National 5 Mathematics Assessment Practice Topic 4: The straight line, solving equations and inequations and simultaneous equations

Transcription

1 SCHOLAR Study Guide National 5 Mathematics Assessment Practice Topic 4: The straight line, solving equations and inequations and simultaneous equations Authored by: Margaret Ferguson Heriot-Watt University Edinburgh EH14 4AS, United Kingdom.

2 First published 2014 by Heriot-Watt University. This edition published in 2016 by Heriot-Watt University SCHOLAR. Copyright 2016 SCHOLAR Forum. Members of the SCHOLAR Forum may reproduce this publication in whole or in part for educational purposes within their establishment providing that no profit accrues at any stage, Any other use of the materials is governed by the general copyright statement that follows. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, without written permission from the publisher. Heriot-Watt University accepts no responsibility or liability whatsoever with regard to the information contained in this study guide. Distributed by the SCHOLAR Forum. SCHOLAR Study Guide Assessment Practice Topic 4: National 5 Mathematics 1. National 5 Mathematics Course Code: C747 75

3 Acknowledgements Thanks are due to the members of Heriot-Watt University's SCHOLAR team who planned and created these materials, and to the many colleagues who reviewed the content. We would like to acknowledge the assistance of the education authorities, colleges, teachers and students who contributed to the SCHOLAR programme and who evaluated these materials. Grateful acknowledgement is made for permission to use the following material in the SCHOLAR programme: The Scottish Qualifications Authority for permission to use Past Papers assessments. The Scottish Government for financial support. The content of this Study Guide is aligned to the Scottish Qualifications Authority (SQA) curriculum. All brand names, product names, logos and related devices are used for identification purposes only and are trademarks, registered trademarks or service marks of their respective holders.

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5 1 Topic 1 The straight line, solving equations and inequations and simultaneous equations Contents 4.1 Learning points Assessment practice

6 2 TOPIC 1. THE STRAIGHT LINE, SOLVING EQUATIONS AND INEQUATIONS AND SIMULTANEOUS EQUATIONS By the end of this topic, you should have identified your strengths and areas for further revision. Read through the learning points before you attempt the assessments and go back to the Course Materials unit if you need more help. You should be able to: determine the gradient of a straight line; determine the equation of a straight line; identify the gradient and y-intercept from the equation of a straight line; solve linear equations; interpret inequality symbols; solve inequations; solve simultaneous equations graphically; solve simultaneous equations algebraically; solve simultaneous equations from a context.

7 TOPIC 1. THE STRAIGHT LINE, SOLVING EQUATIONS AND INEQUATIONS AND SIMULTANEOUS EQUATIONS Learning points The straight line The formula for the gradient of a straight line requires two points (x 1,y 1 ) and (x 2,y 2 ). The formula is m = y 2 y 1 x 2 x 1 A horizontal straight line has gradient 0. A vertical straight line has gradient undefined. Parallel lines have the same gradient. The equation of a straight line with gradient m and y-intercept (0,c)is y = mx + c. The equation of a horizontal straight line with y-intercept (0,c) is x = k. The equation of a vertical straight line with x-intercept (k,0) is x = k. The equation of a straight line with gradient m and passes through the point (a, b) is y b = m(x a). To be able to identify the gradient or y-intercept, the equation of a straight line must be in the form y = mx + c. If the equation is not in that form it must be re-arranged. Solving linear equations Show every step as a new line in your solution. To get rid of a term to the other side of an equation change it s sign: add becomes subtract; subtract becomes add; multiply becomes divide; divide becomes multiply; powers become roots; roots become powers. If the expression has brackets multiply them out first. If the expression has a fraction multiply every term in the equation by the denominator or common denominator if there are more than one fraction. Solving inequations An inequation will contain <, >, or Reverse the inequality when dividing or multiplying by a negative.

8 4 TOPIC 1. THE STRAIGHT LINE, SOLVING EQUATIONS AND INEQUATIONS AND SIMULTANEOUS EQUATIONS Solving simultaneous equations graphically Take each equation in turn and substitute a couple of values for x to calculate y then plot the points and draw in the straight line. The solution is the coordinates of the point where the lines cross (point of intersection). Remember to state what x equals and y equals. Solving simultaneous equations algebraically The method of elimination requires the same quantity of one letter in both equations and one of them positive and the other negative. One or both of the equations may have to be multiplied to achieve this before you can add the two equations together to eliminate one of the terms. The method of substitution requires one or both of the equations to be in the form y =...or x =... This allows you to substitute for x or y in the other equation. Solving simultaneous equations from a context You will be given two statements in a context which can be written algebraically. If possible use appropriate letters to represent each item e.g. a for apple. Then solve algebraically as before. Remember to explain your solution e.g. if a = 0 55 then state an apple costs 0 55.

