Presentation on theme: "A1 Algebraic expressions"— Presentation transcript:
1A1 Algebraic expressions KS3 MathematicsThe aim of this unit is to teach pupils to:Use letter symbols and distinguish their different roles in algebraKnow that algebraic operations follow the same conventions and order as arithmetic operations; use index notation and the index lawsSimplify or transform algebraic expressionsMaterial in this unit is linked to the Key Stage 3 Framework supplement of examples ppA1 Algebraic expressions
3Using symbols for unknowns Look at this problem:+ 9 = 17The symbolstands for an unknown number.We can work out the value ofIntroduce the idea of using symbols to represent unknowns in mathematics.= 8because8 + 9 = 17
4Using symbols for unknowns Look at this problem:–= 5The symbolsstand for unknown numbers.andIn this example, and can have many values.For example,12 – 7 = 53.2 – –1.8 = 5orandare called variables because their value can vary.
5Using letter symbols for unknowns In algebra, we use letter symbols to stand for numbers.These letters are called unknowns or variables.Sometimes we can work out the value of the letters and sometimes we can’t.For example,We can write an unknown number with 3 added on to it asn + 3This is an example of an algebraic expression.
6Writing an expressionSuppose Jon has a packet of biscuits and he doesn’t know how many biscuits it contains.He can call the number of biscuits in the full packet, b.If he opens the packet and eats 4 biscuits, he can write an expression for the number of biscuits remaining in the packet as:Talk through the example. Stress that b stands for the number of biscuits in the full packet. We could choose any letter to stand for the unknown number. We chose b because b stands for biscuit. We could also have n for number or any other letter.Stress that the expression b – 4 describes the relationship between the number of biscuits in the full packet and the number of biscuits in the packet after 4 biscuits have been eaten.For example, if there were 20 biscuits in the original packet, then there are 20 – 4 = 16 biscuits after 4 biscuits have been eaten. If there were 32 biscuits in the original packet, then there are 32 – 4 = 28 biscuits after 4 biscuits have been eaten.b – 4
7Writing an equationJon counts the number of biscuits in the packet after he has eaten 4 of them. There are 22.He can write this as an equation:b – 4 = 22We can work out the value of the letter b.Stress the difference between an expression and an equation. An expression does not contain an equals sign.In an equation we can often work out the value of the letter symbol. In an expression we cannot.b = 26That means that there were 26 biscuits in the full packet.
8Writing expressionsWhen we write expressions in algebra we don’t usually use the multiplication symbol ×.For example,5 × n or n × 5 is written as 5n.The number must be written before the letter.When we multiply a letter symbol by 1, we don’t have to write the 1.Introduce these algebraic conventions.When we write an algebraic expression we try to use as few numbers, letters and symbols as necessary. If we know that 5n means 5 lots of n, then we don’t need to write 5 × n.You may like to mention that leaving out the multiplication sign × avoids confusing it with the letter symbol x which is often used in algebra.You may also like to stress to pupils that when they write the letter x in algebra it should be written in script form to distinguish it from a multiplication sign.For example,1 × n or n × 1 is written as n.
9Writing expressionsWhen we write expressions in algebra we don’t usually use the division symbol ÷. Instead we use a dividing line as in fraction notation.For example,n ÷ 3 is written asn3When we multiply a letter symbol by itself, we use index notation.Tell pupils that n2 is read as ‘n squared’ or ‘n to the power of 2’.For example,n squaredn × n is written as n2.
