MATHEMATICS FOR ENGINEERING BASIC ALGEBRA


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1 MATHEMATICS FOR ENGINEERING BASIC ALGEBRA TUTORIAL 3 EQUATIONS This is the one of a series of basic tutorials in mathematics aimed at beginners or anyone wanting to refresh themselves on fundamentals. The tutorial contains the following. On completion of this tutorial you should be able to do the following. Manipulate and solve simple algebraic equations. Solve simultaneous equations by graphical methods. Solve simultaneous equations by the method of elimination. Solve simultaneous equations by the method of matrices. Define the roots of a quadratic equation. Factorise quadratic equations. Solve quadratic equations by graphical methods. Solve quadratic equations by the method of completing the root. Solve quadratic equations by use of the quadratic equation. Solve cubic equations by use of the graphical method. D.J.Dunn 1
2 In tutorial 1 you learned some of the elements of algebra and how to manipulate formulae. In this tutorial we will look at some of the types of equations you will have to deal with in your studies. 1. SIMPLE EQUATIONS A simple equation contains one unknown quantity with powers no higher than 1. It should be easy to solve the unknown by applying the algebraic rules in tutorial 1. The solution of a simple equation is called the root. WORKED EXAMPLE No.1.1 The subscription for membership of an organisation is x Euros. 54 members joint from Ireland, 76 from Germany, 32 from France and 48 from Spain. The total sum received is 420 Euros. Write out an equation for the sum and determine the subscription. Adding the subscriptions must produce 54x + 76x + 32x + 48 x = 420 Simplify by adding all the x terms. 210x = 420 Divide both sides by 210 x = 420/210 = 2 Euros. WORKED EXAMPLE No. 1.2 A restaurant owner buys 24 boxes of tea bags. Each box contains an unknown number of tea bags. He uses 10 boxes on one day and 8 boxes the next day. He then checks and finds he has 150 tea bags left. How many tea bags were there in each box? Let there be n tea bags in each box. The total number of tea bags bought must be 24n The number of boxes used was = 18 The number of tea bags used was 18n The number of tea bags left must be 24n 18n = 150 Evaluate 6n = 150 Divide both sides by 6 and n = 150/6 = 25 There were 25 bags in each box. SELF ASSESSMENT EXERCISE No Solve the unknown in each of the following. 8x  2x + 3 = 3x +12 5m 5 + 2m = m A cinema sells 25 tickets for screen one, 68 tickets for screen two and 51 tickets for screen three. The price is the same for all tickets and the takings come to 504. What is the price of the tickets? D.J.Dunn 2
3 HARDER EXAMPLES WORKED EXAMPLE No. 1.3 Solve x in the following equation. 5 x x 3 12 Add 7 to both sides 5 4 x x Subtract 5 4 x x x Simplify x 19 Simplify x Simplify x 19 Finally solve x WORKED EXAMPLE No. 1.4 Solve x in the following equation. 2(x 5)  3(x + 7) = x + 12 Multiply out brackets (2x 10)  (3x + 21) = x + 12 Remove brackets noting signs 2x 103x  21 = x + 12 Gather all x terms on one side and all numbers on the other = 2x + 3x + x 43 = 2x x = 43/2 = 21 ½ WORKED EXAMPLE No. 1.5 A rectangular plate is 200 mm long. A strip 50 mm wide is cut off one end. A second strip 15 mm wide is cut off the other end. The remaining plate weight 2025 g. Find the width of the plate if the plate weighs 0.12 g per square mm. Let the plate be x mm wide. Area = 200x The area cut off is 4.5x and 11.5x Area left is 200x 50x 15x The weight of the remaining plate is (200x 50x 15x)(0.12) Equate to known weight (200x 50x 15x)(0.12) = 2025 Divide both sides by 0.09 (200x 50x 15x) = 2025/0.12 = Add the x terms 135x =16875 x = 125 mm D.J.Dunn 3