9 TOPIC 1. THE STRAIGHT LINE, SOLVING EQUATIONS AND INEQUATIONS AND SIMULTANEOUS EQUATIONS Assessment practice Make sure that you have read through the learning points or completed some revision before attempting these questions. Tailor your practice by choosing the most appropriate questions. The straight line: Questions 1 to 15 Solving linear equations: Questions 16 to 21 Solving inequations: Questions 22 to 26 Solving simultaneous equations: Questions Key point Questions 15, 32, 33 and 34 to 37 also assess your reasoning skills. Assessment practice: The straight line, solving equations and inequations and Simultaneous equations Gradient of straight lines Q1: Calculate the gradient of the line which passes through G(-2,8) and H(10,-2). Go online Q2: Calculate the gradient of the line which passes through K(-1,7) and L(-1,-7). Q3: Calculate the gradient of the line which passes through P (3,-5) and (-7,-5). Q4: Is the line GH parallel to the line KL? (answer yes or no) Equations of straight lines

10 6 TOPIC 1. THE STRAIGHT LINE, SOLVING EQUATIONS AND INEQUATIONS AND SIMULTANEOUS EQUATIONS Q5: Find the equation of the blue line in the diagram above. Q6: Find the equation of the red line in the diagram. Q7: Find the equation of the blue line in the diagram above. Q8: Find the equation of the red line in the diagram. Q9: Find the equation of the straight line passing through (-3,7) and (2,-3). Gradients and y-intercepts Q10: What is the gradient of the line with equation y = 7x 3? Q11: What are the coordinates of the y-intercept of the line with equation 2y = 5x 8? Q12: What is the gradient of the line with equation 4y 3x +7=0? Equations of straight lines Q13: Find the equation of the straight line passing through (3,-1) which is parallel to the line with equation y = 4x 3. Q14: Find the equation of the straight line passing through (5,6) which is parallel to the line with equation 3y 9x +5=0.

11 Q20: 2 3 y 10 = 8 TOPIC 1. THE STRAIGHT LINE, SOLVING EQUATIONS AND INEQUATIONS AND SIMULTANEOUS EQUATIONS 7 Solving linear equations Solve these linear equations. Q15: 4k 7 = 25 Q16: 2g +11 = 7 Q17: 3(m 4) = 12 Q18: 9r +8 = 4r 17 Q19: 7(w 2) + 5 = 4 (w +3) Q21: 1 2 (x + 6) = 4x 1 Solving inequations Solve these inequations. Q22: 4a + 1 > 13 Q23: 6f Q24: 12 4g < 18 2a Q25: 5(12 j) 3(2 j) Q26: t +9 > 1 3 (6t 15) A small bottle of shampoo contains n ml and a large bottle contains (3n + 50) ml. The difference between the amounts in a small bottle and a large bottle is 500 ml. Q27: How many millimetres does the small bottle contain?

12 8 TOPIC 1. THE STRAIGHT LINE, SOLVING EQUATIONS AND INEQUATIONS AND SIMULTANEOUS EQUATIONS Q28: How many millimetres does the large bottle contain? Solving simultaneous equations graphically Q29: Solve y = x +4and y = 6 x graphically. Q30: Solve y = 2x 4 and y = 2 x graphically. Solving simultaneous equations algebraically Q31: Solve the following system of equations What is a? What is b? Q32: Solve What is c? What is d? 5c 5d = 15 c 4d = 0 4a 3b = 18 a +3b = 3 Q33: Solve the following pair of simultaneous equations y = 5 x y = 3x +9 What is x? What is y? Solving simultaneous equations Q34: Three burgers and two hotdogs cost Four burgers and three hotdogs cost a) How much is one burger? b) How much is one hotdog?

13 TOPIC 1. THE STRAIGHT LINE, SOLVING EQUATIONS AND INEQUATIONS AND SIMULTANEOUS EQUATIONS 9 Q35: Three ice-cream cones and four ice-lollies cost Two ice-cream cones and Three ice-lollies cost How much would it cost for five ice-cream cones and three ice-lollies? Q36: A cinema sells standard seats and premier seats. Last night it sold 128 tickets for the 8:25pm film on screen 5. Let s be the number of standard seats and p the number of premier seats. From the information above write down an equation connecting s and p. Q37: A standard seat costs 8 95 and a premier seat costs The total takings were From the information above write down an equation connecting s and p. Q38: How many standard seats were sold? Q39: How many premier seats were sold?