10Writing expressions Here are some examples of algebraic expressions: a number n plus 75 – n5 minus a number n2n2 lots of the number n or 2 × n6n6 divided by a number n4n + 54 lots of a number n plus 5Explain that algebra is very much like a language. It follows special rules, like the rules of grammar in a language. Like a language we have to keep to these rules so any mathematician in the world can understand it.Algebra is very important in mathematics because it describes the relationships between numbers.Explain that there is a difference between an unknown and a variable. An unknown usually has a unique value which we can work out given enough information. A variable can have many different values.We can use any letter in the alphabet to stand for unknowns or variables but some letters are used more than others. For example, we often use a, b, n, x or y. But we try not to use o (because it looks like 0).Explain that in algebra we do not need to write the multiplication sign, ×, and so 2 lots of n is written as 2n. 3 lots of a, or 3 times a would be written as 3a. 5 lots of t, or 5 times t would be written as 5t.Give some examples of possible values for 2n. If n is worth 5 then 2n is equal to 10 (not 25). If n is worth 20 then 2n is worth 40.When we divide in algebra we write the number we are dividing by underneath, like a fraction. In this example, if n was worth 2, 6/n would be equal to 3.Tell pupils that n3 is read as ‘n cubed’ or ‘n to the power of 3’. Give some examples, if n is worth 2 then n3 is 2 × 2 × 2, 8.a number n multiplied by itself twice or n × n × nn33 × (n + 4) or 3(n + 4)a number n plus 4 and then times 3.
11Writing expressionsMiss Green is holding n number of cubes in her hand:Write an expression for the number of cubes in her hand if:She takes 3 cubes away.n – 3Discuss the examples on the slides and give other examples from around the classroom.For example,Suppose Joe has p number of pencils in his pencil case. If Harry, sitting next to him, gives him two more pencils, what expression could we write for the number of pencils in his pencil case? (p + 2)We don’t know what p is (without counting) but we can still write an expression for the number of pencils in the case.If p, the original number of pencils in the pencil case was 9, Joe would now have 11. If p was 15, Joe would now have 17.Now, Joe gives Harry back his pencils, so he has p pencils again. Suppose he shares his pencil equally between himself and two of his friends. How many pencils will they have each? (p ÷ 3)Tell pupils that if they are not sure whether or not an expression works, they should try using numbers in place of the letters to check. For example,If Joe had 18 pencils and shared them between himself and his 2 friends they would get 6 each, 18 ÷ 3, so our expression works.Suppose Mary has p pencils and Julia has q pencils. How many pencils do they have altogether? (p + q)She doubles the number of cubes she is holding.2 × nor2n
13x + x + x is identically equal to 3x IdentitiesWhen two expressions are equivalent we can link them with the sign.x + x + x is identically equal to 3xFor example,x + x + x 3xThis is called an identity.In an identity, the expressions on each side of the equation are equal for all values of the unknown.Differentiate between the = sign meaning ‘is equal to’ and the sign meaning ‘is identically equal to’.An equation is true for particular values of the unknown. An identity is true for all values of the unknown.The expressions are said to be identically equal.
14A1.2 Collecting like terms ContentsA1 Algebraic expressionsA1.1 Writing expressionsA1.2 Collecting like termsA1.3 Multiplying termsA1.4 Dividing termsA1.5 Factorizing expressionsA1.6 Substitution
15Like termsAn algebraic expression is made up of terms and operators such as +, –, ×, ÷ and ( ).A term is made up of numbers and letter symbols but not operators.For example,3a + 4b – a + 5 is an expression.Explain that we can think of algebra as the language of mathematics.Expressions are like sentences in a language and terms are like individual words.3a, 4b, a and 5 are terms in the expression.3a and a are called like terms because they both contain a number and the letter symbol a.
16Collecting together like terms Remember, in algebra letters stand for numbers, so we can use the same rules as we use for arithmetic.In arithmetic,= 4 × 5In algebra,a + a + a + a = 4aIn arithmetic, a number plus the same number plus the same number plus the same number = 4 × the number.Emphasize that a can be any number.In this example, we are using algebra to generalize the result.The a’s are like terms.We collect together like terms to simplify the expression.
