# Equations, Inequalities & Partial Fractions

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1 Contents Equations, Inequalities & Partial Fractions.1 Solving Linear Equations 2.2 Solving Quadratic Equations 1. Solving Polynomial Equations 1.4 Solving Simultaneous Linear Equations 42.5 Solving Inequalities 50.6 Partial Fractions 60 Learning outcomes In this Workbook you will learn about solving single equations, mainly linear and quadratic, but also cubic and higher degree, and also simultaneous linear equations. Such equations often arise as part of a more complicated problem. In order to gain confidence in mathematics you will need to be thoroughly familiar with these basis topics. You will also study how to manipulate inequalities. You will also be introduced to partial fractions which will enable you to re-express an algebraic fraction in terms of simpler fractions. This will prove to be extremely useful in later studies on integration.

2 Solving Linear Equations.1 Introduction Many problems in engineering reduce to the solution of an equation or a set of equations. An equation is a type of mathematical expression which contains one or more unknown quantities which you will be required to find. In this Section we consider a particular type of equation which contains a single unknown quantity, and is known as a linear equation. Later Sections will describe techniques for solving other types of equations. Prerequisites Before starting this Section you should... Learning Outcomes On completion you should be able to... be able to add, subtract, multiply and divide fractions be able to transpose formulae recognise and solve a linear equation 2 HELM (2008): Workbook : Equations, Inequalities & Partial Fractions

3 1. Linear equations Key Point 1 A linear equation is an equation of the form ax + b = 0 a 0 where a and b are known numbers and x represents an unknown quantity to be found. In the equation ax + b = 0, the number a is called the coefficient of x, and the number b is called the constant term. The following are examples of linear equations x + 4 = 0, 2x + = 0, 1 2 x = 0 Note that the unknown, x, appears only to the first power, that is as x, and not as x 2, x, x 1/2 etc. Linear equations often appear in a non-standard form, and also different letters are sometimes used for the unknown quantity. For example 2x = x + 1 t 7 = 17, 1 = z + 1, 1 1 y = 2α 1.5 = 0 2 are all examples of linear equations. Where necessary the equations can be rearranged and written in the form ax + b = 0. We will explain how to do this later in this Section. Task Which of the following are linear equations and which are not linear? (a) x + 7 = 0, (b) t + 17 = 0, (c) x = 0, (d) 5p = 0 The equations which can be written in the form ax + b = 0 are linear. (a) (b) (c) (d) (a) linear in x (b) linear in t (c) non-linear - quadratic in x (d) linear in p, constant is zero To solve a linear equation means to find the value of x that can be substituted into the equation so that the left-hand side equals the right-hand side. Any such value obtained is known as a solution or root of the equation and the value of x is said to satisfy the equation. HELM (2008): Section.1: Solving Linear Equations

4 Example 1 Consider the linear equation x 2 = 10. (a) Check that x = 4 is a solution. (b) Check that x = 2 is not a solution. (a) To check that x = 4 is a solution we substitute the value for x and see if both sides of the equation are equal. Evaluating the left-hand side we find (4) 2 which equals 10, the same as the right-hand side. So, x = 4 is a solution. We say that x = 4 satisfies the equation. (b) Substituting x = 2 into the left-hand side we find (2) 2 which equals 4. Clearly the left-hand side is not equal to 10 and so x = 2 is not a solution. The number x = 2 does not satisfy the equation. Task Test which of the given values are solutions of the equation 18 4x = 26 (a) x = 2, (b) x = 2, (c) x = 8 (a) Substituting x = 2, the left-hand side equals = 10. But so x = 2 is not a solution. (b) Substituting x = 2, the left-hand side equals: 18 4( 2) = 26. This is the same as the right-hand side, so x = 2 is a solution. (c) Substituting x = 8, the left-hand side equals: 18 4(8) = 14. But and so x = 8 is not a solution. 4 HELM (2008): Workbook : Equations, Inequalities & Partial Fractions

5 Exercises 1. (a) Write down the general form of a linear equation. (b) Explain what is meant by the root or solution of a linear equation. In questions 2-8 verify that the given value is a solution of the given equation. 2. z 7 = 28, z = 7. 8x = 11, x = s + = 4, s = x + 4 = 2, x = t + 7 = 7, t = x 1 = 10, x = t 1 = 0, t = 100. s 1. (a) The general form is ax + b = 0 where a and b are known numbers and x represents the unknown quantity. (b) A root is a value for the unknown which satisfies the equation. 2. Solving a linear equation To solve a linear equation we make the unknown quantity the subject of the equation. We obtain the unknown quantity on its own on the left-hand side. To do this we may apply the same rules used for transposing formulae given in Workbook 1 Section 1.7. These are given again here. Key Point 2 Operations which can be used in the process of solving a linear equation add the same quantity to both sides subtract the same quantity from both sides multiply both sides by the same quantity divide both sides by the same quantity take the reciprocal of both sides (invert) take functions of both sides; for example cube both sides. HELM (2008): Section.1: Solving Linear Equations 5

6 A useful summary of the rules in Key Point 2 is whatever we do to one side of an equation we must also do to the other. Example 2 Solve the equation x + 14 = 5. Note that by subtracting 14 from both sides, we leave x on its own on the left. Thus x = 5 14 x = 9 Hence the solution of the equation is x = 9. It is easy to check that this solution is correct by substituting x = 9 into the original equation and checking that both sides are indeed the same. You should get into the habit of doing this. Example Solve the equation 19y = 8. In order to make y the subject of the equation we can divide both sides by 19: 19y = 8 19y 19 = 8 19 cancelling 19 s gives y = 8 19 so y = 2 Hence the solution of the equation is y = 2. 6 HELM (2008): Workbook : Equations, Inequalities & Partial Fractions

