1 LINEAR EQUATIONS Old-fashioned scales like these are still used in many markets around the world. Just like linear equations, you must do the same to both sides to keep them balanced. s In this chapter you will: solve simple equations solve linear equations containing brackets and fractions solve linear equations in which the unknown appears on both sides of the equation set up and solve simple linear equations distinguish between the words equation, formula, identity and expression. Before you start You need to be able to: collect like terms in an algebraic expression expand or multiply out brackets calculate using directed numbers apply the rules of BIDMAS. 211
hapter 1 Linear equations 1.1 Solving simple equations You can solve simple equations. Why do this? If you have a number of albums on your mp player, as well as some individual tracks, you can solve a simple equation to find out how many songs you have in total. 1. Simplify a 4 p 2q p 7q b (1 2z) c 4m(m 9) Key Points In any equation, the value of the left-hand side must always be equal to the value of the right-hand side. So whatever operation is applied to the left-hand side must also be applied to the right-hand side. Two children sit on a see-saw, equally distant from the middle. The see-saw is level which means that the weight of each child is the same. A B Weight of child A on the left-hand side weight of child B on the right-hand side. The weight of child A is 5x 12. The weight of child B is 57. So, 5x 12 57. This is a linear equation. Example 1 Solve 5x 12 57 5x 12 57 5x 12 12 57 12 5x 45 5x 5 45 5 x 9 Subtract 12 from both sides of the equation. Divide both sides by 5. Watch Out! Always apply the same operation to both sides of the equation. Exercise 1A Questions in this chapter are targeted at the grades indicated. Solve 1 2a 5 1 2 b 4 17 2c 6 11 4 5d 1 8 5 1 4e 7 6 1 2f 9 7 1 g 5 8 5 1 8h 9 2 5k 15 10 17 4m 212 linear equation
1.2 Solving linear equations containing brackets 1.2 Solving linear equations containing brackets You can solve linear equations that require prior simplification of brackets. 1. Solve a 2x 7 19 b b 8 20 c 16 40 q Key Points Follow the rules of BIDMAS when solving a linear equation. (See Section 1. on BIDMAS). Solutions can be written as mixed fractions, improper fractions or as decimals. Example 2 Solve 4(x 1) 11 4(x 1) 11 4x 4 11 4x 4 4 11 4 4x 7 4x 4 7 4 x 7 4 x 1 or 1.75 4 Expand the left-hand side by multiplying out the brackets. Solve the equation as in Section 1.1. Examiner s Tip Your answer can be written as a fraction or as a decimal. Exercise 1B Solve 1 2(x ) 12 2 5(y 4) 5 4(x 1) 5 4 (2y 1) 9 D 5 1 4(x ) 6 2(1 w) 10 7 5( 4z) 20 8 2 (1 x) 9 2(2x ) 1 11 10 17 6 (5 2y) solution 21
hapter 1 Linear equations 1. Solving linear equations with the unknown on both sides You can solve linear equations, with integer coefficients, in which the unknown appears on both sides of the equation. Why do this? Knowing how to solve equations helps you solve other problems such as finding one weight when given another. 1. Solve a 4 2(x 6) b (a 4) c 4 8 2(8 b) Key Point ollect terms so that the ones involving the unknown are on one side of the equation. Example Solve 5x 5 x 5x 5 x 5x 5 x x x 8x 5 8x 5 5 5 8x 2 8x 8 2 8 x 1 or 0.25 4 Add x to both sides of the equation. Remember x x 0. Solve the equation as in Section 1.1. Watch Out! Always show each stage of your working. Example 4 Solve 6x 7 x Here both terms in x have a negative coefficient. Examiner s Tip 6x 7 x 6x 6x 7 x 6x 7 x 7 x 4 x x 4 Add 6x to both sides of the equation. ollect the terms in x on the side of the equation that gives them a positive coefficient. 214 coefficient
