KS3 Mathematics A3 Formulae The aim of this unit is to teach pupils to: Use letter symbols and distinguish their different roles in algebra Use formulae from mathematics and other subjects Material in this unit is linked to the Key Stage 3 Framework supplement of examples pp 138-143. A3 Formulae

A3.1 Introducing formulae Contents A3 Formulae A A3.1 Introducing formulae A A3.2 Using formulae A A3.3 Changing the subject of a formula A A3.4 Deriving formulae

How can we work out the number of days in a given number of weeks? Formulae What is a formula? In maths a formula is a rule for working something out. The plural of formula is formulae. For example: Ask pupils where they have met the word formula before, in maths or other subjects. A formula is a rule for working something out. For example, the rule for finding the area of a rectangle is: the area of a rectangle is equal to its length times its width. Discuss briefly how we could find how many days there are in a given number of weeks. So what do we do to the number of weeks to find the number of days? (Times the number of days by 7) Click to reveal the formula in words. How can we work out the number of days in a given number of weeks? number of days = 7 × number of weeks

Writing formulae in words What formula would we write to work out the number of table legs in the classroom? Number of table legs = number of tables____ × 4 We can now work out the number of table legs for any given number of tables. In this slide we are assuming that each table has four legs. Lead pupils to the formula by asking how we could work out the number of table legs in the classroom. Test the formula for different numbers of tables. What if I told you there were 52 table legs in the room. How could you work out the number of tables? If there are 16 tables in the classroom: Number of table legs = 16 × 4 = 64

Writing formulae in words Mark is 5 years older than his sister, Kate. What formula would we write to work out Mark’s age if we are given Kate’s age? Mark’s age = Kate’s age____ + 5 If Kate is 8 years old: Mark’s age = 8 + 5 = 13 Emphasize that we should write the amount we are working out (Mark’s age) in front of the equals sign. The amount that we are given (Kate’s age) is used in the formula. Go through the examples. What formula would we use to work out Kate’s age if we are given Mark’s age? This is Kate’s age = Mark’s age – 5. Repeat that we should test formulae by substituting in values. For example, if Mark is 17, Kate is 17 – 5, 12. If Kate is 49 years old: Mark’s age = 49 + 5 = 54

Writing formulae in symbols Formulae are normally written using letters instead of words. Each letter in a formula represents a numerical amount. For example: The perimeter of a square = 4 × the length of one of the sides. s We can write this as, s s Point out that the letters in formulae represent numbers. Formulae never contain units. We can call every side of the square s because s is the same numerical value for each side. If we were writing the formula for the perimeter of a rectangle we would have to use different letters for the lengths of the opposite sides (For example, p = 2l + 2w, where l is the length and w is the width.) When we write a formula using letters we should state what the letters stand for. Tell pupils that when we write a formula using letters, we often choose letters from the beginning of one of the words. Point out that if you choose a capital letter to represent an amount, you have to stick to a capital letter. You can’t change to a small letter at any point nor can you change a small letter to a capital letter. p = 4s s Where p is the perimeter of the square and s is the length of a side.

Writing formulae in symbols Sally earns a basic salary of £200 a week working in an office. She also earns £7 an hour for each hour of overtime she does. What formula could we use to work out Sally’s weekly income? Weekly income, in pounds = 7 × number of extra hours + 200 We can write this in symbols as: When we write a formula using letter symbols we do not include units, in this case pounds. Many units are represented by letters, m for metres, for example. In a formula every letter represents an amount, so we can’t use letters to represent units. This should not cause any problems as long as all units in the formula are the same. Set pupils written practice exercises involving: Inserting integers into formulae expressed in words. Substituting integers into formulae expressed in symbols. Deriving simple algebraic formulae from given situations. W = 7h + 200 How much does Sally earn if she works 9 hours overtime? W = 7 × 9 + 200 = £263