4 SELF ASSESSMENT EXERCISE No H H 1. Solve the unknown in the following equation. 12 H Solve the unknown in the following equation. 6(w + 2) 3(w 2) = 2w 3. A train travels from its starting station to another station x km distance along the track at an average speed of 50 km/h. It waits half an hour in the station before making the return trip at an average speed of 80 km/h. The total time taken for journey is 4 hours. What is the distance between the two stations? 4. A metal strip is x mm wide. It is used to make the four sides of a rectangular box measuring 200 mm by 100 mm. The area of the strip mm 2. What is the width of the strip? D.J.Dunn 4
5 2. SIMULTANEOUS EQUATIONS Consider a vehicle that starts moving from the starting post as shown and accelerates at a constant rate of a m/s 2. After travelling a short distance it passes point 0 with velocity u 0 and a stop watch is started. It accelerates a further distance s 1 and passes point 1 with velocity u 1 and the time is noted as t 1. It accelerates to the next timing point and passes it at velocity u 2 and the time is t 2. The distance travelled from the start of timing is s 2. The equation relating these quantities is known to be s = u 0 t + at 2 /2 We can use the equation twice and have: s 1 = u 0 t 1 + at 1 2 /2 and s 2 = u 0 t 2 + at 2 2 /2 This is an example of simultaneous equations where the same equation is used twice with two sets of values to solve two unknown values. The unknown values are u 0 and a. For a given value of time, the distance travelled is a function of u 0 and a so we could say s = f(u 0, a). We will do a solution for this problem later on (worked example 4). In general simultaneous equations take the form Ax + By = f 1 (x,y) Cx + Dy = f 2 (x,y) Where A, B, C and D are known coefficients. This tutorial only deals with two unknown variables. If we had three unknown variables we would need three simultaneous equations and so on. GRAPHICAL We can rearrange our equation to make x the subject and then y the subject as follows. x = (f 1 (x,y) By) A or x = (f 2 (x,y) Dy) C y = (f 1 (x,y) Ax) B or y = (f 2 (x,y) Cx) D We could plot y against x for either pair of equations and determine the point where graphs cross and hence have the same values of x and y. Graphical solutions are laborious and only really suitable for two simultaneous equations but the following example is useful. D.J.Dunn 5
6 WORKED EXAMPLE No. 2.1 Solve x and y given the following simultaneous equation f(x,y) 1 = 39 = x + 7y...(1) f(x,y) 2 = 23 = 2x + 3y..(2) Rearrange to make y the subject (you could make x the subject). y = (39 x)/7.(3) y = (23 2x)/3 (4) Plot x against y using both equations and we get: We find that x = 4 and y = 5 satisfies both equations. SUBSTITUTION METHOD This is a suitable method when there are only two unknown variables but difficult for three or more. We use one equation to find x in terms of y and substitute it into the other equation to solve y or viceversa. This is best demonstrated by doing the last example again. WORKED EXAMPLE No. 2.2 Solve x and y given the following simultaneous equations: f(x,y) 1 = 39 = x + 7y..(1) f(x,y) 2 = 23 = 2x + 3y..(2) From equation (1) x = 39 7y Substitute this into equation (2) and solve y 23 = 2(39 7y) + 3y 23 = 78 14y + 3y 23 = 78 11y 11y = = 55 y = 55/11 = 5 Now substitute for y into either (1) or (2) to find x. Doing both is a check on the answer. 39 = x + 7y = x + 7(5) = x + 35 x = 4 23 = 2x + 3y = 2x + 3(5) = 2x x = 8 x = 4 D.J.Dunn 6
