 4to both sides and then divide both sides by 3. 6 > 3x 2 > x ∴x < 2 Make x the subject. Switching sides means the inequality symbol must be reversed. x −1 0 1 2 3 Use an open circle since xdoes not include 2. Now you try Solve the following inequalities and graph their solutions on a number line. a 2x + 5 > 11 x ⩽ 4 b 2 − _ c 4x + 1 > 7x − 2 3 Essential Mathematics for the Australian Curriculum ISBN 978-1-009-37313-5 © Greenwood et al. 2024 Cambridge University Press Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 1C Linear inequalities 21 Using calculators to solve inequalities 3 − 2x . Solve the inequality 5 < _ 3 Using the TI-Nspire: Using the ClassPad: In a Calculator page use menu >Algebra>Solve and type Tap solve( and type the inequality as shown. the inequality as shown. Hint: the inequality symbols (e.g. 6 2x + 3 > 3 g _ h x + 4 ⩽ 6 _ 3 3 2 5 _x i + 6 < 4 j x > 5 − 3 + _ k 3(3x − 1) ⩽ 7 l 2(4x + 4) < 5 9 4 Example 10b 3 Solve the following inequalities. Remember: if you multiply or divide by a negative number, you must reverse the inequality sign. a − 5x + 7 ⩽ 12 b 4 − 3x > − 2 c − 5x − 7 ⩾ 18 d _3 − x ⩾ 5 2 e _5 − 2x > 7 f 4 _ − 6x ⩽ − 4 x _ g 3 − ⩽ 8 _x h − − 5 > 2 3 5 2 3 Essential Mathematics for the Australian Curriculum ISBN 978-1-009-37313-5 © Greenwood et al. 2024 Cambridge University Press Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 22 Chapter 1 Algebra, equations and linear relationships Example 10c 4 Solve the following inequalities. a x + 1 < 2x − 5 b 5x + 2 ⩾ 8x − 4 c 7 − x > 2 + x d 3(x + 2) ⩽ 4(x − 1) e 7(1 − x) ⩾ 3(2 + 3x) f − (2 − 3x) < 5(4 − x) PROBLEM–SOLVING 5, 6 5, 6 6, 7 5 For the following situations, write an inequality and solve it to find the possible values for x. a 7more than twice a number xis less than 12. b Half of a number xsubtracted from 4 is greater than or equal to − 2. c The product of 3and one more than a number xis at least 2. d The sum of two consecutive even integers, of which the smaller is x, is no more than 24. e The sum of four consecutive even integers, of which xis the largest, is at most 148. 6 The cost of a satellite phone call is 30cents plus 20cents per minute. a Find the possible cost of a call if it is: i shorter than 5 minutes ii longer than 10 minutes. b For how many minutes can the phone be used if the cost per call is: i less than $2.10? ii greater than or equal to $3.50? 7 Solve these inequalities by first finding the lowest common denominator. a _x − _ x > 1 5x < − 2 b _x − _ 3 2 4 8 _x _3x c − > − 4 2x d − _ _ 9x < 0 6 4 5 20 Essential Mathematics for the Australian Curriculum ISBN 978-1-009-37313-5 © Greenwood et al. 2024 Cambridge University Press Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 1C Linear inequalities 23 REASONING 8 8, 9 9, 10 8 How many whole numbers satisfy these inequalities? Give a reason. a x > 8 b 2 < x ⩽ 3 9 Solve these families of inequalities by writing xin terms of a. Consider cases where a > 0and a < 0. a 10x − 1 ⩾ a + 2 b _2 − x > 4 c a(1 − x) > 7 a 10 Describe the sets (in a form like 2 < x ⩽ 3or − 1 ⩽ x < 5) that simultaneously satisfy these pairs of inequalities. a x − 9.5 x ⩾ 10 ENRICHMENT: Mixed inequalities − − 11−12(1/2) 11 Solve the inequalities and graph their solutions on a number line. Consider this example first. Solve −2 ⩽ x − 3 ⩽ 6 1 ⩽ x ⩽ 9 (add 3 to both sides) x 1 9 a 1 ⩽ x − 2 ⩽ 7 b − 4 ⩽ x + 3 ⩽ 6 c 2 ⩽ x + 7 < 0 − d 0 ⩽ 2x + 3 ⩽ 7 e − 5 ⩽ 3x + 4 ⩽ 11 f − 16 ⩽ 3x − 4 ⩽ − 10 g 7 ⩽ 7x − 70 ⩽ 14 h − 3 < _x − 2 < 0 4 12 Solve these inequalities as per question 11 but noting the negative coefficient of x. a 3 ⩽ 1 − x < 5 b − 1 ⩽ 4 − x ⩽ 8 c − 3 < 2 − x < 2 d 1 < 5 − 2x < 11 4 _ e − 4 < − x < − 2 f − 7 ⩽ _1 − 3x ⩽ − 2 2 2 Essential Mathematics for the Australian Curriculum ISBN 978-1-009-37313-5 © Greenwood et al. 2024 Cambridge University Press Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 24 Chapter 1 Algebra, equations and linear relationships 1D Linear equations involving more complex algebraic fractions OPTIONAL LEARNING INTENTIONS To know how to find the lowest common denominator