Examples of formulae Remember, a formula is a special type of equation that links physical variables. Here are some examples of well-known formulae from maths. The area, A, of rectangle of length, l, and width, w, is given by the formula: A = lw Start by asking pupils what a formula (plural formulae) is and if they can give you any examples from mathematics or anywhere else. Explain that a formula is a special kind of equation that links ‘real-life’ physical variables. A formula is a rule for working something out. Remind pupils that the letters stand for numerical amounts and should not contain units. Also remind pupils that we do not generally use the multiplication symbol in algebra and that lw means l × w. Examples of other formulae that pupils may have met before include: A = ½bh (The area A of a right-angled triangle with base b and height h). S = D/T (Speed = Distance ÷ Time). F = ma (Force = mass × acceleration). E = mc2 (Energy = mass × the speed of light squared). The perimeter, P, of a rectangle is given by the formula: P = 2(l + w)

Writing formulae Write a formula to work out: The cost, c, of b boxes of crisps at £3 each. c = 3b The distance left, d, of a 500 km journey after travelling k km. d = 500 – k For each example substitute some actual values to help pupils determine the correct operations in each formula. The cost per person, c, if a meal costing m pounds is shared between p people. c = m p

Writing formulae The number of seats in a theatre, n, with 25 seats in each row, r. n = 25r The age of a boy Andy, a, if he is 5 years older than his sister Betty, b. a = b + 5 The average weight, w, of Alex who weighs a kg, Bob who weighs b kg and Claire who weighs c kg. w = a + b + c 3 1 of 20

Castle entrance prices Adult £3 Stony Castle Child £2 Entrance fee What formula could we use to work out the total cost in pounds, T, for a number of adults, A and a number of children, C, to visit the castle? Again, pupils will find it easier to derive this formula if they are given some actual examples: How much would it cost for 2 adults and 5 children to visit the castle? How did you work it out? How much would it cost for 4 adults and 10 children? Establish that to find the total cost we need to add the number of adults × £3 to the number of children × £2. From there derive the formula using letter symbols and reveal it on the board. Point out that we don’t need to include units (£s) in the formula because T stands for the total cost in pounds. T = 3A + 2C

Castle entrance prices 3A + 2C Using this formula, how much would it cost for 4 adults and a class of 32 children to visit the castle? We substitute the values into the formula: T = 3 × 4 + 2 × 32 Talk through this example using the formula. Link: A1 Algebraic expressions – substitution. = 12 + 64 = 76 It will cost £76.

Newspaper advert To place an advert in a local newspaper costs £15. There is then an additional charge of £2 for each word used. Write a formula to work out the total cost in pounds, C, to place an advert containing n words. C = 15 + 2n How much would it cost to place an advert containing 27 words? Again give pupils some example cases to help them to derive the formula. Establish that to find the total cost we need to add 15 to 2 × the number of words. From there derive the formula using letter symbols and reveal it on the board. (An alternative would be C = 2n + 15). C = 15 + 2 × 27 = 15 + 54 = 69 It will cost £69.

A3 Formulae Contents A A3.1 Introducing formulae A A3.2 Using formulae A3.3 Changing the subject of a formula A A3.4 Deriving formulae

Substituting into formulae h w l The surface area S of a cuboid is given by the formula S = 2lw + 2lh + 2hw Discuss the problem involved in using different units in a formula. Emphasize that before we can substitute physical quantities into a formula we must ensure that they are written in the same units. Links: A1 Algebraic expressions – formulae. S8 Perimeter area and volume – volume. where l is the length, w is the width and h is the height. What is the surface area of a cuboid with a length of 1.5 m, a width of 32 cm and a height of 250 mm?

Substituting into formulae What is the surface area of a cuboid with a length of 1.5 m, a width of 32 cm and a height of 250 mm? Before we can use the formula we must write all of the amounts using the same units. l = 150 cm, w = 32 cm and h = 25 cm Next, substitute the values into the formula without the units. S = 2lw + 2lh + 2hw Point out that 18 700 cm2 is equal to 1.87 m2. Remind pupils that 1 m2 = 1 m × 1 m = 100 cm × 100 cm = 10000 cm2. Link: S7 Measures – Converting units. = (2 × 150 × 32) + (2 × 150 × 25) + (2 × 25 × 32 ) = 9600 + 7500 + 1600 = 18 700 cm2 Don’t forget to write the units in at the end.

