Equasions

EQUASIONS CLASS ASSIGNMENT 1 Step by step, discover more about... what an equation is and how to solve it.... 1. What do you understand by the term “equation” ? 2. What do you understand by the term “inspection” ? 3. What does it mean if we say: “solve the equation” ? 4. Give an example of an equation. 5. Now solve the equation in (4). (Get your teacher to help you). 6. Solve each of the following equations by inspection.(i.e. determine the value of a) 6.1 1 - a = 7 6.2 5a = 50 6.3 a/7 = 6 6.4 66/a = 6 6.5 -a/2 = 15 6.6 5a - 4 = 26 7. How do you solve an equation if not by inspection? 7.1 Here are a few tips: 2a + 6 = -5a - 9 Step 1: Identify all the unknowns (in this case “ a ”) and get it on the left-hand side of the “=” sign. Step 2: Identify all the constants and get it on the right-hand side of the “=” sign. Step 3: Add up similar terms. Step 4: Identify unknown alone (by multiplying or dividing with coefficient of unknown) Step 1: +5 a (must do the same on the other side) +5 a (to get rid of “ a ’s”) 2a + 6 = -5a- 9 We get the following: 2a+ 5a+ 6 = -9 Step 2: -6 (to get rid of +6) -6 (must do the same on the other side) 2a + 5a + 6 = -9 We get the following: 2a + 5a= -9 - 6 Step 3: Add up the similar terms 7a = -15 Step 4: Get “a” alone: ÷ 7 left and right of the “=” sign 7a7 size 12{ { {7a} over {7} } } {} = -157 size 12{ { {"-15"} over {7} } } {} a = −217 size 12{ - 2 { { size 8{1} } over { size 8{7} } } } {} 7.2 What about ? : a/₃ = 7 a/₃= ⁷/₁(multiply cross-wise) a x 1 = 7 x 3 a = 21 7.3 Remember that if there are brackets, the brackets must first be removed.e.g. 2(2a - 6) + 7 = 9a - 3(a - 2) 4a - 12 + 7 = 9a - 3a + 6 4a- 3a - 9a = 6 - 7 + 12 -8a = 11 a = ¹¹/_-8 a = - 1 ³/₈ 8. Solve the following equations: 8.1: 5 - 3(4 - a) = 5(a + 1) + 2 8.2: 2a - 24 = 3a 8.3: a/₄ + 5 = 10 (tip: fractions and cross-wise multiplication) 8.4 5a - 3a - 7 = 9 8.5: -8a = 72 8.6: 3² + 3a+ 3 = 2³ + 4(-2a) 8.7: ^-12/a = -24 8.8: 5(2a + 1) = 4(2a + 3) 8.9: -2(3a - 3) = 6a + 24 8.10: Write down any equation of your own and solve it. Conditions: it must contain brackets, fractions and negative numbers. CLASS ASSIGNMENT 2 Step by step, discover more about ... the solving of word problems, which can become a feast if you can present them as equations ... To share in this feast, you need the following basic knowledge: ... 1. Words like ... more (means +), less (means -) and times (means ×), consecutive numbers (first number: x, second number: x + 1 and third number: x + 2) 1.1 How would you represent the following consecutive even numbers? First number: …………… Second number: …………… Third number: …………… 1.2 How would you represent a two-digit number if the digits are unknown? How would you write 24 in extended notation: (2 x 10) + (4 x 1) If you can say that the tens digit is half the ones digit, you would do the following: tens ones a 2a(tens digit is half the ones digit) Thus: (a x 10) + (2a x 1) = 10a+ 2a = 12a (12a is the number) 2. Your first question must always be: “About which one do I know nothing?”- This is then represented by e.g. x and the other unknowns in terms of x. 3. Write down an equation. 4. Solve the equation. 5. Answer the question. Let us look at a few examples: Example 1: Problem: the length of a rectangle is 5 cm more than its width. Solution: 1) Draw a rectangle 2) Ask yourself: about which one, length or width, do I knownothing? In this case it is width = x cm

3) Now express the length in terms of x : in this case: (x + 5)(remember 5 more: i.e. x + 5) 4) The question can also say that the perimeter is 80 cm, so calculate the length and width of the rectangle.