Presentation on theme: "KS3 Mathematics A2 Equations"— Presentation transcript:
1 KS3 Mathematics A2 Equations The aim of this unit is to teach pupils to:Construct and solve linear equations, selecting an appropriate methodUse systematic trial and improvement and ICT tools to find solutions, or approximate solutions, to non-linear equationsSet up and use equations to solve word and other problems involving direct proportionMaterial in this unit is linked to the Key Stage 3 Framework supplement of examples ppA2 Equations
2 A2.1 Solving simple equations ContentsA2 EquationsA2.1 Solving simple equationsA2.2 Equations with the unknown on both sidesA2.3 Solving more difficult equationsA2.4 Equations and proportionA2.5 Non-linear equations
3 EquationsAn 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 = 13x = 6When we solve an equation we always line up the equals signs.
4 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 = 7In algebra, if a + b = 10, we can use inverse operations to write:10 – b = aand10 – a = ba = 10 – bandb = 10 – aor
5 Using inverse operations In arithmetic, if 3 × 4 = 12, we can use inverse operations to write:12 ÷ 4 = 3and12 ÷ 3 = 4In algebra, if ab = 12, we can use inverse operations to write:12b=a12a=band12b=a12a=borand
6 Using inverse operations to solve equations We can use inverse operations to solve simple equations.For example,x + 5 = 13x = 13 – 5x = 8Always 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 have8 + 5 = 13
7 Using inverse operations to solve equations Solve the following equations using inverse operations.5x = 4517 – x = 6x = 45 ÷ 517 = 6 + xx = 917 – 6 = xWe always write the letter before the equals sign.11 = xCheck:x = 115 × 9 = 45Check:17 – 11 = 6
8 Using inverse operations to solve equations Solve the following equations using inverse operations.= 3x73x – 4 = 14x = 3 × 73x =x = 213x = 18x = 18 ÷ 3Check:x = 6= 3217Check:3 × 6 – 4 = 14
9 What number am I thinking of…? Start by asking pupils to solve the problem without using diagrams. If someone finds the answer ask them how they worked it out.They may have used trial and error, trying different numbers until they found one that worked. Stress that although this will give you the correct answer eventually this method can take a long time.Talk about what operations have been performed on the number and reveal each stage of the first diagram.Now, explain that if we start with the answer and ‘undo’ the operations we can find the number we started with.Pupils should know that to ‘undo’ addition we subtract and to ‘undo’ multiplication we divide. These are called inverse operations. Inverse means opposite.Stress that we must also reverse the order.(When you get dressed in the morning you put on your socks and then your shoes. When you ‘undo’ this when you get undressed you have to take off your shoes and then socks. The order is reversed.)Reveal each step in the second diagram to find the mystery number.Repeat the activity several times to ensure that pupils can complete these steps mentally.This activity will also practise multiplication tables to 10 x 10 and the associated division facts.
10 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.Tell pupils that this lesson is about how to solve equations. Ask pupils if they know how to recognise an equation. Establish that an equation is made up of an algebraic expression (or expressions) and a single equals sign.We can write the ‘What number am I thinking of?’ problems from the start of the lesson as equations.We use a letter to stand for the unknown number.Discuss briefly the fact that it doesn’t strictly matter what letter we choose. The letter x is often used to stand for an unknown number. Remind pupils that when we write the letter x in algebra we should use a script x (like a back-to-front c and a normal c joined together) so that it does not look like the symbol for multiplication.It is also common to use n for number. Letters like o, l and s are often avoided. Ask pupils why this might be. (O can look like 0, l can look like 1 and s can look like 5. If we do use these letters we have to be very careful to make them look different to numbers.)Reveal on the board the steps required to write x = 11 as an equation. In particular, stress the fact that when we multiply a letter by a number we don’t use a multiplication sign and that we must write the number before the letter.Ask pupils, How would I write x divided by 2? Remind pupils that in algebra this is written like a fraction x/2 not x ÷ 2.Start with xxMultiply by 22xand add 52x + 5to give you 11.2x + 5 = 11
11 What number am I thinking of …? Start with xxMultiply by 22xand add 52x + 5to give you 11.2x + 5 = 11We can solve this equation using inverse operations in the opposite order.Start with the equation:2x + 5 = 11Subtract 5:Verify that x =3 by multiplying it by 2 and adding 5 to get 11.2x = 11 – 52x = 6Divide by 2:x = 6 ÷ 2x = 3
12 Solving equations by transforming both sides Solve this equation by transforming both sides in the same way:– 1 = 2m4+1Add 1 to both sides.= 3m4×4Multiply both sides by 4.Use this slide to go through as many examples as you need to ensure thatpupils are confident with this written method for solving linear equations. Notethat when we write this out the steps written in orange can be performedmentally. Every other step must be written on a new line with the equals signslined up.Give pupils some written ‘What number am I thinking of?’ problems. Askthem to solve these by constructing an equation and solving it. Set pupils avariety of simple linear equations to solve. Start with equations that can besolved in a single step, for example 4a = 20, and progress to equations thatrequire two steps, for example, 4b + 7 = 9. Include examples such as 6 = 2p –8 and tell pupils to start by writing this as 2p – 8 = 6 (Stress that these two areequivalent).m = 12We can check the solution by substituting it back into the original equation:12 ÷ 4 – 1 = 2
13 Find the missing number The number in each brick is found by adding the two numbers above it. What are the missing values?23?62127n23 + nn + 62We 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.We can now write an equation and solve it to find the value of n.