1.4 Solving linear equations containing fractions Exercise 1 Solve 1 4a 8 2a 2 5b b 7 c 2 5c 8 4 d 7 5d 15 D 5 2e 4 e 6 1 7f f 10 7 2(x 4) x 7 8 2x 5 1 (2 x) 9 (4x 1) 2(1 5x) 2 x 10 6 2(x ) x (1 2x) 1.4 Solving linear equations containing fractions You can solve linear equations containing fractions. 1. Solve a 4x 6 2(4 x) 2 b 2 8x 20 (4 9x) c 5 4(x 1) 2x 4(2 x) Key Point To solve an algebraic equation involving fractions, eliminate all fractions by multiplying each term by the LM of the denominators. Example 5 12 p 2 Solve 12 p 2 12 (p 2) (p 2) p 2 12 (p 2) 12 p 6 12 6 p p 6 p 2 Multiply both sides of the equation by (p 2). The terms in (p 2) on the left-hand side cancel out. Examiner s Tip Always try to remove the fraction first. 215
hapter 1 Linear equations Example 6 Solve x 1 4x 1 2 5 12 x 1 4x 1 5 2 12 12 x 2 1 12 4x 1 12 5 12 6 12 x 2 1 4 12 4x 1 1 1 12 5 1 12 1 6(x 1) 4(4x 1) 5 6x 6 16x 4 5 10x 10 5 10x 5 x 1 2 Multiply each of the three terms by 12. Remember: 4 1 4 Exercise 1D B A Solve 1 p 5 7 2 q 2 4 4 7 8 x 6 1 x 4 4 1 2n 1 n 7 5 (2y 10) 4y 7 2 t 6 5 2t 6 9 2 1 x 10 5 5x 2 9 m 2 m 5 21 6 2 ( x ) 16 10 y 4 2y 1 2 5 1 y 1.5 Setting up and solving simple linear equations You can set up and solve simple linear equations. Why do this? Businesses use linear equations to help them work out how much of a product needs to be produced to make a given profit. 1. Solve a 6a 1 2a 7 b 5 6b 10 4b c ( x (x ) 5 ) 2 2 Key Points When setting up an equation, define all the unknowns used that have not already been defined. Make sure that units are consistent on both sides of the equation. 216
1.5 Setting up and solving simple linear equations Example 7 Daniel makes some drinks to sell at the Summer Fair. From one bottle costing 2.80 he can make 40 drinks. Daniel wants to make a profit of 2 on each bottle. a If c is the price of each drink, write an equation in terms of c. b Solve your equation in a to find what the price of each drink should be. a 40 drinks will cost 40 c 40c pence One bottle costs 2.80 280 pence Daniel s profit 2 200 pence Profit total sales total costs 200 40c 280 hange all amounts to pence. b 200 40c 280 200 280 40c 40c 480 c 480 40 c 12 So the cost of each drink is 12 pence. Example 8 By setting up and solving an equation in terms of x, work out the size of thelargest angle of this quadrilateral. 2x 10 x 40 2x 50 Since the sum of the interior angles of a quadrilateral is 60, adding the angles gives: (2x 10) (x 40) (2x) (50) 60 2x 10 x 40 2x 50 60 5x 80 60 5x 60 80 5x 280 x 56 ollect like terms. Watch Out! Read the question carefully; is the size of the largest angle or the smallest angle required? Angles are: 2x 10 2 56 10 112 10 102 x 40 56 40 96 2x 2 56 112 is the largest angle. Exercise 1E 1 Viv thinks of a number. She multiplies the number by 5 and then subtracts 2. Her answer is 2. If x is the number that Viv was thinking of, work out the value of x. D 2 Michelle is 2 years younger than Angela. If the sum of their ages is 64, work out Michelle s age. AO2 217
hapter 1 Linear equations AO2 In an exam, Jessica scored p%. Mason scored three-quarters of Jessica s score. Zach scored 10% less than Jessica. The total of their scores was 210%. How much did each student score? AO 4 The length of a rectangle is (2x ) cm. The width of the rectangle is 1_ (x ) cm. 2 If the perimeter of the rectangle is 49 cm, find the value of x. 2x 1 2 (x ) AO 5 Joanna works for n hours each week for 5 weeks. In the sixth week she works an extra 4 1_ 2 hours. In these six weeks she works a total of 117 hours. Work out the number of hours Joanna works in the sixth week. B 6 The diagram shows an isosceles triangle AB. The lengths of the sides are given in centimetres and AB B. a Write down an equation in terms of x. b Work out the lengths of AB and B. A x 4 (x 6) 5 B AO 7 In one quarter Stuart uses 2512 units of electricity. Part of his electricity bill is shown below. Units used 2512 ost of the first x units at 12.86p per unit ost of remaining units at 11.8p per unit Total cost of electricity used 298. Work out the number of units, at 12.86p per unit, that Stuart used. Give your answer to the nearest unit. 1.6 Distinguishing between equation, formula, identity and expression You can distinguish between the words equation, formula, identity and expression. Why do this? You need to understand mathematical words to be able to understand exam questions. 1. Evaluate these expressions when x 4 and y 2. a x y b x y c 4 x 2y 218