(You can represent the data as an equation.) Do this now: 5) Solve the equation and answer the question in (4) Example 2: Problem: A mother is four times as old as her daughter. Their joint age is 60, how old is each one? Solution: 1) Ask yourself: about which one do I know nothing? Let this be x. 2) Represent your thoughts thus far: Mother 4 x Daughter x 3) Represent the data as an equation, solve the equation and answer the question. Example 3: The problem about ages is always difficult, but do it step by step and write down the plan of your thoughts, and it becomes very easy..... Problem: Milandre is 30 years older than Filandre. In 15 years’ time Milandre will be twice as old as Filandre. How old are they now? Solution : 1) Begin with a plan and write it down: 2) Ask yourself: About which do I know nothing? Make this x. age: now Milandre x + 30 Filandre x age: in 15 years’ time: (i.e. + 15) (x + 30) + 15 x + 15 3) Now comes the most difficult part: in 15 years’ time Milandre will be twice as old as Filandre. As Milandre will be twice as old as Filandre, you will have to multiply Filandre’s age (in the column in 15 years’ time) by 2 so that you can get an equation (i.e. left-hand side = right-hand side) 4) Write down the equation below, solve it and answer the question. See if you can do it on your own.Remember that you must write down your plan of thinking. Here they are..... 1. The sum of two numbers is 15. Write down the two numbers in terms of x. 2. 140 people attend the Steve Hofmeyer concert. The following tickets are available: Children: R 20 Adults: R 45 If the entrance money amounted to R5 580, calculate how many adults and how many children were present. HOMEWORK ASSIGNMENTS 1 and 2 1. Given: a ∈ {-4; -3; -2; -1; 1; 2; 3}Select your answer from the above set in each case. 1.1 -3a + 20 = 23 1.2 8a = -32 2. The value of x is given in each case. Test the correctness of each equation. 2.1 8x - 2(x - 5) = 28 x = 3 2.2 5x - 10 = 10 x - 10 x ∈ R 3. Solve the following equations. Show all your calculations. 3.1 ¹/z= ¹/₁₈z = ? 3.2 :1 - 5z = 11 3.2.1: where z∈ N 3.2.2: where z ∈ Q 3.3: z + 3[z + 2(z - 6)] = 45 3.4: 4(6z - 8) - 2(z+ 7) = 37 3.5: z - 5(z - 8) = -48 4. Write each of the following as an algebraic equation and solve it. 4.1 Six times a number, reduced by 8, is equal to 55. Calculate the number. A negative number is nine times another number. The sum of the two numbers is –64. Determine the two numbers. 4.3 The sum of three consecutive negative integers is –90.Determine the three negative numbers. 4.4 Jessica buys three times more oranges than bananas. If the oranges cost 45c each and the bananas 18c each, how many of each did she buy if the total cost was R18,36? 4.5 Cameron is eight years older than Liam. Six years ago Cameron was three times as old as Liam. How old are they now? 4.6 You have bought stamps from the post office. These include stamps for R1,20 each and for R2,40 each. If the total value of the stamps is R58,80, determine how many of each type of stamp you bought. 4.7 Divide a piece of hosepipe of 18 m in two, so that one piece is 550 mm longer than twice the other piece. Determine the length of each piece of hosepipe. 4.8 18 women and 25 girls have a total mass of 3 792 kg.The girls all have the same mass and each woman is three times heavier than a girl. Determine the mass of a woman and a girl. 4.9 The ones digit of a two-digit number is double the tens digit of the number. If the two digits are swapped around, you get a number that is 36 higher than the original number. Calculate the original two-digit number. 4.10 There are 25 more learners in grade 10 than and grade 9 and 32 more learners in grade 8 than in grade 9. If the total number of learners form grade 8 to grade 10 is 732, calculate how many learners there are in each grade. Assessment Assessment of myself: by myself: Assessment by Teacher: I can…    1 2 3 4 Critical Outcomes 1 2 3 4 solve simple equations by means of inspection; (Lo 1.8); Critical and creative thinking solve simple equations and choose the correct answer from a given set; (Lo 1.8); Collaborating remove brackets and then solve an equation; (Lo 2.5; 2.8.4); Organising en managing solve comparisons containing fractions; (Lo 1.8; 2.5) Processing of information write word problems in algebraic equations, (Lo 2.8.6); Communication and then solve comparisons. (Lo 2.5). Problem solving Independence

 good  average  not so good Comments by the learner: My plan of action: My marks: I am very satisfied with the standard of my work. < Date: I am satisfied with the steady progress I have made. Out of: I have worked hard, but my achievement is not satisfactory. Learner: I did not give my best. >