Substituting into formulae The distance D, in metres, that an object falls after being dropped is given by the formula: D = 4.9t2 where t is the time in seconds. Suppose a boy drops a rock from a 100 metre high cliff. How far will the rock have fallen after: a) 2 seconds b) 3 seconds c) 5 seconds? When t = 2, When t = 3, When t = 5, Explain that this formula is true for any object (whatever its mass) that is dropped on earth (disregarding air resistance). This is because gravity causes objects to fall with the same acceleration regardless of their mass. Talk through each substitution. Ask pupils if they know what it is wrong with the answer to part c. (The cliff is only 100 metres high so the rock cannot fall 122.5 metres. We can therefore conclude that the rock would have hit the ground by then.) D = 4.9 × 22 D = 4.9 × 32 D = 4.9 × 52 = 4.9 × 4 = 4.9 × 9 = 4.9 × 25 = 19.6 metres = 44.1 metres = 122.5 metres

Substituting into formulae and solving equations w l The formula used to find the perimeter P of a rectangle with length l and width w is: We are told the perimeter and the width and we need to find the length. Explain that we can do one of two things to solve this problem. We can either rearrange the formula using inverse operations so that only length l is at the front of the formula (this is called changing the subject of the formula) or we can substitute the values we are given into the formula to make an equation with one unknown, l. P = 2l + 2w What is the length of a rectangle with a perimeter of 20 cm and a width of 4 cm?

Substituting into formulae and solving equations Substitute P = 20 and w = 4 into the formula: P = 2l + 2w 20 = 2l + (2 × 4) Solve this equation: Simplifying: 20 = 2l + 8 Subtracting 8: 12 = 2l This slide shows how to solve the problem by substituting the given values to make an equation. Check the solution by substituting l = 6 and w = 4 into the original formula to make sure that it gives a perimeter of 20 cm. Dividing by 2: 6 = l l = 6 So the length of the rectangle is 6 cm.

Substituting into formulae and solving equations h b The area A of a triangle with base b and perpendicular height h is given by the formula: A = 1 2 bh Again, we can do one of two things to solve this problem. We can either rearrange the formula using inverse operations so that only base length b is at the front of the formula (this is called changing the subject of the formula) or we can substitute the values we are given into the formula to make an equation with one unknown, b. What is the length of the base of a triangle with an area of 48 cm2 and a perpendicular height of 12 cm?

Substituting into formulae and solving equations Substitute A = 48 and h = 12 into the formula: A = 1 2 bh 48 = 1 2 b × 12 Solve this equation. 4 = 1 2 b Dividing by 12: Talk through the solution of the equation. Check the solution by substituting b = 8 and h = 12 into the original formula to make sure that it gives an area of 48 cm2. Multiplying by 2: 8 = b b = 8 So the base of the triangle measures 8 cm.

A3.3 Changing the subject of a formula Contents A3 Formulae A A3.1 Introducing formulae A A3.2 Using formulae A A3.3 Changing the subject of a formula A A3.4 Deriving formulae

Using inverse operations Andy is 5 years older than his brother, Brian. Their ages are linked by the formula: A = B + 5 where A is Andy’s age in years and B is Brian’s age in years. Using this formula it is easy to find Andy’s age given Brian’s age. Suppose we want to find Brian’s age given Andy’s age. Explain that we are going to look at how we can use inverse operations to rearrange formulae. Go through each step on the slide and then ask pupils questions such as: How old was Brian when Andy was 12? Which formula did you use? How old was Andy, when Brian was 28? Using inverse operations, we can write this formula as: B = A – 5

The subject of a formula Look at the formula, V = IR where V is voltage, I is current and R is resistance. V is called the subject of the formula. The subject of a formula always appears in front of the equals sign without any other numbers or operations. Sometimes it is useful to rearrange a formula so that one of the other variables is the subject of the formula. Suppose, for example, that we want to make I the subject of the formula V = IR.