7 Example 4 Solve the equation 4x + 12 = 0. Starting from 4x + 12 = 0 we can subtract 12 from both sides to obtain 4x = 0 12 so that 4x = 12 If we now divide both sides by 4 we find 4x = cancelling 4 s gives x = So the solution is x =. Task Solve the linear equation 14t 56 = 0. t = 4 Example 5 Solve the following equations: (a) x + = 7, (b) x + = 7. (a) Subtracting from both sides gives x = 7. (b) Subtracting from both sides gives x = 7. Note that when asked to solve x + = ± 7 we can write the two solutions as x = ± 7. It is usually acceptable to leave the solutions in this form (i.e. with the 7 term) rather than calculate decimal approximations. This form is known as the surd form. HELM (2008): Section.1: Solving Linear Equations 7

8 Example 6 Solve the equation 2 (t + 7) = 5. There are a number of ways in which the solution can be obtained. The idea is to gradually remove unwanted terms on the left-hand side to leave t on its own. By multiplying both sides by we find (t + 7) = 2 5 = and after simplifying and cancelling, t + 7 = 15 2 Finally, subtracting 7 from both sides gives t = = = 1 2 So the solution is t = 1 2. Example 7 Solve the equation (p 2) + 2(p + 4) = 5. At first sight this may not appear to be in the form of a linear equation. Some preliminary work is necessary. Removing the brackets and collecting like terms we find the left-hand side yields 5p + 2 so the equation is 5p + 2 = 5 so that p = 5. Task Solve the equation 2(x 5) = (x + 6). (a) First remove the brackets on both sides: 2x 10 = x 6. We may write this as 2x 10 = x. 8 HELM (2008): Workbook : Equations, Inequalities & Partial Fractions

9 (b) Rearrange the equation found in (a) so that terms involving x appear only on the left-hand side, and constants on the right. Start by adding 10 to both sides: 2x = x + 7 (c) Now add x to both sides: x = 7 (d) Finally solve this to find x: x = 7 Example 8 Solve the equation 6 1 2x = 7 x 2 This equation appears in an unfamiliar form but it can be rearranged into the standard form of a linear equation. By multiplying both sides by (1 2x) and (x 2) we find (1 2x)(x 2) 6 1 2x = (1 2x)(x 2) 7 x 2 Considering each side in turn and cancelling common factors: 6(x 2) = 7(1 2x) Removing the brackets and rearranging to find x we have 6x 12 = 7 14x Further rearrangement gives: 20x = 19 The solution is therefore x = HELM (2008): Section.1: Solving Linear Equations 9

10 Example 9 Figure 1 shows three branches of an electrical circuit which meet together at x. Point x is known as a node. As shown in Figure 1 the current in each of the branches is denoted by I, I 1 and I 2. Kirchhoff s current law states that the current entering any node must equal the current leaving that node. Thus we have the equation I = I 1 + I 2 I x I 2 I 1 Figure 1 (a) Given I 2 = 10 A and I = 18 A calculate I 1. (b) Suppose I = 6 A and it is known that current I 2 is five times as great as I 1. Find the branch currents. (a) Substituting the given values into the equation we find 18 = I Solving for I 1 we find I 1 = = 8 Thus I 1 equals 8 A. (b) From Kirchhoff s law, I = I 1 + I 2. We are told that I 2 is five times as great as I 1, and so we can write I 2 = 5I 1. Since I = 6 we have 6 = I 1 + 5I 1 Solving this linear equation 6 = 6I 1 gives I 1 = 6 A. Finally, since I 2 is five times as great as I 1, we have I 2 = 5I 1 = 0 A. 10 HELM (2008): Workbook : Equations, Inequalities & Partial Fractions

11 Exercises In questions 1-24 solve each equation: x = x = 6. x = 7 4. x = t = t = t = t = 4 x 9. 6 = 10. x = 11. 7x + 2 = x + 2 = x + 1 = x + 1 = t = x = 2x + 8 x 17. x = 8 + x = x = x = y = y = y = = 4γ. In questions solve each equation: 25. y 8 = 1 y 26. 7t 5 = 4t x + 4 = 4x x = 4x x + 7 = 7x (x + 7) = 7(x + 2) 1. 2x 1 = x 2. 2(x + 4) = 8. 2(x ) = (x ) = 6 5. (x 1) = (2t + 1) = 4(t + 2) 7. 5(m ) = m = 5(m ) + 2m 9. 2(y + 1) = (x 2) + (x 1) = x 41. (x + ) = m = m = x + = x + 10 = 1 m x + 4 = x 4 = If y = 2 find x if 4x + y = If y = 2 find x if 4x + 5y = 50. If y = 0 find x if 4x + 10y = If x = find y if 2x + y = If y = 10 find x when 10x + 55y = If γ = 2 find β if 54 = γ 4β In questions 54-6 solve each equation: x 5 2x 1 x = x 2 x 6 = x 2 + 4x = 2x 7 5 m + 2 = m + 1 x 2 = 5 x 59. x 1 x + 1 = 4 x + 1 x = y y + = 2 4x x 1 = x 6 2s s + 1 = Solve the linear equation ax + b = 0 to find x Solve the linear equation ax + b = 1 (a c) to find x cx + d HELM (2008): Section.1: Solving Linear Equations 11

12 s /6 5. 1/ / / / / /1 21. y = /7 2. y = / / /4 0. 7/ /9 6. 7/6 7. 2/ / /4 4. 5/ / / / / / 60. 1/ /6 6. 2/5 64. b/a 65. (d b) (a c) 12 HELM (2008): Workbook : Equations, Inequalities & Partial Fractions

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