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Tutorial 2: (Equations) Total: 50 1. Given: 5x = 1 1.1 Does the above represent an equation? Give a reason for your answer. [2] 1.2 Determine now the possible answer for x by inspection. [1] 2. Determine the value for a in each of the following ( by INSPECTION). Write down the answers only 2.1: a + 7 = 19 2.2: 5a - 7 = 28 2.3: 6a - 7 = 3a 2.4: ⁴a/₃ + 5 = 9 2.5: ⁴/₃a + 5 ²/₃ = 6 [5 x 2= 10] 3. Determine p in each of the following using substitution. Round of your answers to the nearest 3 decimals. 3.1: f = 4(2b - p) , f = 32 and b = 9 [3] 3.2: f = p/r, f= 45,67 and r = 21,3 [3] 4. Determine the value for a in each of the following. Show all your calculations. 4.1: 7a + a/₃ = 5(2 + 3) [3] 4.2: a(4a - 3) = (-2a)² [3] 4.3: -4(a- 2) = 3(a - 4) [3] 4.4: 5(a+ 3) + 4a + 5 = 2(a - 7) [4] 4.5: 5a = -2(a - 3) [3] 5. Write as algebraic equations and solve. 5.1 Nine times a certain number is 28 more than five times the number. What is the number? [2] 5.2 A rectangle, of which the perimeter is 108 cm, has a length which is four more than the breadth. Determine: 5.2.1 the length and the breadth [4] 5.2.2 the area of the rectangle [2] 6. The sum of four consecutive odd numbers is 112.Determine the four numbers. [4] 7. A book is opened, and the product of the page number of the page on the left and that of the page on the right, is determined. The product is 6 162.What is the number of the page on the left? [3] Learning unit 2Assessment 2.2 Assessment of myself: by myself: Assessment by Teacher: I can…    1 2 3 4 Critical Outcomes 1 2 3 4 define a simple comparison; (Lo 1.2.1 ; 1.2.2); Critical and creative thinking solve elementary comparisons by means of inspection; (Lo 1.8 ; 2.5); Collaborating solve any comparison despite the brackets; (Lo 1.8 ; 2.5); Organising en managing write word problems as comparisons and then solve them; (Lo 1.8 ; 2.5; 2.8.6); Processing of information solve intricate word problems. (Lo 1.8 ; 2.5 ; 2.8.6). Communication Problem solving Independence

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Test 1: (Equations) Total: 45 1. Solve for x: 1.1: x + 5 = 39 [2] 1.2: 7 - 3x = 1 [2] 1.3: 2(x + 5) = 18 [3] 1.4: 8 = 40 - 2x [2] 1.6: 2(x - 3) - (x + 1) = 5x - 4 [4] 1.7: 3x + 6 = 15 [2] 1.8: x2 size 12{ { {x} over {2} } } {}= 4 [2] 1.9: 2(3x + 24) = 114 [3] 1.10: 10x + 9 = 7x + 30 [3] 1.11: ¹/₃ (3x - 6) - 2(x + 1½) = 7 [4] [27] 2. Solve the following word problems: 2.1 If 5 is subtracted form a certain number and the answer is divided by 3, the answer is 4. Determine the number. [3] 2.2 The sum of three consecutive even numbers is 66. Determine the numbers. [3] 2.3 The length of a rectangle is 5,5 cm longer than its breath. If the perimeter of the rectangle is 27 cm, determine the breath. [4] 2.4 Jonte’s age is five times the age of his daughter. If the sum of their ages is 60 years, how old is his daughter? [4] 2.5 Gareth is 5 times as old as his son. In 5 years time he will be 3 times as old as his son. How old is his son now? [4] 3. Bonus Given: ¹/₂a + 2a = 7 Determine the value of 14a2+4a2 size 12{ { {1} over {4a rSup { size 8{2} } } } `+`4a rSup { size 8{2} } } {} [3] Enrichment exercise for the quick worker (Learning unit 2) 1. If the product 2¹² x 5⁸is expanded, how many digits will the answer consist of? 2. 6² = 36. How many other positive single-digit numbers are there whose squares also end with the same digit as the number you are squaring? The three digit number 2A3 is added to 326 and gives 5T9. If 5T9 is divisible by 9, then A + T is equal to ..…………………………………………………………………. 4. If from a four digit number starting with 199, you subtract a four-digit number starting with 34, then your largest possible answer is ... 5. If you write 1,2,3,6 in every possible order to form 4-digit numbers, how many of these numbers will be divisible by 4? 6. Let n be any natural number. If the tens digit of n² is equal to 3,what is the last digit of n². 7. The average of three integers is 86. If one number is 70, what is the average of the other two? 8. At ABSA Saretha exchanged a R10 and a R20 note for an equal number of 50 cent, 20 cent and 5 cent coins. How many coins did Saretha receive?