14 Find the missing number The number in each brick is found by adding the two numbers above it. What are the missing values?23n622123 + nn + 62448312712723 + n+ n + 62= 127Simplify:85 + 2n = 127Subtract 85:2n = 42Divide by 2:n = 21Check this solution.
15 Equation dominoesChoose a volunteer to come to the board and move the dominoes into place.Members of the class must tell the volunteer where to place the dominoes so that the correct solutions are connected to the corresponding equation. Encourage pupils to use substitution to check their suggestions.
16 A2.2 Equations with the unknown on both sides ContentsA2 EquationsA2.1 Solving simple equationsA2.2 Equations with the unknown on both sidesA2.3 Solving more difficult equationsA2.4 Equations and proportionA2.5 Non-linear equations
17 Balancing equationsRemind pupils that we can think of equations as a pair of scales that balance. If we add, subtract, multiply or divide both sides of an equation by the same amount the equation will remain true.Today we will look at equations with the unknown on both sides of the equals sign. We can’t solve these equations using inverse operations because different operations are performed on the unknown on either side of the bracket.The way that we solve these equations is by doing the same thing to both sides, step by step, until we reach the solution in the form x = ‘a number’.To reach this solution we must aim to collect unknowns on the left hand side and numbers on the right hand side.Talk through as many examples as necessary on the board, asking pupils to suggest the operations to be performed. Ask pupils to justify the operations and the order in which they are done.Note that if the coefficient of x (the number of x’s) on the right hand side of the equation is greater than the coefficient of x on the left hand side, we will have a negative coefficient of x when we collect the x’s on the left hand side. To ‘undo’ this we must divide by a negative number in the final step. As an alternative, we could collect x’s on the right hand side and swap the left hand side and the right hand side in the final stage or we could swap the left hand side and right hand side at the beginning.
18 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.Talk through the problem on the board and discuss its representation as an equation.We can solve this problem by writing the equation:3n – 11=2n – 3The number of Ben’s sweetsis the same asthe number of Lucy’s sweets.
19 Solving the equationLet’s solve this equation by transforming both sides of the equation in the same way.3n – 11 = 2n – 3Start by writing the equation down.-2n-2nSubtract 2n from both sides.n – 11 = –3Always line up the equals signs.+11+11Add 11 to both sides.n = 8This is the solution.Go through these steps to solve the equation step by step:Write down the equation.We want to have the unknown n on the left hand side of the equals sign only so we should start by subtracting 2n from both sides of the equation. We must always perform the same operation to both sides of the equation. 3n – 11 – 2n = n - 11 and 2n – 3 – 2n = -3. This step is usually done mentally.Remind pupils that we want the equation in the form n = ‘a number’. To remove the – 11 from the left hand side we need to add 11 to both sides.n – = n and – = 8. The solution to the equation is n = 8.Lastly, we can check the solution by substituting it back into the original equation. Make sure that the left hand side of the equation is equal to the right hand side. 3 × 8 – 11 is equal to 13 as is 2 × 8 – 3.We can check the solution by substituting it back into the original equation:3 8 – 11 =2 8 – 3
20 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.Talk through the problem on the board and discuss its representation as an equation.We will need to solve this equation by doing the same thing to both sides of the equation.We can solve this problem by writing the equation:4n=n + 9The number multiplied by 4is the same asthe number plus 9.