17Collecting together like terms Remember, in algebra letters stand for numbers, so we can use the same rules as we use for arithmetic.In arithmetic,(7 × 4) + (3 × 4) = 10 × 4In algebra,7 × b + 3 × b = 10 × borIn arithmetic, 7 × a number plus 3 × the same number = 10 × the number.Pupils may want an ‘answer’ to 7b + 3b = 10b. Explain that what this is telling us is that any number multiplied by 7 plus the same number multiplied by 3 is always equal to the number multiplied by 10.Emphasize that b can be any number.In this example, we are using algebra to generalize a result rather than to give an answer to a specific problem.7b + 3b = 10b7b, 3b and 10b are like terms.They all contain a number and the letter b.
18Collecting together like terms Remember, in algebra letters stand for numbers, so we can use the same rules as we use for arithmetic.In arithmetic,2 + (6 × 2) – (3 × 2) = 4 × 2In algebra,x + 6x – 3x = 4xIn arithmetic, a number plus 6 × the same number minus 3 × the same number = 4 × the number.Discuss the algebraic equivalent of this.Emphasize that x can be any number.Again, we are using algebra to generalize a result rather than to give an answer to a specific problem.x, 6x, 3x and 4x are like terms.They all contain a number and the letter x.
19Collecting together like terms When we add or subtract like terms in an expression we say we are simplifying an expression by collecting together like terms.An expression can contain different like terms.For example,3a + 2b + 4a + 6b= 3a + 4a + 2b + 6bExplain the meanings of each key word and phrase.In the example 3a + 2b + 4a + 6b, explain that it is helpful to write like terms next to each other. (Remember that we can add terms in any order.)Stress that we cannot simplify 7a + 8b any further. We can’t combine a’s and b’s. This says 7 times one number plus eight times another number.= 7a + 8bThis expression cannot be simplified any further.
20Collecting together like terms Simplify these expressions by collecting together like terms.1) a + a + a + a + a =5a2) 5b – 4b =b3) 4c + 3d + 3 – 2c + 6 – d =4c – 2c + 3d – d= 2c + 2d + 9Whenever possible make comparisons to arithmetic by substituting actual values for the letters. If it is true using numbers then it is true using letters. For example, in 1) we could say that is equivalent to 5 × 7.For example 1) and example 2), stress again that in algebra we don’t need to write the number 1 before a letter to multiply it by 1. 1a is just written as a and 1b is just written as b.For example 3), explain that when there are lots of terms we can write like terms next to each other so that they are easier to collect together. The numbers without any letters are added together separately.In example 4) stress that n2 is different from n. They cannot be collected together. 4n – 3n is n and n2 stays as it is.If we can’t collect together any like terms, as in example 5), we write ‘cannot be simplified’.4) 4n + n2 – 3n =4n – 3n + n2 =n + n25) 4r + 6s – tCannot be simplified
21Algebraic perimetersRemember, to find the perimeter of a shape we add together the length of each of its sides.Write an algebraic expression for the perimeter of the following shapes:2a3bPerimeter =2a + 3b + 2a + 3b= 4a + 6bFor the first example, remind pupils that we need to add together the length of each side. The lengths of two of the sides have not been written on. Since this is a rectangle we can deduce that the length of the side opposite the side of length 2a is also 2a and the length of the side opposite the side of length 3b is also 3b.Ask pupils,How could the longer side be represented by 2a and the shorter side by 3b?Deduce that the letter a must represent a bigger number than the letter b.5x4yxPerimeter =4y + 5x + x + 5x= 4y + 11x
22Algebraic pyramidsUse the algebraic pyramid to practise collecting together like terms. Start by revealing all of the expressions along the bottom row of the pyramid. Find the expression in each brick by adding the two expressions below it.Modify the activity by revealing expressions in the bricks above and one of the expressions in the bricks below it. Pupils must then subtract expressions to find those that are hidden.
23Algebraic magic square Remind pupils that in a magic square each row, column and diagonal has the same sum, the ‘magic total’.Start by working out the ‘magic total’ by revealing three expressions in any row column or diagonal. This can also be found by multiplying the expression in the centre by three.Reveal the expressions in one or two more squares so that pupils have enough information to work out the missing expressions.Clicking on each cell will reveal the missing expression.