of algebraic fractions To be able to combine numerators using expansion and addition of like terms To be able to add and subtract algebraic fractions To be able to solve linear equations involving algebraic fractions The sum or difference of two or more algebraic fractions can be simplified in a similar way to numerical fractions with the use of a common denominator. This is often the first step in solving a linear equation involving multiple algebraic fractions. Lesson starter: Spot the difference Electricians, electrical and electronic engineers Here are two sets of simplification steps. One set has one work with algebraic fractions when modelling critical error. Can you find and correct it? the flow of electric energy in circuits. The application of algebra when using electrical 5 =_ 2−_ _ 15 4−_ _ x+1 = _ x−_ 3(x + 1) 2x − _ formulas is essential in these professions. 3 2 6 6 3 2 6 6 −11 2x − 3x + 3 = _ = _ 6 6 =_−x + 3 6 KEY IDEAS Add and subtract algebraic fractions by first finding the lowest common denominator (LCD) and then combining the numerators. Expand numerators correctly by taking into account addition and subtraction signs. For example, −2(x + 1) = −2x − 2 and −5(2x − 3) = −10x + 15. Solving linear equations can involve multiplying by the LCD of algebraic fractions. Essential Mathematics for the Australian Curriculum ISBN 978-1-009-37313-5 © Greenwood et al. 2024 Cambridge University Press Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 1D Linear equations involving more complex algebraic fractions 25 BUILDING UNDERSTANDING 1 Expand and simplify the following. a 2(x − 2) b − (x + 6) c − 6(x − 2) d 5(x + 1) − 3(x + 2) 2 State the lowest common denominator for these pairs of fractions. a _ 7a a , _ 4xy x , _ b _ 3 4 2 6 3xy − 3x 3 2 , _ d _ c _ , _ x 2x 7 14 3 Solve these equations. a _x + 2 = 4 x − 3 = 1 b _ 2x + 1 = 2 c _ 3 5 6 x + 1 = _ 4 Consider the equation _ x + 5 . 2 3 a Multiply both sides by 2. b Multiply both sides by 3. c Solve the resulting equation. Example 11 Adding and subtracting algebraic fractions Simplify the following algebraic expressions. x + 3 + _ a _ x − 2 2x − 1 − _ x − 1 2 5 b _ 3 4 SOLUTION EXPLANATION 5(x + 3) _ 2(x − 2) x + 3 + _ a _ x − 2 = _ LCD of 2and 5is 10. Express each fraction as an + 2 5 10 10 equivalent fraction with a denominator of 10. 5(x + 3) + 2(x − 2) Use brackets to ensure you retain equivalent = _________________ 10 fractions. = 5x + 15 + 2x _______________ − 4 Combine the numerators, then expand the 10 brackets and simplify. = _ 7x + 11 10 b _ x − 1 = _ 2x − 1 − _ 4(2x − 1) _ 3(x − 1) Express each fraction with the LCD of 12. − 3 4 12 12 4(2x − 1) − 3(x − 1) Combine the numerators. = __________________ 12 8x − 4 − 3x + 3 Expand the brackets: 4(2x − 1) = 8x − 4 and = ______________ 12 − 3(x − 1) = − 3x + 3. = _ 5x − 1 Simplify by collecting like terms. 12 Now you try Simplify the following algebraic expressions. x + 1 + _ a _ x − 2 3x − 2 − _ b _ x − 2 3 2 2 5 Essential Mathematics for the Australian Curriculum ISBN 978-1-009-37313-5 © Greenwood et al. 2024 Cambridge University Press Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 26 Chapter 1 Algebra, equations and linear relationships Example 12 Solving more complex equations involving algebraic fractions Solve the following equations and check your solution using substitution. a _ 3x − 1 4x − 2 = _ b _x + 2 − _ 2x − 1 = 4 3 2 3 2 SOLUTION EXPLANATION a _ 3x − 1 4x − 2 = _ Multiply both sides by the LCD of 2and 3, which is 6 . 3 2 62 (4x − 2) 6 3(3x − 1) Cancel common factors. Alternatively, cross multiply for _ = _ 31 21 the same result. 2(4x − 2) = 3(3x − 1) 8x − 4 = 9x − 3 −4 = x − 3 Expand the brackets and gather terms containing x by −1 = x subtracting 8xfrom both sides. ∴ x = −1 Rewrite with xas the subject. Check by seeing if x = − 1 makes the equation true. b x + 2 − _ 2x − 1 = 4 _ Express the algebraic fractions as a single fraction using 3 2 the LCD of 6. 2(x + 2) _ _ 3(2x − 1) Alternatively, multiply both sides by the LCD of 2and 3, − = 4 6 6 which is 6. 2x + 4 − 6x + 3 = 4 ______________ ?