7 WORKED EXAMPLE No. 2.3 Solve x and y given the following simultaneous equation f(x,y) 1 = 9 = 3x + 5y..(1) f(x,y) 2 = 5 = 2x + 3y..(2) From equation (1) x = (9 5y)/3 Substitute this into equation (2) and solve y 5 = 2(9 5y)/3 + 3y 5 = 6 10y/3 + 3y 5 = 6 y/3 y = 3 Now substitute for y into either (1) or (2) to find x. Doing both is a check on the answer. 9 = 3x + 5y = 3x + 5(3) = 3x x = 6 x = 2 5 = 2x + 3y = 2x + 3(3) = 2x + 9 2x = 4 x = 2 WORKED EXAMPLE No. 2.4 The distance travelled by an object s (metres) is related to the initial velocity u 0 (m/s), the acceleration a (m/s 2 ) and time t (s) by the equation s = u 0 t + at 2 /2 A stop watch is started as the object passes a point such that s = 0 when t = 0. It takes 2 seconds for the object to travel 14 meters from the point and 5 seconds to travel 50 m from the point. Determine the initial velocity and the acceleration. First set up two equations using the information supplied. 14 = 2u 0 + a(2 2 )/2 14 = 2u 0 + 2a..(1) 50 = 5u 0 + a(5 2 )/2 50 = 5u a..(2) From equation (1) u 0 = (14 2a)/2 = 7  a Substitute this into equation (2) and solve a 50 = 5(7  a) a = 355a a = a 7.5a = = 15 a = 2 m/s 2 Now substitute for y into either (1) or (2) to find u. Doing both is a check on the answer. 14 = 2u 0 + 2a = 2u = u 0 = 5 m/s 50 = 5u a = 5u (2) = 5u u 0 = 25 u 0 =5 m/s D.J.Dunn 7
8 ELIMINATION METHOD Many people find this method easier than substitution. The method requires that we make the coefficient of one of the variables the same by multiplying each term by a suitable number. We then add or subtract the two equations to eliminate one of the variables. A worked example shows this best. WORKED EXAMPLE No. 2.5 A resistance thermometer has a resistance R = 101Ω at a temperature θ = 20 o C and 103Ω at 60 o C. The law relating resistance and temperature is R = Ro + αθ where Ro is the resistance at 0 o C and R is the resistance at any other temperature. α is the temperature coefficient of resistance. Calculate the temperature Ro and α. Ro + 20α = 101..(1) Ro + 60α = 103..(2) Just subtract the equations Ro + 20α = 101 Ro + 60α = 103 Subtract 040α = 2 α = 2/40 = 0.05 Substitute for α in (1) Ro + 20(0.05) = 101 Ro = = 100Ω The relationship is R = θ We might use the same process after making x or y the subject as shown in the next example. WORKED EXAMPLE No. 2.6 Repeat worked example 1 using the elimination method. 39 = x + 7y..(1) 23 = 2x + 3y..(2) Make x the subject in both cases. x = 39 7y (3) x = (23 3y)/2 x = y (4) Subtract 0 = y Solve y 5.5y = 27.5 y = 5 78 = 2x + 14y Substitute for y in any equation and x is 4 as before. D.J.Dunn 8
9 WORKED EXAMPLE No. 2.7 Solve x and y given. 3x + 2y = 12.. (1) x + 3y = (2) Multiply each term in equation (2) by 3 so that the coefficient of x is the same in both. We then have: 3x + 2y = 12 3x + 9y = 33 Subtract 07 y = 21 y = 21/7 = 3 Substitute for y in (1) and 3x + 2(3) = 12. 3x = 6 x = 2 Check in equation (2) 3x + 9y = 33 3(2) + 9(3) = 33 correct WORKED EXAMPLE No. 2.8 What are the values of x and y that satisfies both the following equations. x/7 y/2 = 3..(1) x/3 + y/4 = 10..(2) Multiply each term in (1) by 2 and each term in (2) by 4. We have: 2x/7 y = 6 4x/3 + y = 40 Add 34x/ = 34 note that 2/7 + 4/3 = 34/21 34x/21 = 34 x = 21 Substitute into (1) 21/7 y/2 = 3 3 y/2 = 3 6 = y/2 y = 12 Check in (2) x/3 + y/4 = 10 21/3 + 12/4 = 10 correct D.J.Dunn 9
10 SELF ASSESSMENT EXERCISE No. 2.1 Solve the variables in the following simultaneous equations. 1. 3x + 5y = 6 and 2x + 3y = 3 (Answers x = 3, y = 3) 2. 7x + 4y = 20 and 3x + 3y = 6 (Answers x = 4, y = 2) 3. 5x + 5y = 40 and x + 3y = 48 (Answers x = 12, y = 20) 4. x + y = 3/4 and 2x + 3y = 7/4 (Answers x = ½, y = 1/4) 5. x + 2 y = 1.9 and 2x + 5y = 4 (Answers x = 3/2, y = 1/5) 6. 