25Multiplying terms together In algebra we usually leave out the multiplication sign ×.Any numbers must be written at the front and all letters should be written in alphabetical order.For example,4 × a =4a1 × b =bWe don’t need to write a 1 in front of the letter.b × 5 =Ask pupils why they think we try not to use the multiplication symbol ×.One reason is that it is easily confused with the letter, xAnother reason is that when we use algebra we try to write thing as simply as possible, only writing what is absolutely necessary. It’s simpler to write 2n than 2 x n.It is also unnecessary to write a 1 in front of a letter to multiply it by 1. Multiplying by 1 has no effect so we can leave it out altogether.5bWe don’t write b5.3 × d × c =3cdWe write letters in alphabetical order.6 × e × e =6e2
26Using index notation Simplify: x + x + x + x + x = 5x x to the power of 5Simplify:x × x × x × x × x= x5This is called index notation.Similarly,Start by asking pupils to simplify x + x + x + x + x. This is 5 lots of x, which have seen is written as 5x.Now, ask pupils how they might simplify x × x × x × x × x. Impress upon pupils the difference between this, and the previous expression, as they are often confused.If x is equal to 2, for example, x + x + x + x + x equals 10, and x × x × x × x × x equals 32.Some pupils may suggest writing xxxxx. This is not strictly incorrect, however, it should be discouraged in favour of using index notation.When we write a number or term to the power of another number it is called index notation. The power, or index, is the raised number, in this case 5. The plural of index is indices. The number or letter that we are multiplying successive times, in this case, x, is called the base.x2 is read as ‘x squared’ or ‘x to the power of 2’.x3 is read as ‘x cubed’ or ‘x to the power of 3’.x4 is read as ‘x to the power of 4’.x × x= x2x × x × x= x3x × x × x × x= x4
27Using index notationWe can use index notation to simplify expressions.For example,3p × 2p =3 × p × 2 × p =6p2q2 × q3 =q × q × q × q × q =q5Discuss each example briefly.In the last example 2t × 2t the use of brackets may need further clarification. We must put a bracket around the 2t since both the 2 and the t are squared. If we wrote 2t2, then only the t would be squared.Give a numerical example, if necessary. If t was 3 then 2t would be equal to 6. We would then have 62, 36. If we wrote 2t2, that would mean 2 × 32 or 2 × 9 which is 18.Remember the order of operations - BIDMAS. Brackets are worked out before indices, but indices are worked out before multiplication.3r × r2 =3 × r × r × r =3r32t × 2t =(2t)2or4t2
28Grid method for multiplying numbers Use this slide to review the grid method for multiplying two numbers.
29Brackets Look at this algebraic expression: 4(a + b) What do do think it means?Remember, in algebra we do not write the multiplication sign, ×.This expression actually means:4 × (a + b)orDiscuss the meaning of 4(a + b). Take suggestions from pupils and correct any misconceptions.The number outside the brackets multiplies every term inside the brackets.Remind pupils again that, in algebra, letters stand for numbers and that algebra follow the same rules as arithmetic.Link this expression with the multiplication method we met at the beginning of the lesson.a could be equal to 30, for example, and b could be equal to 2. This expression would then be equivalent to 4 × 32.Remember, a and b could each be any number.The expression 4(a + b) is equivalent to 4a + 4b. They are different ways of writing the same thing.Explain that since algebra follows the same rules as arithmetic we can use the grid method to help us multiply out brackets in algebra.(a + b) + (a + b) + (a + b) + (a + b)= a + b + a + b + a + b + a + b= 4a + 4b
30Using the grid method to expand brackets Demonstrate as many examples as is necessary for pupils to understand the method.Once pupils are happy with the method they may be able to multiply out brackets without using a grid.