21 Solving the equationLet’s solve this equation by transforming both sides of the equation in the same way.4n = n + 9Start by writing the equation down.-n-nSubtract n from both sides.3n = 9Always line up the equals signs.÷3÷3Divide both sides by 3.n = 3This is the solution.Go through these steps to solve the equation step by step:Write down the equation.We want to have the unknown n on the left hand side of the equals sign only so we should start by subtracting n from both sides of the equation. We must always perform the same operation to both sides of the equation. (We can think of an equation as a pair of balancing scales, whatever we do to one side we must do to the other to keep the two sides of the equals sign equal). 4n - n = 3n and n n = 9. This step is usually done mentally.Remember 3n means 3 n. We must therefore divide both sides of the equation by 3 to ‘undo’ the multiplication by 3.3n divided by 3 is n and 9 divided by 3 is 3.When we get to just a letter symbol equal to a value we stop. This is our solution.Lastly, we can check the solution by substituting it back into the original equation. Make sure that the left hand side of the equation is equal to the right hand side.Tell pupils that the steps shown in orange can be done mentally once they aremore confidant.Set pupils a variety of linear equations to solve with unknowns on both sides.Encourage pupils to check their answers by substituting them back into theoriginal equation.We can check the solution by substituting it back into the original equation:4 3 =3 + 9
22 Constructing an equation Find the value of x.(2x + 5)o(65 – 2x)oRemember, vertically opposite angles are equal.Talk through the problem on the board and discuss its representation as an equation.We can solve this problem by writing the equation:2x + 5 = 65 – 2x
23 Solving the equationLet’s solve this equation by transforming both sides of the equation in the same way.2x = 65 – 2x+2x+2xAdd 2x to both sides.4x = 65-5-5Subtract 5 from both sides.4x = 60Go through the steps to solve the equation.Establish that if x = 15 the unknown angle in the diagram in the previous slide is 35o.÷4÷4Divide both sides by 4.x = 15This is the solution.Check:2 =65 – 2 15
24 Rectangle problem The area of this rectangle is 27 cm2. Calculate the value of x and use it to find the height of the rectangle.8x – 142x + 1Opposite sides of a rectangle are equal.We can use this fact to write an equation in terms of x.
25 Rectangle problem The area of this rectangle is 27 cm2. 8x – 14 = 2x + 1–2x6x – 14 = 1+148x – 142x + 16x = 15÷6x = 2.5If x = 2.5 we can find the height of the rectangle using substitution:8 × 2.5 – 14 =20 – 14 =6 cm
26 Rectangle problemThe area of this rectangle is 27 cm2. What is its width?yLet’s call the width of the rectangle y.If the height of the rectangle is 6 cm and the area is 27 cm2 then we can find the width by writing the equation:8x – 142x + 16y = 27The dimensions of the rectangle are 6 cm by 4.5 cm.÷6÷6y = 4.5
27 Select the correct equation Thomas has £1 and buys 7 chocolate bars.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?Ask pupils to select the equation that can be used to solve the problem. Explain why £1 must be written as 100.Ask pupils to solve the equation and show that if c =14 the amount of change that they both received was 2p.A: 4c + 58 = 7c + 100B: 58 – 4c = 1 – 7cC: 58 – 4c = 100 – 7cC: 58 – 4c = 100 – 7cD: 4c – 58 = 7c – 1
28 Equation matchSelect a volunteer to come to the board to match an equation to a solution. Pupils may chose to partially solve the equation and then use substitution to check the answer. Discuss the methods used.Select a different volunteer to come to the board each time. Classmates may make suggestions. Towards the end the solution can be found by substituting each remaining solution into the required equation until the equation balances.
