# KS3 Mathematics A2 Equations

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KS3 Mathematics A2 Equations
The aim of this unit is to teach pupils to: Construct and solve linear equations, selecting an appropriate method Use systematic trial and improvement and ICT tools to find solutions, or approximate solutions, to non-linear equations Set up and use equations to solve word and other problems involving direct proportion Material in this unit is linked to the Key Stage 3 Framework supplement of examples pp A2 Equations

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

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.

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 = 3 and 10 – 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

Using inverse operations
In arithmetic, if 3 × 4 = 12, we can use inverse operations to write: 12 ÷ 4 = 3 and 12 ÷ 3 = 4 In algebra, if ab = 12, we can use inverse operations to write: 12 b = a 12 a = b and 12 b = a 12 a = b or and

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 8 + 5 = 13

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

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

What number am I thinking of…?

What number am I thinking of …?

What number am I thinking of …?
Start with x x Multiply by 2 2x and add 5 2x + 5 to give you 11. 2x + 5 = 11 We can solve this equation using inverse operations in the opposite order. Start with the equation: 2x + 5 = 11 Subtract 5: Verify that x =3 by multiplying it by 2 and adding 5 to get 11. 2x = 11 – 5 2x = 6 Divide by 2: x = 6 ÷ 2 x = 3

Solving equations by transforming both sides
Solve this equation by transforming both sides in the same way: – 1 = 2 m 4 +1 Add 1 to both sides. = 3 m 4 ×4 Multiply both sides by 4. Use this slide to go through as many examples as you need to ensure that pupils are confident with this written method for solving linear equations. Note that when we write this out the steps written in orange can be performed mentally. Every other step must be written on a new line with the equals signs lined up. Give pupils some written ‘What number am I thinking of?’ problems. Ask them to solve these by constructing an equation and solving it. Set pupils a variety of simple linear equations to solve. Start with equations that can be solved in a single step, for example 4a = 20, and progress to equations that require 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 are equivalent). m = 12 We can check the solution by substituting it back into the original equation: 12 ÷ 4 – 1 = 2

Find the missing number
The number in each brick is found by adding the two numbers above it. What are the missing values? 23 ? 62 127 n 23 + n n + 62 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. We can now write an equation and solve it to find the value of n.

Find the missing number
The number in each brick is found by adding the two numbers above it. What are the missing values? 23 n 62 21 23 + n n + 62 44 83 127 127 23 + n + n + 62 = 127 Simplify: 85 + 2n = 127 Subtract 85: 2n = 42 Divide by 2: n = 21 Check this solution.

Equation dominoes Choose 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.

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

Balancing equations Remind 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.

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 – 3 The number of Ben’s sweets is the same as the number of Lucy’s sweets.

Solving the equation Let’s solve this equation by transforming both sides of the equation in the same way. 3n – 11 = 2n – 3 Start by writing the equation down. -2n -2n Subtract 2n from both sides. n – 11 = –3 Always line up the equals signs. +11 +11 Add 11 to both sides. n = 8 This 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

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 + 9 The number multiplied by 4 is the same as the number plus 9.

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

Constructing an equation
Find the value of x. (2x + 5)o (65 – 2x)o Remember, 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

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

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 – 14 2x + 1 Opposite sides of a rectangle are equal. We can use this fact to write an equation in terms of x.

Rectangle problem The area of this rectangle is 27 cm2.
8x – 14 = 2x + 1 –2x 6x – 14 = 1 +14 8x – 14 2x + 1 6x = 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

Rectangle problem The area of this rectangle is 27 cm2. What is its width? y Let’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 – 14 2x + 1 6y = 27 The dimensions of the rectangle are 6 cm by 4.5 cm. ÷6 ÷6 y = 4.5

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 + 100 B: 58 – 4c = 1 – 7c C: 58 – 4c = 100 – 7c C: 58 – 4c = 100 – 7c D: 4c – 58 = 7c – 1

Equation match Select 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.

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

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

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

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.

Solving equations involving division
Linear equations with unknowns on both sides can also involve division. For example, 3x + 2 4 = 11 – x In 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 – 4x Talk 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 = 44 Subtract 2 from both sides: 7x = 42 Divide both sides by 7: x = 6

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 + 15 Talk 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 = 15 Add 20 to both sides: 7x = 35 Divide both sides by 7: x = 5

Equivalent equations Ask 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.

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

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

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

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 10 800 2 4 6 8 160 320 480 640 We can write the ratio of for each pair of values. distance time 160 2 320 4 480 6 640 8 800 10 In 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. = = = = = 80 The ratios are equal. What do you notice about these ratios? The time and the distance are directly proportional.

Distance/time graph We can plot the points from the table onto a graph. Time, minutes Distance, metres 160 320 2 4 480 6 640 8 10 800 Susan’s walking speed The points lie on a straight line through the origin. x y Time, minutes Distance, metres 2 6 8 10 160 320 480 640 800 4 The equation of the line is: y = 80x We 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 metres a) 15 minutes? 1200 metres

Mixing colours Orange paint is made by mixing 4 litres of yellow paint with 6 litres of red paint. 1 l = 10 l of orange 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: 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 red b) 1 litre of orange paint? 0.4 l of yellow and 0.6 l of red c) 7 litres of orange paint? 2.8 l of yellow and 4.2 l of red

Mixing colours 1 l = 10 l of orange paint The ratio is = 1.5
red yellow is 6 4 = 1.5 We can write an equation linking the amount of red paint r to the amount of yellow paint y as r = 1.5y How 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 to write down the answers. After a designated time stop to discuss the answers. Yellow paint: red paint = 4:6 = 2:3 4/10 = 2/5 6/10 = 3/5 Stress that to produce the required shade of orange these ratios and proportions must 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 be steeper or less steep? If we plotted the graph for a lighter shade of orange, would the line be Set pupils a variety of problems involving using graphs and setting up equations to solve simple problems involving direct proportion. r = 1.5 × 14 = 21 l

Mixing colours Plenary – Mixing colours

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 write y  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.

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

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 = 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 = –7 = = 42 and –7 + (–7)2 = – = 42

Non-linear equations In 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.

Solving non-linear equations
Solve 3x2 – 5 = 22 We can solve this non-linear equation using inverse operations. 3x2 – 5 = 22 Adding 5: 3x2 = 27 Dividing by 3: x2 = 9 Square rooting: x = ±9 Stress that the inverse of squaring is square rooting. Also, stress that if x2 = 9 then the equation has two solutions: 3 or –3. This is because both of these solve the equation. If necessary remind pupils that (–3)2 = 9 because when we multiply two negative numbers together we have a positive number. (–3 × –3 = 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.

Solving non-linear equations
Solve m – 6 = 25 m – 6 We must start by multiplying by m – 6. (m – 6)(m – 6) = 25 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 = 5 + 6 or m = –5 + 6 m = 11 m = 1

Using a calculator to solve non-linear equations
25 m – 6 Solve 4n3 + 5 = 33 We can solve this equation by using inverse operations. 4n3 + 5 = 33 Subtracting 5: 4n3 = 28 Dividing by 4: n3 = 7 Cube rooting: 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.)

Using trial and improvement
Solve h2 + 3h = 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) 5 8 40 too small 6 9 54 too big 5.4 8.4 45.36 too big By 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). 5.3 8.3 43.99 too small 5.37 8.37 too small 5.38 8.36 too big is closer to 45 than so h = 5.37 (to 2 d.p.)

Using trial and improvement

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.

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.

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.