KS4 Mathematics A2 Linear equations

A2.1 Equations, formulae and identities Contents A2 Linear equations A2.1 Equations, formulae and identities A2.2 Balancing equations A2.3 Equations with brackets A2.4 Equations with fractions A2.5 Using equations to solve problems

Expressions and identities Remember in an algebraic expression we usually have a combination of numbers, letters and operations. For example, 3x + 5 + 2x – 6. Expressions do not contain an equals sign. Sometimes an expression is rewritten in an equivalent form. 3x + 5 + 2x – 6 = 5x – 1 This is an example of an identity. An identity is true for all values of x. This is sometime written with the symbol ≡ as, In both of these examples x can be any number, positive negative rational or irrational. Substitute a few values on both sides of the identity to show that it is true for any chosen value of x. 3x + 5 + 2x – 6 ≡ 5x – 1 Is identically equal to

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. Pupils often misunderstand the equals sign as meaning ‘is the answer to’ rather than ‘is equal to’. Make sure that pupils are clear about how the sign is used in equations before proceeding. x + 7 = 13 x = 6 When we solve an equation we should line up the equals signs.

Formulae A formula is a special type of equation that links two or more physical variables. For example in the formula, P = 2(l + w) P represents the perimeter of a rectangle and l and w represent its length and width. We can use this formula to work out the perimeter of any rectangle given its length and width. Physical variables such as length, time, mass and temperature have units attached to them. These are not written in the formula and so it is important that they are consistent.

A2 Linear equations Contents A2.1 Equations, formulae and identities A2.2 Balancing equations A2.3 Equations with brackets A2.4 Equations with fractions A2.5 Using equations to solve problems

Linear equations Linear equations are the easiest type of equation to solve because the unknown is not raised to any power other than 1. We can solve very simple linear equations by inspection. For example, 19 – x = 8 We think of this as, “what number subtracted from 19 gives us an answer of 8?” x = 11 In each of these examples, only one operation has been performed on the unknown. It is therefore relatively straightforward to solve them mentally. Discuss what methods pupils use to work these out in there heads. These methods may involve trial and improvement, inverse operations, flow diagrams or doing the same thing to both sides of an equation. 7x = 42 We think of this as, “what number multiplied by 7 gives us an answer of 42?” x = 6

Linear equations When more than one operation is preformed on the unknown we need to solve the equation in several steps. We can do this by performing the same operations to both sides of the equals sign to keep the equation balanced. For example, 4x + 5 = 29 subtract 5 from both sides, – 5 4x = 24 Establish that for more complex equations a more rigorous method is required. Remind pupils that we are trying to get the unknown x on its own on the left hand side of the equals sign. ÷ 4 divide both sides by 4, x = 6 Check that 4 × 6 + 5 is equal to 29 in the original equation.

Equations with unknown on both sides In some cases the unknown appears on both sides of the equals sign. For this type of equation it is important to work systematically to collect terms with unknowns on the left hand side of the equation and numbers on the right. Always aim to have a single unknown on the left equal to a single number on the right. For example, unknowns numbers 6x – 2 = 2x + 1 Here we’ll label the sides as ‘unknowns’ and ‘numbers’. We are aiming to have the unknown on the left and a number on the right.

Equations with unknown on both sides The equation can then be solved by performing the same operations on both sides until the solution is found. 8x – 2 = 2x + 1 unknowns numbers + 2 add 2 to both sides, 8x = 2x + 3 subtract 2x from both sides, – 2x 6x = 3 divide both sides by 6, Start by explaining that we only want terms containing x’s on the left-hand side. Ask pupils how we could ‘get rid of’ the – 2 from the left-hand side (by adding 2 to both sides of the equation). Next, ask how we could ‘get rid of’ the 2x from the right-hand side (by adding 2x to both sides of the equation). We could write 3 ÷ 6 as a fraction, 3/6. This then cancels to 1/2. It is usually better to write the solution as a vulgar fraction when the equivalent decimal cannot be written exactly. The solution should be substituted into the original equation to make sure that the left-hand side is equal the right-hand side as required. ÷ 6 x = 0.5 Check by substituting x = 0.5 into the expressions in the original equation. Both sides are equal to 2, so the solution is correct.

A2.3 Equations with brackets Contents A2 Linear equations A2.1 Equations, formulae and identities A2.2 Balancing equations A2.3 Equations with brackets A2.4 Equations with fractions A2.5 Using equations to solve problems

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 add 10 to both sides, + 10 + 10 6x = 4x + 10 Talk through this method for solving an equation containing brackets. Check the answer 5 by substituting it back into the original equation and verifying that the left hand side of the equation equals the right hand side. subtract 4x from both sides, - 4x - 4x 2x = 10 divide both sides by 2, ÷ 2 ÷ 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 add 5 to both sides, + 5 3x = 2x + 5 subtract 2x from both sides, - 2x 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. x = 5 In this example, dividing first means that there are fewer steps.

