YEAR 11 GENERAL MATHS (ADVANCED) - UNITS 1 & 01 General Information Further Maths 3/4 students are now allowed to take one bound reference into the end of year eams. In line with this, year 11 students will be able to use one binder book of notes and eamples for skills SACs and end of semester eams. It is recommended that several pages be left blank at the end of each topic to allow the student to add any etra notes or eamples they may require to assist them in completing the SACs and eams. They can use one bound reference for application SACs. Students need to have a second binder book in which to do their course work. The appropriate use of technology is required so students must have a TI nspire graphics calculator. The end of semester eam will count for 30% of the Unit Grade.
YEAR 11 GENERAL MATHS (ADVANCED) - UNITS 1 & 01 To successfully complete a unit (gain an S) the following are required: A graded average greater than UG for all tests. The binder book of notes must be kept to use during SACs for each topic. Outcome 1 Outcome Necessary to achieve a graded average greater than UG for all application SACs. If a student scores a graded average of UG for either Outcome 1 or, an opportunity will be given at the end of each semester for a student to redeem the subject. If the student is successful in the replacement assessment, the unit result will change from N to S, but the UG grade will remain the same. UNIT 1 Assessment Outcome1 Outcome Outcome 3 TOPICS Test (multiple choice/short answer) Assignment under test conditions. Use of Technology to be covered in Outcomes 1 and Linear Relations and Week 4- test SAC Week 4- application SAC Equations Shape and Measurement Week 9 - test SAC Week 9 - application SAC Number Systems Week 1 test SAC Week 1 - application SAC Linear Graphs and Modelling Week 1 - test SAC Week 1 - application SAC
UNIT TOPICS Assessment Outcome1 Outcome Outcome 3 Test (multiple choice/short answer) Assignment under test conditions. Use of Technology to be covered in Outcomes 1 and Univariate Data and Term 3 test SAC application SAC Bivariate Data Networks test SAC application SAC Trigonometry Term 4 test SAC application SAC Matrices test SAC application SAC
Topic Linear Equations Lesson Outcome Content Eamples Resources 3 period Define a linear equation Solve a linear equation Back tracking and inverse operations Show back tracking as a method of solution Using inverse operations to solve as alternative method Solve simple and comple as shown in the tet. Do simple checks for solutions Using TI nspire to show the use of the solve function Tet E A
A B Solving problems using linear equations C Substitution and transposition in linear relations D Linear recursion relationships E Simultaneous equations F Solving problems using simultaneous equations Linear equations AREAS OF STUDY The solution of linear equations including literal linear equations Developing formulas from word descriptions, substitution of values into formulas Solution of worded problems involving linear equations Substitution and transposition in linear relations, such as temperature conversion The construction of tables of values from a given formula using technology Linear relations defined recursively and simple applications The algebraic and graphical solution of simultaneous linear equations with two variables Solution of worded problems involving simultaneous linear equations with two variables ebookplus A Digital doc 10 Quick Questions A linear equation is an equation which contains a pronumeral (unknown value) raised to the power of 1. Such an equation may also be called an equation of the fi rst degree. Eamples of linear or first degree equations include: 4 8, y 1 and y + 5 3. Equations of the type: y 1, y, 4 8, + y 4 and y 3 8 are not linear since they contain pronumerals which are raised to powers other than 1; in these cases, 1, 1,, and 3 respectively. A linear equation is an equation which contains a pronumeral raised to the power of 1. It may also be called an equation of the first degree. Solving linear equations When we are asked to solve an equation, we are to find the value of the pronumeral so that when it is substituted into the original equation, it will make the equation a true statement. Equations are solved by performing a number of inverse operations to both sides of the equation until the value of the unknown is found. Chapter 1