29 A2.3 Solving more difficult equations ContentsA2 EquationsA2.1 Solving simple equationsA2.2 Equations with the unknown on both sidesA2.3 Solving more difficult equationsA2.4 Equations and proportionA2.5 Non-linear equations
30 Equations with brackets Equations can contain brackets. For example:2(3x – 5) = 4xTo solve this we canMultiply out the brackets:6x–10= 4x+ 10+ 10Add 10 to both sides:6x = 4x + 10Talk through this method for solving an equation containing brackets.Check the answer of 5 by substituting it back into the original equation and verifying that the left hand side of the equation equals the right hand side.Explain that generally, if an equation contains brackets, we should start by multiplying them out.- 4x- 4xSubtract 4x from both sides:2x = 10÷ 2÷ 2Divide both sides by 2:x = 5
31 Equations with brackets Sometimes we can solve equations such as:2(3x – 5) = 4xby first dividing both sides by the number in front of the bracket:Divide both sides by 2:3x – 5 = 2x+ 5+ 5Add 5 to both sides:3x = 2x + 5Explain that we could also solve this equation by first dividing both sides by the number in front of the bracket. Note that we can only do this if the term on the right hand side is divisible by the number in front of the bracket on the left hand side.Point out that using this method in this example involves fewer steps and is therefore more efficient. However, if pupils are unsure of what to do it may be best to multiply out any brackets first.- 2x- 2xSubtract 2x from both sides:x = 5In this example, dividing first means that there are fewer steps.
32 Balancing equations with brackets Go through some more examples using the pen tool to record each stage.Ask pupils to tell you each transformation to be performed on both sides of the equation and to justify their choice in each case.
33 Solving equations involving division Linear equations with unknowns on both sides can also involve division.For example,3x + 24= 11 – xIn this case we must start by multiplying both sides of the equation by 4.3x + 2 = 4(11 – x)Multiply out the brackets:3x + 2 = 44 – 4xTalk about the equation on the board. Explain that when we have (3x + 2)/ 4 the dividing line acts as a bracket – that means that we add before we divide. To undo this, then we must start by multiplying by 4. If we multiply the left hand side by 4 we must also multiply the right hand side by 4.Ensure that all pupils are happy with this step before moving on to solve the equation.Set pupils a variety of equations to solve, with and without brackets, and with the unknown on either or both sides.Add 4x to both sides:7x + 2 = 44Subtract 2 from both sides:7x = 42Divide both sides by 7:x = 6
34 Solving equations involving division Sometimes the expressions on both sides of the equation are divided.For example,4(x + 3)5(3x – 5)=In this example, we can multiply both sides by (x + 3) and (3x – 5) in one step to give:4(3x – 5) = 5(x + 3)Multiply out the brackets:12x – 20 = 5x + 15Talk about the equation on the board. Explain that when we have (3x + 2)/ 4 the dividing line acts as a bracket – that means that we add before we divide. To undo this, then we must start be multiplying by 4. If we multiply the left hand side by 4 we must also multiply the right hand side by 4.Ensure that all pupils are happy with this step before moving on to solve the equation.Set pupils a variety of equations to solve, with and without brackets, and with the unknown on either or both sides.Subtract 5x from both sides:7x – 20 = 15Add 20 to both sides:7x = 35Divide both sides by 7:x = 5
35 Equivalent equationsAsk pupils to tell you which equations they believe to be equivalent to the one in the window.Clicking on an expression will reveal whether of not it is equivalent to the required expression. Each time an equivalent expression is found ask pupils to tell you what has been done to both sides of the equation to make the selected equivalent expression.Ask pupils to solve the equation in the window and substitute this solution into the equivalent equations found to verify the solution.
36 A2.4 Equations and proportion ContentsA2 EquationsA2.1 Solving simple equationsA2.2 Equations with the unknown on both sidesA2.3 Solving more difficult equationsA2.4 Equations and proportionA2.5 Non-linear equations
37 Sale! Footballs! Were £4, Now only £3! Original Price Sale Price £3 £4 £8£12£16£20£6£9£12£15We can write the ratio of for each pair of values.sale priceoriginal priceDiscuss the fact that if we buy twice as many footballs we will pay £6 instead of £8. If we buy three times as many footballs we will pay £12 instead of £9.Show how these values can be entered into a table.Discuss the ratio of the sale price:original price for each pair of values.Pupils should note that the ratio of sale price:original price is always equal to 3:4. Stress that if the ratio is constant then the two amounts must be directly proportional.346891212161520=====0.75The ratios are equal.What do you notice about these ratios?
