A1 Introduction to algebra

Presentation on theme: "A1 Introduction to algebra"— Presentation transcript:

A1 Introduction to algebra
S1–S4 Mathematics A1 Introduction to algebra

A1 Introduction to algebra
Contents A1 Introduction to algebra A A1.1 Writing expressions A A1.2 Collecting like terms A A1.3 Multiplying terms and expanding brackets A A1.4 Dividing terms A A1.5 Factorizing expressions A A1.6 Substitution

Using symbols for unknowns
Look at this problem: + 9 = 17 The symbol stands for an unknown number. We can work out the value of Introduce the idea of using symbols to represent unknowns in mathematics. = 8 because 8 + 9 = 17

Using symbols for unknowns
Look at this problem: = 5 The symbols stand for unknown numbers. and In this example, and can have many values. For example, 12 – 7 = 5 3.2 – –1.8 = 5 or and are called variables because their value can vary.

Using 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 as n + 3 This is an example of an algebraic expression.

Writing an expression Suppose 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 a. 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. Make clear that a stands for the number of biscuits in the full packet. We could have chosen any letter to stand for the unknown number. Explain that the expression a – 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. a – 4

Writing an equation Jon counts the number of biscuits in the packet after he has eaten 4 of them. There are now 22. He can write this as an equation: a – 4 = 22 We can work out the value of a: Make clear 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. a = 26 That means that there were 26 biscuits in the full packet.

Writing expressions When 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. Suggest that when pupils 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.

Writing expressions When 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 as n 3 When 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 squared n × n is written as n2.

Writing expressions Here are some examples of algebraic expressions:
a number n plus 7 5 – n 5 minus a number n 2n 2 lots of the number n or 2 × n 6 n 6 divided by a number n 4n + 5 4 lots of a number n plus 5 Explain that algebra is very much like a language. It follows special rules, like the rules of grammar in a language. We have to keep to these rules so that 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 avoid 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 and by itself again or n × n × n n3 3 × (n + 4) or 3(n + 4) a number n plus 4 and then times 3.

Writing expressions Miss 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 – 3 Discuss the examples on the slides and give other examples from around the classroom. For example: Suppose Joe has p 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 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 pencils 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 × n or 2n

Equivalent expression match

A1.2 Collecting like terms
Contents A1 Introduction to algebra A A1.1 Writing expressions A A1.2 Collecting like terms A A1.3 Multiplying terms and expanding brackets A A1.4 Dividing terms A A1.5 Factorizing expressions A A1.6 Substitution

Like terms An 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. 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.

Collecting 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 × 5 In algebra, a + a + a + a = 4a In arithmetic, a number plus the same number plus the same number plus the same number = 4 × the number. Emphasize that a represents any number. The a’s are like terms. We collect together like terms to simplify the expression.

Collecting 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 × 4 In algebra, 7 × b + 3 × b = 10 × b or In 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 = 10b 7b, 3b and 10b are like terms. They all contain a number and the letter b.

Collecting 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 × 2 In algebra, x + 6x – 3x = 4x In 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.

Collecting 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 + 6b Explain 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 + 8b This expression cannot be simplified any further.

Collecting together like terms
Simplify these expressions by collecting together like terms. 1) a + a + a + a + a = 5a 2) 5b – 4b = b 3) 4c + 3d + 3 – 2c + 6 – d = 4c – 2c + 3d – d = 2c + 2d + 9 Whenever 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 + n2 5) 4r + 6s – t Cannot be simplified

Algebraic perimeters Remember, to find the perimeter of a shape we add together the lengths of each of its sides. Write algebraic expressions for the perimeters of the following shapes: 2a 3b Perimeter = 2a + 3b + 2a + 3b = 4a + 6b For 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. 5x 4y x Perimeter = 4y + 5x + x + 5x = 4y + 11x

Algebraic pyramids Use 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.

Algebraic magic square
Remind pupils that in a magic square each row, column and diagonal has the same sum, called 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.

A1.3 Multiplying terms and expanding brackets
Contents A1 Introduction to algebra A A1.1 Writing expressions A A1.2 Collecting like terms A A1.3 Multiplying terms and expanding brackets A A1.4 Dividing terms A A1.5 Factorizing expressions A A1.6 Substitution

Multiplying 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 = 4a 1 × b = b We 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 could easily be confused with the letter x. Another reason is that when we use algebra we try to write things 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. 5b We don’t write b5. 3 × d × c = 3cd We write letters in alphabetical order. 6 × e × e = 6e2