When solving equations, the order of operations process, BODMAS (that is, Brackets Of Division, Multiplication, Addition, Subtraction) is reversed. We may therefore apply the SAMDOB process (BODMAS in reverse). This means that the operations of subtraction and addition are taken care of first, followed by multiplication and division. Brackets are dealt with last. WORKED EXAMPLE 1 Solve the following equations. a 3 4 b 3 + + 8 c 10 5 THINK WRITE a 1 Write the given equation. a 3 4 (Optional step.) Rule up a table with two columns to the side of the equation. In the first column, note each of the operations performed on in the correct order. In the second column, write the corresponding inverse operation. The arrows indicate which operation to begin with. Operation 3 + 3 3 Solve the equation by making the subject. Add 3 to both sides of the equation. 3 4 3 + 3 4 + 3 4 Divide both sides of the equation by. 5 Simplify. 3 1 (or 3.5) b 1 Write the given equation. b (Optional step.) Rule up a table with two columns. In the first column, note each of the operations performed on in the correct order. In the second column, write the corresponding inverse operation. The arrows indicate which operation to begin with. 3 Solve the equation by making the subject. Subtract from both sides of the equation. + 8 Operation + 8 + 8 4 Multiply both sides of the equation by. 5 Simplify. 3 c 1 Write the given equation. c 10 3 5 + Inverse Inverse (Optional step.) As in a and b above. Operation Inverse 3 3 + 10 10 Maths Quest 11 Standard General Mathematics for the TI-Nspire
3 Solve the equation by making the subject. Subtract 10 from both sides of the equation. 4 Multiply both sides of the equation by. 5 Divide both sides of the equation by 3. 10 3 5 10 3 10 5 5 10 3 5 3 5 3 10 3 10 3 3 10 3 Simplify. 3 1 3 Step is an optional step which may be used initially to help you become familiar with the process of solving equations. The answers may be checked by substituting the values obtained back into the original equation or using a calculator. If the pronumeral appears in the equation more than once, we must collect terms containing the unknown on one side of the equation and all other terms on the other side. WORKED EXAMPLE Solve for in the equation: 4 4 +. THINK WRITE 1 Write the given equation. 4 4 + Transpose 4 to the LHS of the equation by subtracting it from both sides of the equation. 4 4 4 4 + 4 3 Add 4 to both sides of the equation. 4 + 4 + 4 10 4 Divide both sides of the equation by. 10 5 Simplify. 5 To check whether your answer is correct, substitute it back into the equation to see if it will make a true statement: LHS ( 5) 4 14 RHS 4 ( 5) + 14 as LHS RHS, the solution is correct. If the equation contains brackets, they should be epanded first. In some cases, however, both sides of the equation can be divided by the coefficient in front of the brackets instead of epanding. WORKED EXAMPLE 3 Solve for : a ( + 5) 3( ) b 4( + ) 0. Chapter 3
THINK WRITE a 1 Write the given equation. a ( + 5) 3( ) Epand each of the brackets on both sides of the equation. 3 Transpose to the LHS of the equation by subtracting it from both sides of the equation. 4 Subtract 10 from both sides of the equation. 5 Divide both sides of the equation by 4. + 10 18 + 10 18 4 + 10 18 4 + 10 10 18 10 4 8 4 8 4 4 Simplify. b 1 Write the given equation. b 4( + ) 0 On a Calculator page, press: MENU b 3: Algebra 3 1: Solve 1 Complete the entry line as: solve (4 ( + ) 0, ) Then press ENTER. 3 Write the solution. Solving 4( + ) 0 for gives 1 If an equation contains a fraction, we should first remove the denominators by multiplying each term of the equation by the lowest common denominator (LCD). WORKED EXAMPLE 4 Find the value of which will make the following a true statement: + 5 3. THINK WRITE 1 Write the given equation. + 5 3 Determine the LCD of and 3. LCD of and 3 is. 3 Multiply each term of the equation by the LCD. ( + ) 5 3 ( + ) 4 Simplify both sides of the equation. 30 3 ( + ) 30 3 4 Maths Quest 11 Standard General Mathematics for the TI-Nspire