Changing the subject of the formula V is the subject of this formula The formula: V = IR can be written as: I × R V The inverse of this is: I ÷ R V We can write the equation V = IR using functions. Ask pupils what do we do to I to get V and establish that we times it by R. Reveal the first diagram showing the operation × R. Compare this to a function diagram. Ask pupils how we can find the inverse of this. Reveal the second diagram corresponding to V ÷ R = I which gives us the formula I = V/R. Give numerical example. For example, ask pupils to give you the value of I when V = 12 and R = 3. Ask pupils how we could make R the subject of the formula (R = V/I). or I is now the subject of this formula I = V R

Matchstick pattern Look at this pattern made from matchsticks: Number, n 1 2 3 4 Number of Matches, m 3 5 7 9 The formula for the number of matches, m, in pattern number n is given by the formula: Go through each step on the slide and then ask: If we are given m, in this case m = 47, how can we find n? Encourage pupils to consider what have we done to n? (We’ve multiplied it by 2 and added 1.) How do we ‘undo’ times 2 and add 1? Remember, we have to reverse the order of the operations as well as the operations themselves. Establish that we need to subtract 1 and divide by 2. (47 – 1) ÷ 2 is 23. We can check that this is correct by verifying that 2 x 23 + 1 = 47. m = 2n + 1 Which pattern number will contain 47 matches?

Changing the subject of the formula m is the subject of this formula The formula: m = 2n + 1 can be written as: n × 2 + 1 m The inverse of this is: n ÷ 2 – 1 m We can rearrange this formula using inverse operations. Writing the formula as n = (m – 1)/2 allows us to find the pattern number given the number of matches. n is the subject of this formula or n = m – 1 2

Changing the subject of the formula To find out which pattern will contain 47 matches, substitute 47 into the rearranged formula. n = m – 1 2 n = 47 – 1 2 n = 46 2 Check the solution by substituting 23 into the original formula m = 2n + 1 to get 47. n = 23 So, the 23rd pattern will contain 47 matches.

Changing the subject of the formula To make C the subject of the formula: F = + 32 9C 5 F – 32 = 9C 5 Subtract 32: Multiply by 5: 5(F – 32) = 9C This formula converts degrees Celsius to degrees Fahrenheit. This slide demonstrates how to change the subject of the formula using inverse operations. Ask pupils how we could write the formula using functions. We could start with the input C, multiply it by 9, divide it by 5 and add 32. The inverse of this is to start with F, subtract 32, multiply by 5 and divide by 9. Remind pupils that we are trying to rearrange the formula so that the C appears in front of the equals sign without any numbers or operations. 5(F – 32) 9 = C Divide by 9: 5(F – 32) 9 C =

Equivalent formulae The aim of this activity is to correctly identify four equivalent forms of the same function. The correct function appears in four different forms. An example, c = 36 – 9d and d = 36 – c/9 c = 9(4 – d) and d = 4 – c/9 Equivalent formulae can be found by rearranging the given formula. If required ask pupils to find just one of the equivalent formulae. For example, ask pupils to find the formula which gives the other variable as the subject. Pressing the reset button will generate a new set of equivalent formulae.

A3 Formulae Contents A A3.1 Introducing formulae A A3.2 Using formulae A3.3 Changing the subject of a formula A A3.4 Deriving formulae

Connecting Dots When you click draw lines all red dots will be connected to all blue dots using straight lines. Show two red dots and three blue dots on the board. Ask pupils, how many lines do I need to draw to connect each red dot to each blue dot? Press the “Draw lines” button. L, the number of lines = 6. Demonstrate a few more examples. Ask pupils for a formula to find, L, the number of lines, given R, the number of red dots and B, the number of blue dots. Take some suggestions, asking pupils to justify their formulae in the context of the problem, before pressing the show formula button. Ask pupils for a justification of this formula. Link: A4 Sequences – sequences from practical contexts.

Tiling patterns By changing n and recording the results in the table ask pupils to derive a formula to give the number of tiles given the pattern number. Ask pupils to justify the formula in the context from which it is generated. Each time n increases a square n tiles by n tiles is produced, this is where the n² part of the formula comes from. There are always 4 more tiles around the outside of the pattern, this is where the + 4 part of the formula comes from.