31Expanding brackets then simplifying Sometimes we need to multiply out brackets and then simplify.For example,3x + 2(5 – x)We need to multiply the bracket by 2 and collect together like terms.3x+ 10– 2xShow pupils that it is not necessary to construct a grid to multiply out a bracket. If required pupils can use lines, as shown here in orange, to make sure that every term inside the bracket is multiplied by the term outside the bracket.= 3x – 2x + 10= x + 10
32Expanding brackets then simplifying We need to multiply the bracket by –1 and collect together like terms.4– 5n+ 3In this example, we have 4 and then –(5n – 3).This is equivalent to –1 × (5n – 3).Repeat that a minus sign in front of a bracket has the effect of changing the sign of every term inside the bracket. 5n becomes – 5n and – 3 becomes + 3.Let’s write the like terms next to each other.When we collect the like terms together we have 7 – 5n.= – 5n= 7 – 5n
33Expanding brackets then simplifying 2(3n – 4) + 3(3n + 5)We need to multiply out both brackets and collect together like terms.6n– 8+ 9n+ 15In this example, we have two sets of brackets. The first set is multiplied by 2 and the second set is multiplied by 3.We don’t need to use a grid as long as we remember to multiply every term inside the bracket by every term outside it.Talk through the multiplication of (3n – 4) by 2 and (3n + 5) by 3.Let’s write the like terms next to each other.When we collect the like terms together we have 6n + 9n which is 15n and – = 7.= 6n + 9n –= 15n + 7
34Expanding brackets then simplifying 5(3a + 2b) – 2(2a + 5b)We need to multiply out both brackets and collect together like terms.15a+ 10b– 4a–10bHere is another example with two sets of brackets. The first set is multiplied by 5 and the second set is multiplied by minus 2.Talk through the multiplication of (3a + 2b) by 5.Next stress that we need to multiply (2a + 5b) by negative 2. Talk through this.Let’s write the like terms next to each other.When we collect together the like terms we have 15a – 4a which is 11a. The b terms cancel out.= 15a – 4a + 10b – 10b= 11a
35Algebraic multiplication square Complete this activity as a group exercise.Reveal the expressions in the blue cells and ask pupils to find the expressions in the white cells by multiplying.Ask pupils how the multiplications could be written using brackets.
36Pelmanism: Equivalent expressions Work together as a class to match equivalent expressions.
37Algebraic areasThis exercise may be used as a whole class activity to practice multiplying and dividing terms.Start by finding areas by multiplying the lengths of the sides.Remind pupils that the letters stand for numbers, so if a rectangle has sides of length 3a and 6a, for example, one side is twice as long as the other no matter what a is. The corresponding area will be 18a2. For example, if a was 2 the sides would be 6 and 12 and the area would be 18 x 22, 18 × 4, 72.It is not necessary to substitute for every example. The aim is to demonstrate the power of algebra to generalize relationships.Practise dividing terms by finding missing side lengths. Pupils may find this easier to think of if you ask ‘what do I multiply this length by to get this area?’ rather than asking pupils to divide the area by the side length.
39Dividing termsRemember, in algebra we do not usually use the division sign, ÷.Instead we write the number or term we are dividing by underneath like a fraction.For example,Point out that we do not need to write the brackets when we write a + b all over c. Since both letters are above the dividing line we know that it is the sum of a and b that is divided by c. The dividing line effectively acts as a bracket.(a + b) ÷ cis written asa + bc
40Dividing termsLike a fraction, we can often simplify expressions by cancelling.For example,n3n26p23pn3 ÷ n2 =6p2 ÷ 3p =1121n × n × nn × n6 × p × p3 × p==In the first example, we can divide both the numerator and the denominator by n. n ÷ n is 1. We can divide the numerator and the denominator by n again to leave n. (n/1 is n).If necessary, demonstrate this by substitution.For example, 3 cubed, 27, divided by 3 squared, 9, is 3. Similarly 5 cubed, 125, divided by 5 squared, 25, is 5.In the second example, we can divide the numerator and the denominator by 3 and then by p to get 2p.Again, demonstrate the truth of this expression by substitution, if necessary.For example, if p was 5 we would have 6 × 5 squared, 6 × 25, which is 150, divided by 3 × 5, divided by 15 is 10, which is 2 times 5. This power of algebra is such that this will work for any number we choose for p.Pupils usually find multiplying easier than dividing. Encourage pupils to check their answers by multiplying (using inverse operations). For example, n × n2 = n3. And 2p × 3p = 6p21111= n= 2p
41Algebraic areasThis exercise may be used as a whole class activity to practise multiplying and dividing terms.Start by finding areas by multiplying the lengths of the sides.Practise dividing terms by finding missing side lengths. Pupils may find this easier to think of if you ask ‘what do I multiply this length by to get this area?’ rather than asking pupils to divide the area by the side length.