14 10 ANSWERS: TOPIC 4 Answers to questions and activities 4 The straight line, solving equations and inequations and simultaneous equations Assessment practice: The straight line, solving equations and inequations and Simultaneous equations (page 5) Q1: 5 6 Q2: undefined Q3: 0 Q4: no (because the gradients are not equal) Q5: x = 2 Q6: y = 2x 1 Q7: y = 4 Q8: y = 3x + 2 Q9: Hint: Calculate the gradient then use y b = m(x b) to find the equation. Answer: y = 2x +1 Q10: 7 Q11: (0,-4) Q12: 3 / 4 Q13: Steps: What is the gradient of the line with equation y = 4x 3? 4 Use the gradient and the point (3,-1) to find the equation of the line. Answer: y = 4x 13 Q14: Steps: What is the equation of the 3y 9x + 5 = 0 when rearranged into the form y = mx + c? y = 3x 5 / 3 What is the gradient of the line with equation 3y 9x + 5 = 0?3 Use the gradient and the point (5,6) to find the equation of the line.

15 ANSWERS: TOPIC 4 11 Answer: y = 3x 9 Q15: k = 8 Q16: g = 9 Q17: m = 8 Q18: r = 5 Q19: w = 7 Q20: y = 12 Q21: x = 1 Q22: a > 3 Q23: f 7 Q24: g > 3 Q25: j 27 Q26: t < 14 Q27: Hint: (3n + 50) - n = 500 Answer: 225 ml Q28: 725 ml Q29: Steps: Find a couple of points on y = x +4. When x = 0, y = so y = 4 giving (0,4). When x = 2, y = so y = 6 giving (2,6). Plot the points and draw the line.

16 12 ANSWERS: TOPIC 4 Find a couple of points on y = 6 x. When x = 0, y = 6 0 so y = 6 giving (0,6). When x = 4, y = 6 4 so y = 2 giving (4,2). Plot the points and draw the line. Find the point of intersection The point of intersection is (1,5) and the solution is x = 1 and y = 5. Answer: (1,5) Q30: Steps: Find a couple of points on y = 2x 4. When x = 0, y = so y = 4 giving (0,-4). When x = 3, y = so y = 2 giving (3,2). Plot the points and draw the line. Find a couple of points on y = 2 x. When x = 0, y = 2 0 so y = 2 giving (0,2). When x = 3, y = 2 3 so y = 0 giving (3,-1).

17 ANSWERS: TOPIC 4 13 Plot the points and draw the line. Find the point of intersection The point of intersection is (2,0) and the solution is x = 2 and y = 0. Answer: (2,0) Q31: Hints: If we add the two equations together we will eliminate b. Divide by 5 to find a. Substitute a into the first equation to solve and find b. Answer: a = 3 and b = 2 Q32: Hints: This time we need to have a common multiple of 4 and -5 so we have an 2 extra steps: 5c 5d = 15, multiply this equation by 4 giving 20c 20d = 60. c 4d = 0, multiply this equation by -5 giving 5c +20d = 0. If we add the two equations together we will eliminate d. Divide by 5 to find c. Substitute c into the first equation to solve and find d. Answer: c = 4 and d = 1 Q33: Hints: If we replace y in equation 1 with 5 x from equation 2 we get 5 x = 3x +9. Rearrange and solve to equation to find x. Substitute in x to find y.

18 14 ANSWERS: TOPIC 4 Answer: x = 2 and y = 3 Q34: Hints: Let b represent the number of burgers and h represent the number of hotdogs. 3b +2h = 6 45, multiply this equation by 3 giving 9b +6h = b +3h = 8 93, multiply this equation by -2 giving 8b 6h = If we add the two equations together we will eliminate h and are left with b. Substitute b into the first equation to solve and find h. Answer: a) Burger = 1 49 b) Hotdog = 0 99 Q35: Steps: How much is 1 ice-cream cone? 1 75 How much is 1 ice-lolly? 1 39 Use these answers to find the cost of five ice-cream cones and three ice-lollies. Answer: Q36: s + p = 128 Q37: 8 95s p = Q38: 103 Q39: 25