3x  5 y = 204 and 4x + 5y = 412 (Answers x = 88, y = 12) 7. x  5 y = 95 and 2x + y = 245 (Answers x = 120, y = 5) MATRIX METHOD This method should only be used if you have a good understanding of matrix theory. The main use of this method is that it enables the solution of problems with more than two unknown variables and because the method is based on strict rules, it is suitable for use in computer programmes. Here is a basic example that might serve as an introduction to matrix theory. WORKED EXAMPLE No. 2.9 What are the values of x and y that satisfies both the following equations. 3x + 2y = 22..(1) 2x + 3y =23..(2) Create a matrix of the coefficient and column vectors based on the variables and the function values. The object is to solve the column vector X. The simultaneous equations may be represented as A X = b Next we need to find the inverse matrix A 1 For a 2 x 2 matrix the rule is : D.J.Dunn 10
11 Put in the numbers The solution is given by the product X = A 1 b The rule for multiplying is a 4 x 4 matrix by a single column is Putting in the numbers we get: The solution is x = 4 and y = 5 as before. The matrix method looks difficult but matrix theory is very important in modern Engineering methods and should be thoroughly studies for advanced courses. The example is used only as an introduction. The rules are much more complicated for larger arrays. You might care to solve some of the previous examples with the above model. D.J.Dunn 11
12 3. QUADRATIC EQUATIONS Most relationships in Engineering and Science are anything but proportional and many are quadratic. This means that the function contains a highest power of 2. Take a simple case y = f(x) = 64 = x 2 To find the value of x that satisfies the equation we simply take the square root. x = 64 Now there are two possible answers 8 and 8 because a minus number squared is positive. An immediate problem arises when we look at the case y = 64 = x 2 Taking the square root this time gives x = 64 and you probably don t know how to deal with the roots of negative numbers and we wont cover it here. QUADRATIC EQUATIONS and ROOTS Quadratic equations have the general form: y = f(x) = 0 = ax 2 + bx + c a, b and c are the coefficient of x 2 x 1 and x o respectively. Of course we don t normally bother writing x 1 as x 1 = x and we don t normally write x o as x o =1. If y is not zero, we only need to subtract y from both sides of the equation to produce the required form. There are normally two values of x that satisfy the function and this may be seen by plotting a graph. Consider the case y = f(x) = 2x 24x 6 If we plot x against y over a suitable range we get the graph shown. From the graph we can pick off the two values of x that correspond to any value of y but it is of particular interest to find the values of x when y is zero. These are called the ROOTS OF THE EQUATION and in the example they are x = 1 and x = 3. It is always a good idea to check that putting these values into the equation produces y = 0. x = 1 y = 2(1) 24(1) 6 = = 0 x = 3 y = 2(3) 24(3) 6 = = 0 It is nice when the roots are whole numbers (integers) but this is not usually the case. Note that quadratic equations do not always cross the x axis and the solution to these requires advanced studies. D.J.Dunn 12
13 FACTORISATION You should know that an integer will factorise such that the product of the factors give the original number. For example the factors of 8 may be (8)(1) = 8 or (4)(2) = 8. In the case of equations, the factors are not numbers but expressions. A quadratic equation often factorises into two parts such that the sum of the two parts produces the original equation. Knowing how to factorise an equation takes practice. The first step in understanding this is see how multiplying two factors gives the original equation. Suppose that the two factor are (2x + 2) and (x  3). Let s multiply them. Here s how. The two factors of y = 2x 24x 6 are hence (2x + 2) and (x  3). The trouble is that we need to do the reverse. Given