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© Boardworks Ltd 2004 1 of 62 A2 Equations KS3 Mathematics.

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1 © Boardworks Ltd of 62 A2 Equations KS3 Mathematics

2 © Boardworks Ltd of 62 A2.1 Solving simple equations Contents A2 Equations A2.5 Non-linear equations A2.4 Equations and proportion A2.2 Equations with the unknown on both sides A2.3 Solving more difficult equations

3 © Boardworks Ltd of 62 Equations An equation links an algebraic expression and a number, or two algebraic expressions with an equals sign. For example, x + 7 = 13 is an equation. In an equation the unknown usually has a particular value. Finding the value of the unknown is called solving the equation. x + 7 = 13 x = 6 When we solve an equation we always line up the equals signs.

4 © Boardworks Ltd of 62 Using inverse operations In algebra, letter symbols represent numbers. Rules that apply to numbers in arithmetic apply to letter symbols in algebra. For example, in arithmetic, if = 10, we can use inverse operations to write: 10 – 7 = 3and10 – 3 = 7 In algebra, if a + b = 10, we can use inverse operations to write: 10 – b = a and 10 – a = b a = 10 – b and b = 10 – a or

5 © Boardworks Ltd of 62 Using inverse operations In arithmetic, if 3 × 4 = 12, we can use inverse operations to write: 12 ÷ 4 = 3and12 ÷ 3 = 4 In algebra, if ab = 12, we can use inverse operations to write: and or 12 b = a a = b b = a a = b

6 © Boardworks Ltd of 62 Using inverse operations to solve equations We can use inverse operations to solve simple equations. For example, x + 5 = 13 x = 13 – 5 x = 8 Always check the solution to an equation by substituting the solution back into the original equation. If we substitute x = 8 back into x + 5 = 13 we have = 13

7 © Boardworks Ltd of 62 Using inverse operations to solve equations Solve the following equations using inverse operations. 5 x = 45 x = 45 ÷ 5 x = 9 Check: 5 × 9 = – x = 6 17 = 6 + x 17 – 6 = x Check: 17 – 11 = 6 11 = x x = 11 We always write the letter before the equals sign.

8 © Boardworks Ltd of 62 Using inverse operations to solve equations Solve the following equations using inverse operations. x = 3 × 7 x = 21 Check: 3 x – 4 = 14 3 x = x = 18 Check: 3 × 6 – 4 = 14 x = 18 ÷ 3 x = 6 = 3 x

9 © Boardworks Ltd of 62 What number am I thinking of…?

10 © Boardworks Ltd of 62 What number am I thinking of …? I’m thinking of a number. When I multiply the number by 2 and add 5 the answer is 11. What number am I thinking of …? We can write this as an equation. Instead of using ? for the number I am thinking of, let’s use the letter x. Start with x x Multiply by 2 2x2x and add 5 2 x + 5 to give you x + 5 = 11

11 © Boardworks Ltd of 62 What number am I thinking of …? Start with x x Multiply by 2 2x2x and add 5 2 x + 5 to give you x + 5 = 11 We can solve this equation using inverse operations in the opposite order. Start with the equation: 2 x + 5 = 11 Subtract 5:2 x = 11 – 5 Divide by 2: 2 x = 6 x = 6 ÷ 2 x = 3

12 © Boardworks Ltd of 62 Solving equations by transforming both sides Solve this equation by transforming both sides in the same way: Add 1 to both sides. Multiply both sides by 4. m = 12 We can check the solution by substituting it back into the original equation: 12 ÷ 4 – 1 = 2 – 1 = 2 m 4 = 3 m 4 +1 ×4

13 © Boardworks Ltd of 62 Find the missing number 23?62 ?? 127 The number in each brick is found by adding the two numbers above it. What are the missing values? We can start by calling the unknown number in the top brick n. The unknown numbers in the next two bricks can be written in terms of n. n 23 + nn + 62 We can now write an equation and solve it to find the value of n.