42Hexagon PuzzleThe numbers in the squares are found by multiplying the terms in the circles on either side.Start by selecting one of the blank circles. Indicate the relevant circle and square and askWhat would I have to multiply this term by to get this term?(It is probably easier for pupils to find the circle terms in this way initially. Later, you may ask pupils to divide.)Next ask pupils to find the term in the adjacent square.Continue until the puzzle is complete.Discuss briefly the fact that when we multiply two terms together with the same base (the letter n, in this case) we add the powers. When we divide two terms with the same base (the letter n) we subtract the powers.
44Factorizing expressions Some expressions can be simplified by dividing each term by a common factor and writing the expression using brackets.For example, in the expression5x + 10the terms 5x and 10 have a common factor, 5.We can write the 5 outside of a set of brackets and mentally divide 5x + 10 by 5.We can write the 5 outside of a set of brackets(5x + 10) ÷ 5 = x + 2This is written inside the bracket.5(x + 2)5(x + 2)
45Factorizing expressions Writing 5x + 10 as 5(x + 2) is called factorizing the expression.Factorize 6a + 8Factorize 12 – 9nThe highest common factor of 6a and 8 isThe highest common factor of 12 and 9n is2.3.(6a + 8) ÷ 2 =3a + 4(12 – 9n) ÷ 3 =4 – 3n6a + 8 =2(3a + 4)12 – 9n =3(4 – 3n)
46Factorizing expressions Writing 5x + 10 as 5(x + 2) is called factorizing the expression.Factorize 3x + x2Factorize 2p + 6p2 – 4p3The highest common factor of 3x and x2 isThe highest common factor of 2p, 6p2 and 4p3 isx.2p.(2p + 6p2 – 4p3) ÷ 2p=(3x + x2) ÷ x =3 + x1 + 3p – 2p23x + x2 =x(3 + x)2p + 6p2 – 4p3=2p(1 + 3p – 2p2)
47Algebraic multiplication square Complete this activity as a group exercise.Use factorization to find the values of the missing terms across the top and along the side.Ask pupils how some of the multiplications could be written using brackets.
48Pelmanism: Equivalent expressions Work as a class to match equivalent expressions.
50Work it out!4 + 3 ×0.6–74385The aim of this exercise is to revise the use of the correct order of operations and to introduce the concept of substitution.The following values are inserted in turn:58430.6–7Pupils could be asked to write the answers down on paper or write them down on individual whiteboards.= 133= –17= 5.8= 28= 19
51Work it out!7 ×0.4–322692The following values are inserted in turn:69220.4–3= –10.5= 31.5= 1.4= 21= 77
52Work it out!0.2–412932 + 6The following values are inserted in turn:39120.2–4= 6.04= 150= 22= 87= 15
53Work it out!2( )–133.669187The following values are inserted in turn:7183.6–13= 23.2= 154= –10= 30= 52
54What does substitution mean? Ask pupils where they have heard the word substitution before. One example would be in team games when one player is replaced by another.In algebra, substitution means to replace letters with numbers.In algebra, when we replace letters in an expression or equation with numbers we call it substitution.