Memorandum LEARNING UNIT 2 1. - 5. General 6.1 1 – a = 7 a = –6 a = 10 42 = a a = 11 –30 = a a = 6 5 – 12 + 3a = 5a + 5 + 2 3a – 5a = 5 + 2 + 12 – 5 –2a = 14 a = –7 2a – 3a = 24 –a = 24 ∴ size 12{∴} {}a = –24 8.3 a4 size 12{ { {a} over {4} } } {} = 51 size 12{ { {5} over {1} } } {}a = 20 8.4 2a = 16 a = 8 a = –9 9 + 3a + 3 = 8 – 8a 11a = –4 ∴ size 12{∴} {}a = −411 size 12{ { { - 4} over {"11"} } } {} (–2 38 size 12{ { {3} over {8} } } {}) 8.7 −12a size 12{ { { - "12"} over {a} } } {} = −241 size 12{ { { - "24"} over {1} } } {} –24a = –12 a = 12 size 12{ { {1} over {2} } } {} 10a + 5 = 8a + 12 2a = 7 a = 3 12 size 12{ { {1} over {2} } } {} 8.9 –6a + 6 = 6a + 24 –12a = 18 a = −1812 size 12{ { { - "18"} over {"12"} } } {}−32 size 12{ { { - 3} over {2} } } {} (–1 12 size 12{ { {1} over {2} } } {}) Own choice CLASSWORK ASSIGNMENT 2 20: 1. Number 1: x Number 2: 15 – x 2. Children: x 20 x Adults: (140 – x) 45(140 – x) 20 x + 45(140 – x) = 5 580 20 x + 6 300 – 45 x = 5 580 –25 x = –720 x = 28,8 28 / 29 Adults: 140 – 28 = 112 or 140 – 29 = 111 HOMEWORK ASSIGNMENT 1 AND 2 –3a = 3 a = –1 a = –4 8x – 21x + 10 = 28 8(3) – 2(3) + 10 = 28 – 6 + 10 = 28 √ 2.2 5x – 10 = 10x – 10 5x – 10x = –10 + 10 –5x = 0 x = 0 x size 12{ in } {} 1R √ 3.1 1z size 12{ { {1} over {z} } } {} = 118 size 12{ { {1} over {"18"} } } {}z = 18 3.2 1 – 5z = 11 –5 z = 10 z = –2 3.2.1 No solution 3.2.2 z = –2 3.3 z + 3[z + 2 z – 12] = 45 z + 3[3 z – 12] = 45 z + 9 z – 36] = 45 10 z = 81 z = 8,1 3.4 24 z – 32 – 2 z – 14 = 37 22 z = 83 z size 12{ approx } {} 3,77 3.5 z – 5 z + 40 = 48 –7 z = –88 z = 22 6x – 8 = 55 6 x = 63 x = 10 12 size 12{ { {1} over {2} } } {} (10,5) 4.2 x + 9 x = –64 10 x = –64 x = –6,4 4.3 x + x 1 x + 2 = –90 3 x = –93 x = –31 –31; –30; 29 4.4 Oranges: 3 x x 45 36 Bananas: x x 18 12 18 x + 135 x = 1 836 153 x = 1 836 x = 12 Now –6 Cameron: x + 8 [ 18 ] x + 8 –6 Liam: x [ 10 ] x – 6 3(x – 6) = x + 2 3 x – 18 = x + 2 2 x = 20 x = 10 or4.6 Stamps: R1,20 : x 50 – x R2,40 : 50 – xx 120 x + 240(50 – x) = 5 880 1,20(50 – x) + 2,40 x = 58,8 or120 x + 12 000 – 240 x) = 5 880 60 – 1,20 x) + 2,40 x = 58,8 –120 x = –6 120 1,20 x = 1,20 x = 51 x = 1 R1,20 4 4.7 One part: x Other part: x + 550 x + 2(x + 550) = 18 000 x 2 x + 1 100 = 18 000 3 x = 16 900 4.8 Mass 3 792 kgWomen: 18 3 x Girls: 25 x 25x + 3 x (18) = 3 792 25x + 54 x = 3 792 79 x = 3 792 x = 48 Girls: 48 kg each Women: 144 kg each Number 12 x4.9 Units: 2 x 8 Tens x x 10 4 converted: 21 x 2 x – 36 = 12 x 9 x = 36 x = 4 4.10 Gr. 8: x + 3² 257 Gr. 