A2.4 Equations with fractions Contents A2 Linear equations A2.1 Equations, formulae and identities A2.2 Balancing equations A2.3 Equations with brackets A2.4 Equations with fractions A2.5 Using equations to solve problems

Solving equations with fractional coefficients Sometimes the coefficient of an unknown is a fraction. For example, 3 4 x – 5 = 9 – x We can remove the 4 from the denominator by multiplying both sides of the equation by 4. 3 4 4( x – 5) = 4(9 – x) expand the brackets, 3x – 20 = 36 – 4x Explain that we multiply by the number in the denominator to cancel it out. Point out that 3/4 x can also be written as 3x/4. Ask pupils to check the solution by substituting x = 8 into the original equation. add 4x to both sides, 7x – 20 = 36 add 20 to both sides, 7x = 56 divide both sides by 7. x = 8

Solving equations with fractional coefficients If an equation contains more than one fraction these can be removed by multiplying throughout by the lowest common multiple of the two denominators. For example, 2 3 x = x + 1 1 What is the lowest common multiple of 3 and 2? The lowest common multiple of 3 and 2 is 6. Multiplying both sides by 6, Talk through the multiplication of each term by 6. for example we can think of 6 × 2/3 as 6 × 2 ÷ 3 or as 6 ÷ 3 × 2. 2 3 6( x) = 6( x + 1) 1 expand the brackets, 4x = 3x + 6 subtracting 3x from both sides: x = 6

Solving equations involving division In this example the whole of one side of the equation is divided by 5. 2x + 7 5 = x – 1 To remove the 5 from the denominator we multiply both sides of the equation by 5. 2x + 7 = 5(x – 1) expand the brackets, 2x + 7 = 5x – 5 swap sides, 5x – 5 = 2x + 7 Talk through the equation on the board. Explain that when we have (2x + 7)/ 5 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 5. If we multiply the left hand side by 5 we must also multiply the right hand side by 5. add 5 to both sides, 5x = 2x + 12 subtract 2x from both sides, 3x = 12 divide both sides by 3. x = 4

Solving equations involving division When both sides of an equation are divided by a number we must remove these by multiplying both sides by the lowest common multiple of the two denominators. For example, 5x – 3 4 = 2x – 1 3 What is the lowest common multiple of 4 and 3? The lowest common multiple of 4 and 3 is 12. Multiplying every term by 12 gives us: 3 4 12(5x – 3) 4 = 12(2x – 1) 3 1 1 which simplifies to: 3(5x – 3) = 4(2x – 1)

Solving equations involving division We can then solve the equation as usual. 3(5x – 3) = 4(2x – 1) expand the brackets, 15x – 9 = 8x – 4 add 9 to both sides, 15x = 8x + 5 subtract 8x from both sides, 7x = 5 x = 5 7 divide both sides by 7 Although this answer could be written as a rounded decimal, it is more exact left as a fraction.

Simplifying equations by cross-multiplication

Solving equations involving division 5x – 3 4 = 2x – 1 3 We have seen that Simplifies to 3(5x – 3) = 4(2x – 1) How could we perform this simplification in one step? Multiplying both sides by 4 cancels out the 4 on the left hand side and multiplies the expression on the right hand side by 4. Cross multiplication is an efficient way to rearrange an equation or formula in this form. Make sure that pupils understand why it works in terms of doing the same thing to both sides of the equation. Multiplying both sides by 3 cancels out the 3 on the right hand side and multiplies the expression on the left hand side by 3. Doing this in one step in often called cross multiplication.

Solving equations involving division Sometimes the unknowns appear in the denominator. 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) expand 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 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 question. 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 tell you which form of the equation is the easiest to solve (the form with no denominators). Ask pupils to solve the equation in the window and substitute this solution into the equivalent equations found to verify the solution.

A2.5 Using equations to solve problems Contents A2 Linear equations A2.1 Equations, formulae and identities A2.2 Balancing equations A2.3 Equations with brackets A2.4 Equations with fractions A2.5 Using equations to solve problems

Constructing an equation I’m thinking of a number. When I subtract 9 from the number and double it, I get the same answer as dividing the number by 5. What number am I thinking of? Let’s call the unknown number n. We can solve this problem by writing the equation: 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. n 5 2(n – 9) = The number with 9 subtracted and doubled is the same as the number divided by 5.

Solving the equation We can solve this equation by performing the same operations on both sides, 2(n – 9) = n 5 multiply both side by 5, 10(n – 9) = n expand the bracket, 10n – 90 = n add 90 to both sides, 10n = n + 90 subtract n from both sides, 9n = 90 Go through these steps to solve the equation step by step. divide both sides by 9, n = 10 We can check the solution by substituting it back into the original equation: 2 × (10 – 9) = 10 ÷ 5

Using equations to solve problems The sum of the ages of Ben and his daughter, Alice, is 66 years. In one year’s time Ben will be three times Alice’s age. How old is Alice now? Let’s call Alice’s age a. Looking at the first piece of information we can write Ben’s current age as 66 – a Using the second piece of information we can also write Ben’s current age as 3(a + 1) – 1 To solve this problem we can write Bens age in two different ways using the information given. Equating these two expressions will give us an equation in a. This is three times Alice’s age in one year time, minus one to give Ben’s age now. This simplifies to, 3(a + 1) – 1 = 3a + 3 – 1 = 3a + 2

Using equations to solve problems The expressions 66 – a and 3a + 2 both give Ben’s current age. This means that they are equal and so we can write them in an equation: 66 – a = 3a + 2 swap sides, 3a + 2 = 66 – a subtract 2 from each side, 3a = 64 – a add a to both sides, 4a = 64 divide both sides by 4. a = 16 Alice is 16, so Ben is 50. In one year Alice will be 17 and Ben will be 51. 51 is three times 17 and so our solution is correct.

Find the value of x Discuss the equation that need to be written to find x. Write this equation on the board using the pen tool and ask a volunteer to solve it. Chose level 2 to give the perimeter in term of x. Reveal the solution and ask pupils to use it to find the actual width and height of the rectangle and so check that the solution is correct.