14 © Boardworks Ltd of 62 Find the missing number 23 n nn The number in each brick is found by adding the two numbers above it. What are the missing values? 23 + n+ n + 62= n = 127Simplify: 2n = 42Subtract 85: Divide by 2:n = 21 Check this solution

15 © Boardworks Ltd of 62 Equation dominoes

16 © Boardworks Ltd of 62 A2.2 Equations with the unknown on both sides Contents A2 Equations A2.5 Non-linear equations A2.4 Equations and proportion A2.1 Solving simple equations A2.3 Solving more difficult equations

17 © Boardworks Ltd of 62 Balancing equations

18 © Boardworks Ltd of 62 Constructing an equation Ben and Lucy have the same number of sweets. Ben started with 3 packets of sweets and ate 11 sweets. Lucy started with 2 packets of sweets and ate 3 sweets. How many sweets are there in a packet? Let’s call the number of sweets in a packet, n. We can solve this problem by writing the equation: 3 n – 11 The number of Ben’s sweets = is the same as the number of Lucy’s sweets. 2 n – 3

19 © Boardworks Ltd of 62 Solving the equation Let’s solve this equation by transforming both sides of the equation in the same way. 3 n – 11 = 2 n – 3 Start by writing the equation down. -2 n Subtract 2 n from both sides. n – 11 = –3 Always line up the equals signs. +11 Add 11 to both sides. n = 8 This is the solution. We can check the solution by substituting it back into the original equation: 3  8 – 11 =2  8 – 3

20 © Boardworks Ltd of 62 Constructing an equation I’m thinking of a number. When I multiply the number by 4, I get the same answer as adding 9 to the number. What number am I thinking of? Let’s call the unknown number n. We can solve this problem by writing the equation: The number multiplied by 4 is the same as 4n4n = n + 9 the number plus 9.

21 © Boardworks Ltd of 62 Solving the equation Let’s solve this equation by transforming both sides of the equation in the same way. 4n = n + 9Start by writing the equation down. -n-n-n-n Subtract n from both sides. 3n = 9Always line up the equals signs. ÷3 Divide both sides by 3. n = 3This is the solution. We can check the solution by substituting it back into the original equation: 4  3 = 3 + 9

22 © Boardworks Ltd of 62 Constructing an equation Remember, vertically opposite angles are equal. We can solve this problem by writing the equation: Find the value of x. (2 x + 5) o (65 – 2 x ) o 2 x + 5 = 65 – 2 x

23 © Boardworks Ltd of 62 Solving the equation Let’s solve this equation by transforming both sides of the equation in the same way. 2 x + 5 = 65 – 2 x +2 x Add 2 x to both sides. 4 x + 5 = Subtract 5 from both sides. 4 x = 60 Divide both sides by 4. ÷4 x = 15 Check: 2  =65 – 2  15 This is the solution.

24 © Boardworks Ltd of 62 Rectangle problem The area of this rectangle is 27 cm 2. 8 x – 142 x + 1 Calculate the value of x and use it to find the height of the rectangle. Opposite sides of a rectangle are equal. We can use this fact to write an equation in terms of x.

25 © Boardworks Ltd of 62 Rectangle problem The area of this rectangle is 27 cm 2. 8 x – 142 x x – 14 = 2 x + 1 –2x–2x –2x–2x 6 x – 14 = x = 15 ÷6 x = 2.5 If x = 2.5 we can find the height of the rectangle using substitution: 8 × 2.5 – 14 =20 – 14 =6 cm

26 © Boardworks Ltd of 62 Rectangle problem The area of this rectangle is 27 cm 2. What is its width? 8 x – 142 x + 1 Let’s call the width of the rectangle y. y If the height of the rectangle is 6 cm and the area is 27 cm 2 then we can find the width by writing the equation: 6 y = 27 ÷6 y = 4.5 The dimensions of the rectangle are 6 cm by 4.5 cm.

27 © Boardworks Ltd of 62 Select the correct equation Veronica has 58p and buys 4 chocolate bars. They both receive the same amount of change. If c is the cost of one chocolate bar, which equation could we use to solve this problem? A: 4 c + 58 = 7 c + 100B: 58 – 4 c = 1 – 7 c D: 4 c – 58 = 7 c – 1C: 58 – 4 c = 100 – 7 c Thomas has £1 and buys 7 chocolate bars.