Using index notation Simplify: a + a + a + a + a = 5a
a to the power of 5 Simplify: a × a × a × a × a = a5 This is called index notation. Similarly, Start by asking pupils to simplify a + a + a + a + a. This is 5 lots of a, which we have seen is written as 5a. Now, ask pupils how they might simplify a × a × a × a × a. Impress upon pupils the difference between this, and the previous expression, as they are often confused. If a is equal to 2, for example, a + a + a + a + a equals 10, and a × a × a × a × a equals 32. Some pupils may suggest writing aaaaa. 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, a, is called the base. a² is read as ‘a squared’ or ‘a to the power of 2’. a³ is read as ‘a cubed’ or ‘a to the power of 3’. a4 is read as ‘a to the power of 4’. a × a = a2 a × a × a = a3 a × a × a × a = a4

Using index notation We can use index notation to simplify expressions. For example, 3p × 2p = 3 × p × 2 × p = 6p2 q2 × q3 = q × q × q × q × q = q5 Discuss 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. When an expression is evaluated, brackets are worked out before indices, but indices are worked out before multiplication. See N10.1 Order of operations. 3r × r2 = 3 × r × r × r = 3r3 2t × 2t = (2t)2 or 4t2

Grid method for multiplying numbers
Use this slide to review the grid method for multiplying two numbers. See N10.3 Multiplication and division.

Brackets Look at this algebraic expression: 4(a + b)
What do you think it means? Remember, in algebra we do not write the multiplication sign ×. So this expression means: 4 × (a + b) or: Discuss 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

Using 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 can begin to multiply out brackets without using a grid.

Expanding expressions with brackets
Look at this algebraic expression: 3y(4 – 2y) This means 3y × (4 – 2y), but we do not usually write × in algebra. To expand or multiply out this expression we multiply every term inside the bracket by the term outside the bracket. The orange lines show which terms we are multiplying together. 3y(4 – 2y) = 12y – 6y2

Expanding expressions with brackets
Look at this algebraic expression: –a(2a2 – 2a + 3) When there is a negative term outside the bracket, the signs of the multiplied terms change. –a(2a2 – 3a + 1) = –2a3 + 3a2 – a In general, –x(y + z) = –xy – xz –x(y – z) = –xy + xz –(y + z) = –y – z –(y – z) = –y + z

Expanding 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 – 2x Show 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. Explain that it is sometimes helpful to rewrite the expression with like terms next to each other before collecting them together. Stress that when this is done the plus or minus sign before the term must be moved as well. = 3x – 2x + 10 = x + 10

Expanding brackets and simplifying
Expand and simplify: 4 – (5n – 3) We need to multiply the bracket by –1 and collect together like terms. 4 – (5n – 3) = 4 – 5n + 3 In 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. Again, explain the intermediate step where the expression is rewritten with like terms next to each other = – 5n = 7 – 5n

Expanding brackets and simplifying
Expand and simplify: 2(3n – 4) + 3(3n + 5) We need to multiply out both brackets and collect together like terms. 2(3n – 4) + 3(3n + 5) = 6n – 8 + 9n + 15 In 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

Expanding brackets and simplifying
Expand and simplify: 5(3a + 2b) – a(2 + 5b) We need to multiply out both brackets and collect together like terms. 5(3a + 2b) – a(2 + 5b) = 15a + 10b – 2a – 5ab Here is another example with two sets of brackets. The first set is multiplied by 5 and the second set is multiplied by minus a. Talk through the multiplication of (3a + 2b) by 5. Next we need to multiply (2 + 5b) by -a. Talk through this. = 15a – 2a + 10b – 5ab = 13a + 10b – 5ab

Algebraic 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.

Pelmanism: Equivalent expressions
Work together as a class to match equivalent expressions. This activity can also be used to practice factorizing expressions.

Algebraic areas This 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 6 × 12 = 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.

A1 Introduction to algebra
Contents A1 Introduction to algebra A A1.1 Writing expressions A A1.2 Collecting like terms A A1.3 Multiplying terms and expanding brackets A A1.4 Dividing terms A A1.5 Factorizing expressions A A1.6 Substitution

Dividing terms Remember, 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 c (a + b) ÷ c is written as

Dividing terms As with fractions, we can often simplify expressions by cancelling. For example, n3 n2 6p2 3p n3 ÷ n2 = 6p2 ÷ 3p = 1 1 2 1 n × n × n n × n 6 × p × p 3 × 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 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. 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 = 6p² 1 1 1 1 = n = 2p

Hexagon puzzle The numbers in the squares are found by multiplying the terms in the circles on either side. Reveal each term until the puzzle is complete. The order can be modified to practice both multiplication and division of indices. 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.

A1.5 Factorizing expressions
Contents A1 Introduction to algebra A A1.1 Writing expressions A A1.2 Collecting like terms A A1.3 Multiplying terms and expanding brackets A A1.4 Dividing terms A A1.5 Factorizing expressions A A1.6 Substitution

Factorizing expressions
Factorizing an expression is the opposite of expanding it. Expanding or multiplying out Factorizing a(b + c) ab + ac Often: When we expand an expression we remove the brackets. When we factorize an expression we write it with brackets.