9: x 225 Gr. 10: x + 25 250 x + 32 + x + x + 25 = 732 3 x = 675 x = 225 TUTORIAL 2 Yes √ = can work out the value for x [5 x = 5⁰ . x = 0] x = 0 √ a = 12 √√ a = 7 √√ a = 73 size 12{ { {7} over {3} } } {} (2 13 size 12{ { {1} over {3} } } {}) √√ a = 3 √√ a = 4 √√ 32 = 4(2(9) – p) √ 32 = 4(18 – p) 8 = 18 – p √ p = 10 √ 45,67 = p21,3 size 12{ { {p} over {"21",3} } } {} √ p = 972,771 √ 4.1 7a1 size 12{ { {7a} over {1} } } {} + a3 size 12{ { {a} over {3} } } {} = 30 22a3 size 12{ { {"22"a} over {3} } } {} = 301 size 12{ { {"30"} over {1} } } {} √ 22a = 90 √ a = 9022 size 12{ { {"90"} over {"22"} } } {} = 4 222 size 12{ { {2} over {"22"} } } {} = 4 111 size 12{ { {1} over {"11"} } } {} √ 4a² – 3a = 4a^{2 √} √ 0 = 3a √ = a √ –4a + 8 = 3a – 12 √ –7a = –20 √ a = 207 size 12{ { {"20"} over {7} } } {} = 2 67 size 12{ { {6} over {7} } } {} (or size 12{ approx } {} 2,86) √ 5a + 15 + 4a + 5 = 2a – 14 √ 7a = –34 √ a = 4,9 √ 5a = –2a + 6 √ 7a = 6 √ a = 67 size 12{ { {6} over {7} } } {} (0,86) √ 9x – 28 = 5x √ 4x = 28 x = 7 √ 5.2.1 x + 4 2x + 2(x + 4) = 108 √ 2x + 2 x + 8 = 108 x 4x = 100 x = 25√ √ √ ∴ size 12{∴} {} l = 29 cmb = 25 cm 5.2.2 A = 29 x 25 √ = 725 cm² √ 6. x + x + 2 + x + 4 + x + 6 = 112 √ 4 x = 112 – 12 4 x = 100 x = 25 √ ∴ size 12{∴} {} Numbers : 25; 27; 29; 31 √√ 7. 6162 size 12{ sqrt {"6162"} } {} size 12{ approx } {} 78 √ ∴ size 12{∴} {} 6 162 ÷ 78 = 79 ∴ size 12{∴} {} 78; 79 √√ TEST 1 x = 34 –3 x = 1 –3 x = –6 ∴ size 12{∴} {}x = 2 2 x + 10 = 18 2 x = 8 ∴ size 12{∴} {}x = 4 2 x = 40 – 8 2 x = 32 ∴ size 12{∴} {}x = 16 2 x – 6 – x – 1 = 5 x – 4 2 x – x – 5 x = –4 + 1 + 6 –4 x = 3 x = −34 size 12{ { { - 3} over {4} } } {} 3 x = 15 – 6 3 x = 9 ∴ size 12{∴} {}x = 3 1.7 x = 4 x 2 = 8 1.8 6 x + 48 = 114 6 x = 66 ∴ size 12{∴} {}x = 11 10 x – 7 x = 30 – 9 3 x = 21 ∴ size 12{∴} {}x = 7 1.10 x – 2 – 2 x 3 = 7 – x = 7 + 3 + 2 – x = 12 ∴ size 12{∴} {}x = –12 x−53 size 12{ left ( { {x - 5} over {3} } right )} {} = 4 x = (4 x 3) + 5 = 17 x + x 1 + x + 2 = 66 3 x = 63 x = 21 2x + 2(x + 5,5) = 27 2x + 2x + 11 = 27 4x = 16 x 4 cm x = 4 x + 5,5 9.5 cm 2.4 Jonte: 5 x 50 5x + x = 60 Daughter: x 10 6x = 60 x = 10 2.5 Now +5 Gareth 5x 5x + 5 Son xx + 5 ∴ size 12{∴} {} Gareth: 25 ∴ size 12{∴} {} Son: 5 x + 5 = 3(x + 5) 5 x + 5 = 3 x + 15 2 x = 10 x = 5 3. (12a size 12{\( { {1} over {2a} } } {} + 2 a) size 12{a\)} {}² = 7² 14a2 size 12{ { {1} over {4a rSup { size 8{2} } } } } {} + 2 + 4 a size 12{a} {}² = 49 14a2 size 12{ { {1} over {4a rSup { size 8{2} } } } } {} + 4 a size 12{a} {}²= 47