1.6 Distinguishing between equation, formula, identity and expression Key Points An example of an algebraic expression is 2p 10. It is made up of the terms 2p and 10. An example of an equation is 2p 10 9. This can be solved to find the value of p. 2(p 5) 2p 10 is called an identity. The left-hand side is the same as the right-hand side. 2p 10 is called a formula. If the value of p is known, you can substitute it into the equation to work out the value of. Example 9 Write down whether each of the following is an expression, an equation, an identity or a formula. a 5ab 2ab ab b 5p 2q Examiner s Tip c T 2 l d g 1 2x 5 An expression is the only one without an sign. a The right-hand side is the same as the left-hand side and so 5ab 2ab ab is an identity. ollect like terms of 5ab 2ab to give ab. b 5p 2q is an expression. c If the values of l and g are known, the value of T can be worked out using the formula T 2 l_ g. d 1 2x 5 can be solved to find the value of x and is therefore an equation. Exercise 1F Write down whether each of the following is an expression, an equation, an identity or a formula. 1 A 2(l b) 2 m m m m 4m 2a b 4 y 2 24 5 y mx c 6 E mc 2 7 x 2 5x 8 1 x 2 x 4 9 V 4 πr 10 2(x ) x 4 11 x 2 5x x(x 5) 12 x 2 x 5 x 7 D equation identity formula 219
hapter 1 Linear equations hapter review In any equation, the value of the left-hand side is always equal to the value of the right-hand side. So whatever operation is applied to the left-hand side must also be applied to the right-hand side. Follow the rules of BIDMAS when solving a linear equation. Solutions can be written as mixed fractions, improper fractions or as decimals. ollect terms so that the ones involving the unknown are on one side of the equation. To solve an algebraic fraction equation, eliminate all fractions by multiplying each term by the LM of the denominators. When setting up an equation, define all unknowns used that have not already been defined. Make sure that units are consistent on both sides of an equation. An algebraic expression is made up of terms. An equation can be solved to find the value of a term. In an identity the left-hand side is the same as the right-hand side. If the value of a term is known, you can substitute it into a formula to work out the value of another term. Review exercise 1 Solve 4t 1 19 Exam Question Report D 2 Solve 4(x 2) 10 Solve a x 1 x 6 b 4y 2y 8 86% of students answered this question well because they remembered to do the same to both sides of the question. Nov 2008 4 The sizes of the angles, in degrees, of the quadrilaterals are x 10, 2x, x 90 and x 20. Work out the smallest angle of the quadrilateral. x 90 x 20 x 10 2x Diagram NOT drawn accurately Nov 2005 AO2 5 The diagram shows a rectangle. 4x 1 x x Exam Question Report 2x 12 All the measurements are in centimetres. Work out the perimeter of the rectangle. 64% of students answered this question well. June 2009, Adapted 220
hapter review 6 Uzma has x. Hajra has 20 more than Uzma. Mabintou has twice as much as Haira. The total amount of money they have is 12. Find how much money they each have. 7 AO AO x A B Here are boxes. Box A has x. Box B has 4 more than box A. Box has one third of the money in box B. Altogether there is 24 in the boxes. Find the amount of money in each box. 8 Solve x 4 1 12 9 Solve x x 5 June 2005 5 10 Solve y 2 y 1 2 B A 221