28 © Boardworks Ltd of 62 Equation match

29 © Boardworks Ltd of 62 A2.3 Solving more difficult equations Contents A2 Equations A2.5 Non-linear equations A2.4 Equations and proportion A2.1 Solving simple equations A2.2 Equations with the unknown on both sides

30 © Boardworks Ltd of 62 Equations with brackets Equations can contain brackets. For example: 2(3 x – 5) = 4 x To solve this we can Multiply out the brackets: 6x6x –10 = 4 x + 10 Add 10 to both sides: 6 x = 4 x x Subtract 4 x from both sides:2 x = 10 ÷ 2 Divide both sides by 2: x = 5

31 © Boardworks Ltd of 62 Sometimes we can solve equations such as: 2(3 x – 5) = 4 x by first dividing both sides by the number in front of the bracket: Divide both sides by 2:3 x – 5 = 2 x Add 5 to both sides:3 x = 2 x + 5 Subtract 2 x from both sides: x = 5 In this example, dividing first means that there are fewer steps x Equations with brackets

32 © Boardworks Ltd of 62 Balancing equations with brackets

33 © Boardworks Ltd of 62 Solving equations involving division Linear equations with unknowns on both sides can also involve division. For example, 3 x = 11 – x In this case we must start by multiplying both sides of the equation by 4. 3 x + 2 = 4(11 – x ) Multiply out the brackets: 3 x + 2 = 44 – 4 x Add 4 x to both sides:7 x + 2 = 44 Subtract 2 from both sides: 7 x = 42 Divide both sides by 7: x = 6

34 © Boardworks Ltd of 62 Solving equations involving division Sometimes the expressions on both sides of the equation are divided. For example, In this example, we can multiply both sides by ( x + 3) and (3 x – 5) in one step to give: 4(3 x – 5) = 5( x + 3) Multiply out the brackets: 12 x – 20 = 5 x + 15 Subtract 5 x from both sides:7 x – 20 = 15 Add 20 to both sides: 7 x = 35 Divide both sides by 7: x = 5 4 ( x + 3) 5 (3 x – 5) =

35 © Boardworks Ltd of 62 Equivalent equations

36 © Boardworks Ltd of 62 A2.4 Equations and proportion Contents A2 Equations A2.5 Non-linear equations A2.1 Solving simple equations A2.2 Equations with the unknown on both sides A2.3 Solving more difficult equations

37 © Boardworks Ltd of 62 Sale! Footballs! Were £4, Now only £3! Original Price Sale Price£3 £4 £8 £6 £12 £9 £16 £12 £20 £15 What do you notice about these ratios? We can write the ratio of for each pair of values. sale price original price The ratios are equal. = = ===0.75

38 © Boardworks Ltd of 62 Drawing a graph Original Price, x Sale Price, y £3£6 £4£8 £9 £12 £16£20 £15 Original price, £ y Sale price, £ Sale price of footballs The points lie on a straight line through the origin. The equation of the line is: or y = 3 4 x y = 0.75 x x

39 © Boardworks Ltd of 62 Constant speed Susan walks to school at a constant speed. Altogether, it takes her 10 minutes to walk 800 metres. How far would she have walked in two minutes? Time, minutes Distance, metres The time and the distance are directly proportional. What do you notice about these ratios? We can write the ratio of for each pair of values. distance time The ratios are equal. =====80

40 © Boardworks Ltd of 62 Distance/time graph x y Time, minutes Distance, metres Susan’s walking speed We can plot the points from the table onto a graph. Time, minutes Distance, metres The points lie on a straight line through the origin. The equation of the line is: y = 80 x How far would Susan walk in: a) 3 minutes?240 metres a) 15 minutes?1200 metres

41 © Boardworks Ltd of 62 Mixing colours Orange paint is made by mixing 4 litres of yellow paint with 6 litres of red paint. To make the same shade of orange we must keep the amount of yellow paint and red paint in direct proportion. How many litres of each colour do you need to make: a) 5 litres of orange paint? b) 1 litre of orange paint? c) 7 litres of orange paint? 1 l = 10 l of orange paint 2 l of yellow and 3 l of red 0.4 l of yellow and 0.6 l of red 2.8 l of yellow and 4.2 l of red

42 © Boardworks Ltd of 62 We can write an equation linking the amount of red paint r to the amount of yellow paint y as 1 l = 10 l of orange paint r = 1.5 y The ratio red yellow is 6 4 = 1.5 How many litres of red paint would be needed to mix with 14 litres of yellow paint to make the same shade of orange? r = 1.5 × 14 = 21 l Mixing colours

43 © Boardworks Ltd of 62 Mixing colours

44 © Boardworks Ltd of 62 Equations and direct proportion When two quantities x and y are directly proportional to each other we can link them with the symbol . We write y  xy  x We can also link these variables with the equation y = kx where k is a constant value equal to. y x The graph of y = kx will always be a straight line through the origin. y is directly proportional to x.