38 Drawing a graph Original Price, x Sale Price, y £3 £6 £4 £8 £9 £12 £16 £20£15ySale price of footballs15The points lie on a straight line through the origin.12The equation of the line is:9Sale price, £6y =34xExplain to pupils that we can draw a graph of the Sale price against the original price to demonstrate the relationship between the two values.Ask a volunteer to click on the graph to plot the points given by the table. When the points have been correctly positioned the draw line button will become highlighted. Clicking on it will connect the points with a straight line through the origin.Stress that graphs that show direct proportion are always straight lines and that they always pass through the origin. Ask pupils to explain, by referring to the equation of the line, why this is the case.3ory = 0.75xx48121620Original price, £
39 Constant speedSusan 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, minutesDistance, metres108002468160320480640We can write the ratio of for each pair of values.distancetime160232044806640880010In this next example, discuss the fact that if Susan is walking at constant speed, then in two minutes she will have walked a fifth of the total distance, 160 metres. In four minutes she will have walked double this distance, 320 metres.Show how these values can be entered into a table.Discuss the ratio of the sale price:original price for each pair of values.Pupils should note that the ratio of sale price:original price is always equal to 3:4. Stress that if the ratio is constant then the two amounts must be directly proportional.=====80The ratios are equal.What do you notice about these ratios?The time and the distance are directly proportional.
40 Distance/time graphWe can plot the points from the table onto a graph.Time, minutesDistance, metres160320244806640810800Susan’s walking speedThe points lie on a straight line through the origin.xyTime, minutesDistance, metres268101603204806408004The equation of the line is:y =80xWe can draw a graph of the distance against the time to demonstrate the relationship between the two values.Ask a volunteer to click on the graph to plot the points given by the table. When the points have been correctly positioned the draw line button will become highlighted. Clicking on it will connect the points with a straight line through the origin.Again, stress that graphs that show direct proportion are always straight lines and that they always pass through the origin.Ask pupils, Suppose Susan did not walk at a constant speed. Would the time and distance be directly proportional?Discuss how to calculate the distance walked in a given time, using the graph in the first instance and the equation of the graph in the second.How far would Susan walk in:a) 3 minutes?240 metresa) 15 minutes?1200 metres
41 Mixing coloursOrange paint is made by mixing 4 litres of yellow paint with 6 litres of red paint.1 l= 10 l of orange paintTo 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:Discuss the situation involving mixing a particular shade of paint.Conclude that to make 5 litres of the same shade we would need half as much of each colour: 2 litres of yellow and 3 litres of red.To make a tenth of the total amount of paint we would need a tenth of each colour: 400 ml of yellow and 600 ml of red.To make 7 litres of paint we need to multiply the amount needed to make one litre of paint by 7: 2.8 litres of yellow and 4.2 litres of red.a) 5 litres of orange paint?2 l of yellow and 3 l of redb) 1 litre of orange paint?0.4 l of yellow and 0.6 l of redc) 7 litres of orange paint?2.8 l of yellow and 4.2 l of red
42 Mixing colours 1 l = 10 l of orange paint The ratio is = 1.5 redyellowis64= 1.5We can write an equation linking the amount of red paint r to the amount of yellow paint y asr = 1.5yHow many litres of red paint would be needed to mix with 14 litres of yellow paint to make the same shade of orange?Ask pupils to answer the questions on the board individually or in pairs and towrite down the answers.After a designated time stop to discuss the answers.Yellow paint: red paint = 4:6 = 2:34/10 = 2/56/10 = 3/5Stress that to produce the required shade of orange these ratios and proportionsmust remain constant regardless of how much paint is required.Establish that if the number of litres of red goes up in threes, then the number of litres of yellow goes up in twos.Check pupils’ graphs. Remind pupils that the axes should be labelled and the the graph should have a title.The equation of the line can be written as 3y=2x or y = 2/3 x.Ask pupils:If we plotted the graph for a darker shade of orange, would the line besteeper or less steep?If we plotted the graph for a lighter shade of orange, would the line beSet pupils a variety of problems involving using graphs and setting upequations to solve simple problems involving direct proportion.r = 1.5 × 14= 21 l
43 Mixing colours Plenary – Mixing colours Use this activity to review the main points of the lesson.Start by demonstrating how to produce various shades of colour by varying the gradient of the line.Ask pupils,If this line is steeper, will the resulting colour be darker or lighter?Move the end point of the line to the point (5, 10). Ask pupils to justify why a darker colour is produced by interpreting the graph as showing that for 5 parts yellow there are 10 parts blue. For this shade, the ratio of blue to yellow is 10:5 which simplifies to 2:1. The equation of the graph is y=2x.Ask pupils to tell you the proportion of yellow paint required to make this shade of paint (1/3).Ask pupils to tell you how much blue paint and how much yellow paint would be required to make 45 litres of this shade of paint.Repeat for different shades.Hide the ratios and the equation of the line and ask pupils to work these out.Conclude by stressing that when two quantities are directly proportional we can plot points of proportional pairs to produce a straight line passing through the origin. The ratio of the proportional pairs will be constant.