55How can be written as an algebraic expression? SubstitutionHow can be written as an algebraic expression?4 + 3 ×Using n for the variable we can write this as4 + 3nWe can evaluate the expression 4 + 3n by substituting different values for n.When n = 54 + 3n =4 + 3 × 5=Ask pupils to think about the activity they were doing at the start of the lesson. What we were actually doing was a kind of substitution. We were replacing a symbol (the box) with a number each time.Ask pupils how we could write × as an algebraic expression.It doesn’t matter what letter they use but do remind pupils that we don’t write the multiplication sign in algebra.Define the keyword, evaluate – to find the value of.Discuss the substitution and order of operations:When n is 5, what is 3n? (15)So what is 4 + 3n? ( = 19)Suggest to pupils that they may wish to work out the value of 3n before writing anything down. This would avoid errors involving order of operations. In other words, they can write 4 + 3n = and leave out the intermediate step of × 5. This might be incorrectly evaluated to 35.(We have written × 5 on the board to reinforce the meaning of 3n).= 19When n = 114 + 3n =4 + 3 × 11== 37
56Substitution 7 × 2 7n 2 can be written as We can evaluate the expression by substituting different values for n.7n27n2=When n = 47 × 4 ÷ 2= 28 ÷ 2Stress that when we are multiplying and dividing, it doesn’t matter what order we do it in. For example 7 × 4 ÷ 2 will always give the same answer as 4 ÷ 2 × 7.(The order is important when we combine multiplying and dividing with adding and subtracting. If there aren’t any brackets we always multiply or divide before we add or subtract.)= 147n2=When n = 1.17 × 1.1 ÷ 2= 7.7 ÷ 2= 3.85
57Substitution 2 + 6 can be written as n2 + 6 We can evaluate the expression n2 + 6 by substituting different values for n.When n = 4n2 + 6 =42 + 6== 22Remind pupils that 42 is read as ‘4 squared’ and means ‘4 × 4.’Pupils are less likely to make mistakes involving incorrect order of operations if they can be encouraged to square in their heads rather than write down the intermediate step of = 4 ×In particular, expressions such as , may be written as × 2 and then incorrectly evaluated to 10.When n = 0.6n2 + 7 === 6.36
58Substitution 2( + 8) can be written as 2(n + 8) 2( )can be written as2(n + 8)We can evaluate the expression 2(n + 8) by substituting different values for n.When n = 62(n + 8) =2 × (6 + 8)= 2 × 14= 28Remind pupils again that when there are brackets we need to work out the value inside the brackets before we multiply.When n = 132(n + 8) =2 × (13 + 8)= 2 × 21= 41
59Substitution exercise Here are five expressions.1) a + b + c= –1= 62) 3a + 2c= 3 × × –1= 15 + –2= 133) a(b + c)= 5 × (2 + –1)= 5 × 1= 54) abc= 5 × 2 × –1= 10 × –1= –10Tell pupils that expressions can contain many different variables. Remember when we use a letter to represent a number in an expression it can have any value. The value can vary and so we call it a variable.If pupils are ready you may wish to use the above examples as a pupil exercise before revealing the solutions.Alternatively, talk through each example stressing the correct order of operations each time. Then set pupils an exercise made up of similar problems.Edit the slide to make the numbers being substituted more or less challenging.5)ab2 – c522 – –1== 5 ÷ 5= 1Evaluate these expressions when a = 5, b = 2 and c = –1
60Noughts and crosses - substitution Divide the class into two teams, red and blue. Decide which team will start by, for example, flipping a coin.The starting team chooses an expression from the board. Click on the expression to highlight it.The number that appears in the large rectangle must be substituted into the expression.Everyone in the team must try to work out the answer mentally within 5 seconds.Select a pupil from the team to give you their answer.Check their answer by clicking on the notelet.If the answer is correct, select the teams’ symbol. If the answer is incorrect colour the circle in the opposing teams’ symbol.It is then the turn of the opposing team.The game is over when one of the teams gets three of their symbols in a row, horizontally, vertically, or diagonally. (Or when the board is full, in which case, the game ends in a draw).If no mistakes are made and the game is finished quickly, play again, but allow the other team to start.