an equation, find the factors. This rule helps us. Even with this rule, you need to play around with the possibilities before you arrive at the correct answer. The problem is easier if the coefficient a = 1. In the above example, we could simply divide by 2 to get x 22x 3 and the factors are (x + 1)(x  3). Note that the 2 is obtained by adding 1 and 3 and this always works. WORKED EXAMPLE No. 3.1 Solve f(x) = 0 = 6x 2 + 8x + 2 Divide by 6 to get 0 = x 2 + (8/6)x + (2/6) Make two brackets (x + A)( x + B) A B = 2/6 and A + B = 8/6 so make A = 1 and B = 2/6 (x + 1)(x + 2/6) = 0 x = 1 or x = 2/6 = 1/3 Check them out in the original equation x = 1/3 f(x) = 6(1/3) 2 + 8(1/3) + 2 = 6/9 8/3 + 2 = 2/3 8/3 + 2 = 6/3 + 2 = 0 x = 1 f(x) = 6(1) 2 + 8() + 2 = = 0 It takes practice to do this quickly and it only works if the numbers are kind. D.J.Dunn 13
14 WORKED EXAMPLE No. 3.2 Solve f(x) = 0 = 2x 2 + 3x  5 Factorise to get two brackets. We know we must have 2x and x (2x + A)( x  B) = 0 We know AB must be 5 so try 1 and 5 (2x + 1)( x  5) = 0 The middle term must be the sum of x and 10x and this gives 9x which is wrong. Try 5 and 1 (2x + 5)( x  1) = 0 The middle term must be the sum of 5x and 2x and this gives 3x which is correct. (2x + 5)( x  1) = 0 so the roots are x = 5/2 and x = 1 Check it in the original equation. x = 5/2 2(5/2) 2 + 3(5/2)  5 = 25/2 15/2 5 = 0 correct x = 1 2(1) 2 + 3(1)  5 = = 0 correct SELF ASSESSMENT EXERCISE No. 3.1 Solve the following by factorisation. 1. 2x 25x 3 = 0 2. x 2 + 2x 8 = 0 3. x 22x 3 = x 220x 3 = 0 PERFECT SQUARE ROOTS AND COMPLETING THE SQUARE When the two factors are the same, we have a perfect square root. Consider the case x 2 6x + 9 = 0 If we factorise x 2 6x + 9 we get (x 3) (x 3). We could write x 2 6x + 9 = (x 3) 2 = 0 and there is only one solution x = 3, and this is the perfect root. Factorisation is easier if we can change the equation so that it has a perfect square root. This is best demonstrated with an example. WORKED EXAMPLE No. 3.3 Solve x 2 6x + 8 = 0 If the last number was 9 we would have a perfect square root. We can do this by manipulating the equation to: x 2 6x = 0 or x 2 6x + 9 = 1 Factorise the left side into (x  3)(x  3) or (x  3) 2 Now solve by taking the square root. (x  3) = 1 = ±1 so x = 4 or x = 2 Check x = 4 x 2 6x + 8 = (4) + 8 = = 0 so 4 is the root of the original equation. Check x = (2) + 8 = = 0 so 2 is the root of the original equation. D.J.Dunn 14
15 SELF ASSESSMENT EXERCISE No. 3.2 Find the roots of the following by completing the square. 1. x 2 + 6x + 5 = 0 2. x x + 20 = 0 3. x 28x + 11 = 0 5. x 24x  3 = 0 QUADRATIC FORMULAE A method of finding the solution to quadratic equations that is most reliable and widely used is given here without proof. Arrange the equation into the form ax 2 + bx + c = 0 b b 2 4ac The two solutions are then given by x and this is the quadratic formula that 2a should be memorised. Most scientific/engineering calculators can find the solution direct but this formula is useful in advanced studies. WORKED EXAMPLE No. 3.4 Solve 2x 2 4x  5= 0 a = 2 b = 4 c = 5 b x 2 b 4ac 2a (4) (4) 2(2) 4(2)(5) x x or x 2.871or Check the answers 2(2.871) 2 4(2.871)  5= 0 2(0.871) 2 4(0.871)  5= 0 2 SELF ASSESSMENT EXERCISE No. 3.3 Find the roots of the following by completing the square. 1. 3x 2 + 5x + 2 = 0 (Answer 1 and ) 2. 5x 22x  4 = 0 (Answer and ) 3. 2x 22x + 5 = 0 (Answer and 1.158) 4. 5x 2 + 3x + 7 = 0 (Answer and 1.521) D.J.Dunn 15