45 © Boardworks Ltd of 62 A2.5 Non-linear equations Contents A2 Equations A2.4 Equations and proportion A2.1 Solving simple equations A2.2 Equations with the unknown on both sides A2.3 Solving more difficult equations

46 © Boardworks Ltd of 62 Non-linear equations A number added to its square equals 42. What could the number be? If we call the unknown number x we can write the following equation: x + x 2 = 42 This is a non-linear equation. It contains powers greater than 1. One solution to this equation is x = 6. There is another solution to this equation. x = – = = 42and–7 + (–7) 2 = – = 42

47 © Boardworks Ltd of 62 Non-linear equations In a linear equation, unknowns in the equation cannot be raised to any power other than 1. For example, 4 x + 7 = 23 is a linear equation. In a non-linear equation, unknowns in the equation can have indices other than 1. For example, x x = 20 is a non-linear equation. In a quadratic equation, the highest index of any of the unknowns is 2. For example, x 2 – 3 x = 10 is a quadratic equation. Quadratic equations usually have two solutions.

48 © Boardworks Ltd of 62 Solving non-linear equations Solve 3 x 2 – 5 = 22 We can solve this non-linear equation using inverse operations. 3 x 2 – 5 = 22 3 x 2 = 27Adding 5: Dividing by 3: x 2 = 9 Square rooting: x = ±  9 x = 3 or –3 When we find the square root in an equation we must always give both the positive and the negative solution.

49 © Boardworks Ltd of 62 Solving non-linear equations We must start by multiplying by m – 6. ( m – 6)( m – 6) = 25 Solve m – 6 = 25 m – 6 We can write this as: ( m – 6) 2 = 25 Square rooting: m – 6 = 5 or –5 To find both solutions we must add 6 to both 5 and –5. m = or m = –5 + 6 m = 11 m = 1

50 © Boardworks Ltd of 62 Using a calculator to solve non-linear equations 25 m – 6 Solve 4 n = 33 We can solve this equation by using inverse operations. 4 n = 33 Subtracting 5: Dividing by 4: Cube rooting: 4 n 3 = 28 n 3 = 7 n =  7 3 We know that 7 is not a cube number so we can use the key on a calculator.  3 n = 1.91 (to 2 d.p.)

51 © Boardworks Ltd of 62 Using trial and improvement Solve h h = 45 We can factorize this equation to give h ( h + 3) = 45 We can then solve the equation using trial and improvement. h h + 3 h ( h + 3) is closer to 45 than so h = 5.37 (to 2 d.p.) 5840too small 6954too big too big too small too small too big

52 © Boardworks Ltd of 62 Using trial and improvement

53 © Boardworks Ltd of 62 Using a spreadsheet Solve a 3 – 4 a = 2 We can use a spreadsheet to solve this equation by trial and improvement. After heading the columns, put your first guess into cell A2.

54 © Boardworks Ltd of 62 Using a spreadsheet In cell A3 we can type =A2+0.1 without any spaces. Clicking on the bottom right hand corner of this cell and dragging down to cell A12 will enter the numbers 2.1, 2.2, 2.3 … increasing in steps of 0.1 up to 3.

55 © Boardworks Ltd of 62 Using a spreadsheet

56 © Boardworks Ltd of 62 Using a spreadsheet

57 © Boardworks Ltd of 62 Using a spreadsheet

58 © Boardworks Ltd of 62 Using a spreadsheet We can see that the solution lies between 2.2 and 2.3. We can now change the value in cell A2 to 2.2. In cell A3 we can then enter =A without any spaces. Clicking on the bottom right hand corner of this cell and dragging down to cell A12 will enter the numbers 2.21, 2.22, 2.23 … increasing in steps of 0.01 up to 2.3.

59 © Boardworks Ltd of 62 Using a spreadsheet

60 © Boardworks Ltd of 62 Using a spreadsheet

61 © Boardworks Ltd of 62 Solving non-linear equations Using the results from the spreadsheet we can see that for a 3 – 4 a = 2 a = (to 3 d.p.) Could this equation have any more solutions? Use the spreadsheet to look at values of a 3 – 4 a between –2 and 0. Increase the values of a by 0.1. Find every solution to the equation a 3 – 4 a = 2.

62 © Boardworks Ltd of 62 Using a spreadsheet There are two more solutions to a 3 – 4 a = 2 between –1.7 and –1.6 … … and between –0.6 and –0.5.


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