Factorizing expressions
Expressions can be factorized by dividing each term by a common factor and writing this outside a pair of brackets. For example, in the expression 5x + 10 the 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 Encourage pupils to check this by multiplying the expression out to 5x + 10. (5x + 10) ÷ 5 = x + 2 This is written inside the bracket. 5(x + 2) 5(x + 2)

Factorizing expressions
Writing 5x + 10 as 5(x + 2) is called factorizing the expression. Factorize 6a + 8 Factorize 12 – 9n The highest common factor of 6a and 8 is The highest common factor of 12 and 9n is 2. 3. (6a + 8) ÷ 2 = 3a + 4 (12 – 9n) ÷ 3 = 4 – 3n For each example start by asking pupils to give you the highest number or term that will divide into both terms in the expression. 6a + 8 = 2(3a + 4) 12 – 9n = 3(4 – 3n)

Factorizing expressions
Writing 5x + 10 as 5(x + 2) is called factorizing the expression. Factorize 3x + x2 Factorize 2p + 6p2 – 4p3 The highest common factor of 3x and x2 is The highest common factor of 2p, 6p2 and 4p3 is x. 2p. (2p + 6p2 – 4p3) ÷ 2p = (3x + x2) ÷ x = 3 + x Again, for each example ask pupils to give you the highest number or term that will divide into every term in the given expression. 1 + 3p – 2p2 3x + x2 = x(3 + x) 2p + 6p2 – 4p3 = 2p(1 + 3p – 2p2)

Factorization Start by asking pupils to give you the value of the highest common factor of the two terms. Reveal this and then ask pupils to give you the values of the terms inside the brackets. The activities at the end of A1.3 Multiplying terms and expanding brackets can also be used to practice factorization.

A1 Introduction to algebra
Contents A1 Introduction to algebra A A1.1 Writing expressions A A1.2 Collecting like terms A A1.3 Multiplying terms and expanding brackets A A1.4 Dividing terms A A1.5 Factorizing expressions A A1.6 Substitution

Work it out! 4 + 3 × 0.6 –7 43 8 5 The 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: 8 5 43 0.6 –7 Pupils could be asked to write the answers down on paper or write them down on individual whiteboards. = 133 = –17 = 5.8 = 28 = 19

Work it out! 7 × 0.4 –3 22 6 9 2 The following values are inserted in turn: 6 9 22 0.4 –3 = –10.5 = 31.5 = 1.4 = 21 = 77

Work it out! 0.2 –4 12 9 3 2 + 6 The following values are inserted in turn: 3 9 12 0.2 –4 = 6.04 = 150 = 22 = 87 = 15

Work it out! 2( ) –13 3.6 69 18 7 The following values are inserted in turn: 7 18 69 3.6 –13 = 23.2 = 154 = –10 = 30 = 52

What 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. Explain that, 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.

How can be written as an algebraic expression?
Substitution How can be written as an algebraic expression? 4 + 3 × Using n for the variable we can write this as 4 + 3n. We can evaluate the expression 4 + 3n by substituting different values for n. When n = 5 4 + 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). = 19 When n = 11 4 + 3n = 4 + 3 × 11 = = 37

Substitution 7 × 2 7n 2 can be written as
We can evaluate the expression by substituting different values for n. 7n 2 7n 2 = When n = 4 7 × 4 ÷ 2 = 28 ÷ 2 Stress 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.) = 14 7n 2 = When n = 1.1 7 × 1.1 ÷ 2 = 7.7 ÷ 2 = 3.85

Substitution 2 + 6 can be written as n2 + 6
We can evaluate the expression n2 + 6 by substituting different values for n. When n = 4 n2 + 6 = 42 + 6 = = 22 Remind 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 4² + 6 = 4 × In particular, expressions such as 3 + 2², may be written as × 2 and then incorrectly evaluated to 10. When n = 0.6 n2 + 6 = = = 6.36

Substitution 2( + 8) can be written as 2(n + 8)
2( ) can be written as 2(n + 8) We can evaluate the expression 2(n + 8) by substituting different values for n. When n = 6 2(n + 8) = 2 × (6 + 8) = 2 × 14 = 28 Remind pupils again that when there are brackets we need to work out the value inside the brackets before we multiply. When n = 13 2(n + 8) = 2 × (13 + 8) = 2 × 21 = 42

Substitution exercise
Here are five expressions. 1) a + b + c = –1 = 6 2) 3a + 2c = 3 × × –1 = 15 + –2 = 13 3) a(b + c) = 5 × (2 + –1) = 5 × 1 = 5 4) abc = 5 × 2 × –1 = 10 × –1 = –10 Tell 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) a b2 – c 5 22 – –1 = = 5 ÷ 5 = 1 Evaluate these expressions when a = 5, b = 2 and c = –1.

Noughts and crosses – substitution
Divide the class into two teams, noughts and crosses. 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, using a calculator if necessary. 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 mark the cell with the opposing team’s 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.