16 4. CUBIC EQUATIONS Suppose wish to solve x in the equation 2x x 220x 50 = 0 The solution is the roots of f(x) = 2x x 220x 50 To solve x we must plot f(x) = y = 2x x 220x 50 and see if any value of x produces a result y = 0 The plot is shown. The graph crosses the y axis three times so there are three values of x that give a solution and these are x = 6, x = and 2.59 Check them out to see if they are correct. x = 6 2x x 220x 50 = 2 so the answer is not exact but close. x = x x 220x 50 = so the answer is not exact but close. x = x x 220x 50 = 0.03 so the answer is not exact but close. The graphical method will not usually give exact answers depending on the scale of the graph. Many cubic plots do not cross the y axis and there may be three, two, one or no real answers that satisfy the equation. We can rearrange the equation to make x 3 the subject and plot two graphs like before. In this case we have : x 3 = ( 10x x + 50)/2 = ( 5x x + 25) If we let f 1 (x) = x 3 and f 2 (x) = ( 5x x + 25) and plot both functions against x, the point where the graphs cross is the point where the functions are equal and give the answer for x. D.J.Dunn 16
17 WORKED EXAMPLE No. 4.1 Solve the following equation graphically. (4x 3 3x 236x + 27) = 4.67(x+1) Rearrange to make two functions that can be plotted. Let f 1 (x) = (4x 3 3x 236x + 27) and f 2 (x) = 4.67(x + 1) Plot both over a suitable range and pick off the answers from the points where the graphs cross (the common abscissa) x = 3.1,x = 0.56 and x = 3.3 WORKED EXAMPLE No. 4.2 Solve the following equation graphically. (x 2 6)(4 + x 2 ) = 32x If we multiply out the brackets we get a quartic equation x 46x 228x 24 = 0 This example is to show how a graphical method may be used for any equation if the right approach is used. A solution is to rearrange the equation to make two functions that can be plotted. This is the method used here. Rearrange to the form (x 2 6) = 32x/(4 +x 2 ) Let f 1 (x) = (x 2 6) and f 2 (x) = 32x/(4 +x 2 ) Plot both over a suitable range and pick off the answers x = and x = 3.6 D.J.Dunn 17
18 SELF ASSESSMENT EXERCISE No Solve the following equations. x 2 5.2x = 0 (Answers x = 3.2 or 2) 3 x 220 log(x) = 0 (Answer x = 2.15) x x x = 0 (Answers x = 2 or 1.2 or 2.67) 2. A closed metal cylindrical canister has a mean radius R and length L. The surface area of the metal used is given by the formula A = 2π(R 2 + RL) Given that the area is 1000 mm 2 and the length is 20 mm what is the radius? (Answers 6.1 mm, the negative answer is ignored as it is not physically possible) 3. A cylindrical vessel of diameter D and length L has hemispherical ends. The volume of the vessel is hence given by V = π(dl + D 3 /6). If the volume is 200 x 10 3 mm 3 and the length L is 50 mm, what is the diameter? (Answer mm, the other answer mm is obviously only theoretical and not physical) 4. Solve the three roots of the equation y = 5x 3 +10x 22x 6 (Answer , and 0.735) SAE x = 3 m = 3 1/ SAE S SA H H 4H 3H 6H H 1. H H w w + 6 = 2w w = Time taken to complete the first journey is x/50 Time spent waiting = ½ Time taken for the return journey is x/80 Total time = x/50 + ½ + x/80 = 4 x/50 + x/80 = 3 ½ =7/2 x(1/50 + 1/80) = 7/2 x( )/4000 = 7/2 x (130/4000) = 7/2 x = (28000/260) = km mm D.J.Dunn 18
19 S.A.E (Answers x = 3, y = 3) 2. (Answers x = 4, y = 2) 3. (Answers x = 12, y = 20) 4. (Answers x = ½, y = 1/4) 5. (Answers x = 3/2, y = 1/5) 6. (Answers x = 88, y = 12) 7. (Answers x = 120, y = 5) S.A.E (3 and ½) 2. (2 and 4) 3. (3 and 1) 4. (3 and 1/7) S.A.E (Answer 1 and 5) 2. (Answer 2 and 10) 3. (Answer and 6.236) 6. (Answer and ) S.A.E (Answer 1 and ) 2. (Answer and ) 3. (Answer and 1.158) 4. (Answer and 1.521) S.A.E (Answers x = 3.2 or 2) (Answer x = 2.15) (Answers x = 2 or 1.2 or 2.67) 2. (Answers 6.1 mm, the negative answer is ignored as it is not physically possible) 3. (Answer mm, the other answer mm is obviously only theoretical and not physical) 4. (Answer , and 0.735) D.J.Dunn 19
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