44 Equations and direct proportion When two quantities x and y are directly proportional to each other we can link them with the symbol .y is directly proportional to x.We writey xWe can also link these variables with the equationy = kxwhere k is a constant value equal to .yxThe graph of y = kx will always be a straight line through the origin.
45 A2.5 Non-linear equations ContentsA2 EquationsA2.1 Solving simple equationsA2.2 Equations with the unknown on both sidesA2.3 Solving more difficult equationsA2.4 Equations and proportionA2.5 Non-linear equations
46 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 + x2 = 42This 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 = –7= = 42and–7 + (–7)2 = – = 42
47 Non-linear equationsIn a linear equation, unknowns in the equation cannot be raised to any power other than 1.For example, 4x + 7 = 23 is a linear equation.In a non-linear equation, unknowns in the equation can have indices other than 1.For example, x4 + 2x = 20 is a non-linear equation.In a quadratic equation, the highest index of any of the unknowns is 2.The quadratic equation in this example has two solutions: x = 5 and x = –2. this can be verified by substitution.For example, x2 – 3x = 10 is a quadratic equation.Quadratic equations usually have two solutions.
48 Solving non-linear equations Solve 3x2 – 5 = 22We can solve this non-linear equation using inverse operations.3x2 – 5 = 22Adding 5:3x2 = 27Dividing by 3:x2 = 9Square rooting:x = ±9Stress that the inverse of squaring is square rooting.Also, stress that if x2 = 9 then the equation has two solutions: 3 or –3. This isbecause both of these solve the equation. If necessary remind pupils that (–3)2= 9 because when we multiply two negative numbers together we have apositive number. (–3 × –3 = 9)x = 3 or –3When we find the square root in an equation we must always give both the positive and the negative solution.
49 Solving non-linear equations Solve m – 6 =25m – 6We must start by multiplying by m – 6.(m – 6)(m – 6) = 25We can write this as:(m – 6)2 = 25Square rooting:m – 6 = 5 or –5To find both solutions we must add 6 to both 5 and –5.m = 5 + 6orm = –5 + 6m = 11m = 1
50 Using a calculator to solve non-linear equations 25m – 6Solve 4n3 + 5 = 33We can solve this equation by using inverse operations.4n3 + 5 = 33Subtracting 5:4n3 = 28Dividing by 4:n3 = 7Cube rooting:n = 73We know that 7 is not a cube number so we can use thekey on a calculator.3n = 1.91 (to 2 d.p.)
51 Using trial and improvement Solve h2 + 3h = 45We can factorize this equation to give h(h + 3) = 45We can then solve the equation using trial and improvement.hh + 3h(h + 3)5840too small6954too big5.48.445.36too bigBy substituting h = 5 and h = 6, we can see that the solution lies between these two numbers. Since 40 is closer to 45 than 54, we can try 5.3 or 5.4 for our next guess.We see that the answer is between 5.4 and 5.3, but closer to 5.4 (because is closer to 45).5.38.343.99too small5.378.37too small5.388.36too bigis closer to 45 than so h = 5.37 (to 2 d.p.)
53 Using a spreadsheet Solve a3 – 4a = 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 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.
58 Using a spreadsheetWe 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.
61 Solving non-linear equations Using the results from the spreadsheet we can see that fora3 – 4a = 2a = (to 3 d.p.)Could this equation have any more solutions?Use the spreadsheet to look at values of a3 – 4a between –2 and 0. Increase the values of a by 0.1.Find every solution to the equation a3 – 4a = 2.
62 Using a spreadsheetThere are two more solutions to a3 – 4a = 2 between –1.7 and –1.6 …… and between –0.6 and –0.5.