5 Epand the bracket on the LHS of the equation. + 4 30 3 Add 3 to both sides of the equation. + 3 + 4 30 3 + 3 5 + 4 30 Subtract 4 from both sides of the equation. 5 + 4 4 30 4 5 8 Divide both sides of the equation by 5. 5 5 5 9 Simplify. 5 1 (or 5.) 5 Sometimes in equations containing fractions, a pronumeral appears in the denominator. Such equations are solved in the same manner as those in the previous eamples. However, care must be taken to identify the value (or values) for which the pronumeral will cause the denominator to be zero (0). If in the process of obtaining the solution the pronumeral is found to take such a value, it should be discarded. Even though the process of identifying the value of the pronumeral that causes the denominator to be zero is at this stage merely a precaution, this process should be practised as it will prove useful in future chapters. WORKED EXAMPLE 5 For the equation 3 1 + 1 : a state which value(s) of will cause the equation to be undefined b solve for. ebook plus s Tutorial int-085 Worked eample 5 THINK a Identify the values of which will cause the denominator to be zero. Note: Once the equation has been solved, values which cause the denominator to be 0 will be discarded. WRITE b 1 Write the given equation. b 3 1 + 1 On a Calculator page, press: MENU b 3: Algebra 3 1: Solve 1 Complete the entry line as: 3 1 solve +, 1 Then press ENTER. a First fraction: 0 Second fraction: 0 0 Third fraction: 1 0 1 cannot assume the values of 0 and 1, since this will cause the fraction to be undefined. Chapter 5
3 Write the solution. Note: The value of is not 0 or 1. Hence, it is a valid solution. Solving 3 1 + 1 for gives 5 1 5 REMEMBER 1.. 3. 4. 5.. A linear equation is an equation that contains a pronumeral raised to the power of 1. It may also be called an equation of the fi rst degree. are solved by using inverse operations. When solving linear equations the order of operations process, BODMAS, is reversed. If the pronumeral appears more than once, the terms containing the unknown are collected onto one side of the equation and the numbers onto the other. If the equation contains brackets, either epand, or divide both sides by the coefficient in front of the bracket. If an equation contains fractions, multiply each term of the equation by the LCD. To check your solution, substitute it back into the equation to see if it will make a true statement. EXERCISE A ebook plus s Digital doc SkillSHEET.1 Solving linear equations 1 WE 1 Solve the following equations. a 1 b.3 c + 4 d 9 e 3 + 5 f 3 10 g 0. 10 h 5 15 i 1 3 j 4 k 3 1 3 5 l 5 m 1 n 3 + 11 0 o 3 4 4 11 3 4 p + 3 q 8 3 8 5 4 r + 13 1 9 + 3 s 5 t 13 u 3 + 10 v 5 w 5 3 11 + 13 9 5 y 15 z 1 0 3 WE Solve for. a + 9 b + 5 c 15 + d 1 1 5 e 3 + 4 f + 9 9 g 5 + 3 + 18 h 5 3 3 5 i 1 5 + 8 j + 1 k 5 + 3 l 15 + 13 10 m 8 3 4 n 1 + 5 + 11 o 13 3 4 p 9 + 11 3 Maths Quest 11 Standard General Mathematics for the TI-Nspire
3 WE3 Solve for. a 4( 0) 1 b 3(1 ) c 5( + 8) 0 d (3 4) e (5 + ) 13 f ( ) 3 g 3( ) 4( + 3) h 8( + 1) ( 3) i 4( + 3) ( 4) + 5 j 5( 4) 3 + ( ) 0 4 WE4 For each of the following, find the value of which will make the statement true. a + 4 + 1 b 3 4 3 5 3 c 1 d + 3 0 e 1 3 + 1 f 4 5 4 g ( + 5) + h 3 ( 3 ) 5 + 3 4 3 3 8 3 + 1 1 3+ 5 1 i + j 4 3 4 k ( (3 3 ) 4 ( ) ) 4 l + ) 5 3 ( 5 3 9 5 WE5 For each of the following: a state which value (or values) of will cause the equation to be undefined b solve for. 1 4 i + 1 3 4 ii 1 + 3 (3 ) 1 4 iii + ( 1)( 1 )( + 1) 1 5 iv 1 + 1 4 v 3 ( 4) 4 15 11 3 vi 4 8 3 MC Without solving the equation 4 1 3 + + 1 we know that will not be equal to: A 3 B 1 C 0 D 0 or 1 E 1 or 3 or 0 MC To solve 3 ( 1) 4 5, each term of the equation could be multiplied by: 3 A B 3 C 4 D 5 E 8 MC In order to solve the equation 4, the operations which must be performed are: 3 A both sides by, then by 3 B both sides by 3 C both sides by 3 D both sides by 3, then by 4 E both sides by 4, then by 9 Find the value of z, such that the solution to the following equation is 1. 3 z 8 + 1 ( ) ) (+ 1) 10 Solve the following equation. 5 4 1 + ebookplus Digital doc SkillSHEET. Finding the lowest common multiple Chapter
B Solving problems using linear equations can often be used to help us in problem solving. This is usually done in the following way. 1. Identify the unknown and choose any convenient pronumeral (usually ) to represent it.. Use the information given in the problem to compose an equation in terms of the pronumeral. 3. Solve the equation to find the value of the pronumeral. 4. Interpret your result by relating the answer back to the problem. WORKED EXAMPLE If the sum of twice a certain number and 5 is multiplied by 3 and then divided by, the result is 9. Find the number. THINK WRITE 1 Assign the pronumeral to the unknown value. Let the unknown number. Build the equation according to the information given. (a) Twice the number; this means, so write this. (b) The sum of twice the number and 5; this means + 5, so add this on. (c) The sum is multiplied by 3; this means 3( + 5). Add this on. Note: We include brackets to indicate the order of operations. (d) The result is divided by ; this means 3 ( + 5). Add this on. (e) The result is 9; which means that all of the previous computations will equal 9. Write this. 3 Solve for. (a) Multiply both sides of the equation by. (b) Divide both sides of the equation by 3 since they are both divisible by 3. (c) Subtract 5 from both sides of the equation. + 5 3( + 5) 3 ( + 5) 3 ( + 5) 9 3 ( + 5) 9 3( + 5) 3 3 ( + 5) 3 3 3 + 5 1 + 5 5 1 5 1 (d) Divide both sides of the equation by. 1 (e) Simplify. 8 4 Answer the question. The unknown number is 8. 5 Your answer can be easily verified by checking whether it will satisfy the conditions specified in the problem. 8 Maths Quest 11 Standard General Mathematics for the TI-Nspire
b Tet A: (13, 35, 59, 8, 98); Tet B: (3, 55, 3.5, 0, 8) c 0 0 40 0 80100 Result d to f Various answers a 3.8 b Tet A Tet B 8 80 8 84 8 88 90 9 94 9 98 100 c 89.5 d The median height of boys increases with age. CHAPTER Eercise A 1 a - 19 b 8.3 c - 1 d 8 e 1 3 f 3 1 g - 50 h - 3 i 3 j k 0 l 5 5 m n 1 o 11 p 5 q r 1 5 8 u - 45 v - 59 w 13 1 5 y - 18 z 4 1 a - 1 b 1 1 e - 5 f 3 3 5 s 44 t 1 3 c - 1 d 1 g - h 0 i - 10 j k 11 l - m n 1 o - 3 p 1 3 a 4 b 3 c - 1 d 3 e - 3 f g - 33 h -. i 1 4 a 1 1 4 e 13 4 i j j 3 b 15 c - 14 d 1 4 f g 10 11 43 80 k 11 h 149 l 3 5 a i 0, - 1 ii 0, 1 iii 1, - 1 iv 0 v vi 0, 3 b i iv 3 4 1 3 ii 10 19 v 4 5 E E 8 C 9 10 5 or 11 5 iii 5 3 vi 5 43 Eercise B Solving problems using linear equations 1-1 - 1 3 40 4 5 18 red tulips, 10 yellow tulips 8 Width 8 cm, length 0 cm 9 10 cm, 10 cm, 1 cm 10 40, 10, 0 ; obtuse angled 1 1 $4 1 $99 13 a B b D 14 a 10 b 15 15 a 3 b 3 years in America, 4 years in Germany, 8 years in London, 1 year in Australia 1 1 15 Jonathan, 0 Golden Delicious 1 8 5 shorts, 15 T-shirts 1 9 Ale will reach Nat in 1.5 h; that is, at 11.30 am. 0 Bus: 50 km/h, car: 0 km/h 1 4 km/h 10 h 3 1 days. No 4 $.08 5 0 $000 Eercise C Substitution and transposition in linear relations 1 a 95 b 1 c 5 d - 58 e. f 3.4 g 19.4 h 3 a.5 cm,.5 cm, 1.5 cm, 1.5 cm,.5 cm b 15. cm, 4.1 cm, 8.5 cm, 109.9 cm, 141.3 cm 3 A$ US$ US$ A$ 5 3.84 5.51 10.8 10 13.0 15 11.5 15 19.53 0 15.3 0.04 5 19.0 5 3.55 30 3.04 30 39.0 35.88 35 45.5 40 30. 40 5.08 45 34.5 45 58.59 50 38.40 50 5.10 55 4.4 55 1.1 0 4.08 0 8.13 5 49.9 5 84.4 0 53. 0 91.15 5 5.0 5 9. 80 1.44 80 104.1 85 5.8 85 110.8 90 9.1 90 11.19 95.9 95 13.0 100.80 100 130.1 4 Principal ($) Amount ($) 5 000 5 11.5 5 500 5.88 000 139.50 500 51.13 000 1.5 500 4.38 8 000 8 18.00 8 500 8 9.3 9 000 9 09.5 9 500 9 0.88 10 000 10 3.50 1 Answers