MATHEMATICS FOR ENGINEERING BASIC ALGEBRA

Transcription

1 MATHEMATICS FOR ENGINEERING BASIC ALGEBRA TUTORIAL 1 ALGEBRAIC LAWS This tutorial is useful to anyone studying engineering. It uses the principle of learning by example. On completion of this tutorial you should be able to do the following. Explain and use algebraic symbols. Explain the conformity of units in equations. Manipulate algebraic expressions and equations. Explain and use the Distributive Law. Calculate the product of two binomial expressions. D.J.Dunn 1

2 1. BASIC ALGEBRAIC MANIPULATION 1.1 SYMBOLIC REPRESENTATION You should already be familiar with mathematical sums such as: = 17 and 18 = 9 18 The division sign can be replaced by a dividing line so we can write 18 = 9 as = 9 In engineering it is common to use the solidus instead of the dividing line and write 18/ = 9 Assuming that you can do these basic arithmetical operations let s see if you can replace the symbols in the following sums with the missing numbers. For example in the sum 1 + = 17 The apple has replaced the number 5 SELF ASSESSMENT EXERCISE No.1 1. What does the strawberry represent in this sum? + 15 =. What does the cherry represent in this sum? 16 - = 4 3. What does the banana represent in this sum? 4 = 6 4. What does the pineapple represent in this sum? x = 17 If you have got the idea then you can do algebra. The pictures are symbols and in algebra we don t use pictures but letters to represent unknown numbers and the most common ones are a, b, c and x, y, z but any letter can be used. Many symbols are letters from the Greek alphabet such as α (alpha), β Beta) and γ (gamma). It is a good idea to speak these symbols by name as you read them until it is natural to read equations containing them. SELF ASSESSMENT EXERCISE No. 1. x = 180. y 7 = 3 3. z 7 = /α = β = 6. 9 x γ = 7 Note that when using symbols we often leave out the multiplication sign so in the last question we could write 9 γ = 7 instead of 9 x γ = 7 The multiplication sign x is easily confused with the symbol x. In many maths text books you will find a dot is used to show multiplication so that for example, 7 x = 14 becomes 7. = 14 but take care as this is easily confused with the decimal point. D.J.Dunn

3 1.. UNITS AND DIMENSIONS The purpose of algebra is to help us solve unknown quantities in equations that represent real things. It is useful for engineers to realise that the numbers and symbols must have units and that the units must make sense. This is called dimensional compatibility. For example it would be nonsense to write an equation such as: apples + 3 plums = 5 pears It would make sense if the equation was apples + 3 apples = 5 apples To avoid confusion between symbols and units, the units will be written in square brackets. In an engineering equation we might sensibly have 3 [kg] + 5 [kg] =8 [kg] We could not have an equation that mixed the units up so the following is nonsense. 3 [kg] + 5 [m] = 8 In multiplication and division the units must make sense as well. For example [m] x 3[m] = 6 [m ] This is because the product of metres (m) and metres is square metres [m ] We can mix the units so long as the product is correct for example [N] x 3 [m] = 6 [Nm] The product of Newton [N] and metre [m] is Newton Metre [Nm] Here is another example 6[m] [s] = 3 [m/s] or 6[m]/[s] = 3 [m/s] Metres divided by seconds are m/s When you need to work out the units of the answers, it is a good idea to use the square bracket method. Never mix multiples of the same units in equation. For example never mix mm and m or g and kg or seconds and minutes. Be consistent. Although units are vitally important, you don t have to worry about units until you come to replace symbols with numbers in order to work out the answer so most of the equations will not show units SWAPPING SIDES Consider the sum 1 = 6 x It is just as correct to write 6 x = 1 In the same way an equation such as x = y z is just as correct written as y z = x The equality is the same so when ever we want to, we can simple swap the two sides of the equals sign over. D.J.Dunn 3

4 1.4. DIVIDING and MULTIPLYING BOTH SIDES In an equation such as a = b x c a is the subject for obvious reasons. We can make b the subject by dividing both sides of the equation by c. Let s write the equation as a = b c a b c Put a dividing line under both sides and c on the bottom. = c c a c This is the same as = b c c a Since anything divided by itself is one then = b c We can swap sides and write sides by b to make c the subject. a b = and b is now the subject. Similarly we could have divided both c a c = Note that this involved simply changing c with b b What we have done is move the c from above on one side to below on the other and this is called cross multiplication. The rule can be written like this. WORKED EXAMPLE No.1 A rectangle has sides of Depth D and Breadth B. The area is given by A = B x D Make D the subject. If the area is 0 m and B is 5 m, what is the depth? A = B x D divide both sides by B A BD A = = D swap sides and D = B B B 0 Put in the known values D = and so D = 4 and further the units must be metres 5 B Now consider the equation A = C If we multiply both sides by C the equality is maintained and A C = B Swap both sides and B = A C so B is now the subject of the equation. WORKED EXAMPLE No. V Ohm s Law states R = where R is the resistance in Ohms, V the voltage and I the current in I Amps. Make I the subject and determine the current when 0 Volts is applied to 5 Ohm resistor. V R = Multiply both sides by I and R I = V I Divide both sides by R and: V I = Evaluate and I = 0/5 = 4 Amperes. R Note the units are not the same in this equation but the equation defines the Ohm as Volts per Ampere and this makes it correct. D.J.Dunn 4

5 1.5 SUBTRACTING OR ADDING TO BOTH SIDES Consider the sum 0 = Now add to both sides. 0 + = or = = The sum is still correct the equality has been maintained. This is always true if we subtract the same number from both sides. Say we subtract 3 from both sides. 0-3 = or 17 = = 17 Again the sum is still correct. If we add or subtract a number or symbol the same as one already there, we make it disappear on one side. For example if we have 0 = Subtract 5 from both sides. 0 5 = = 15 Here is another example = 1 Add 4 to both sides. 16 = This may put another way. If you move a number or symbol to the other side, it changes sign from plus to minus or minus to plus. WORKED EXAMPLE No.3 In the equation y = x + z make z the subject. Subtract x from both sides Swap sides y x = x + z x = z z = y - x When you are use to this you would go straight to the solution by moving x over and changing the sign. Here is a harder example. WORKED EXAMPLE No.4 The Perimeter of a rectangle of Breadth B and Depth D is P = D + B. Make B the subject. Given that P is 00 mm and B is 40 mm, what is D? First divide both sides by P = D + B = D + B Now subtract B from both sides to get rid of B on the right P B = D + B - B P P B = D Now swap both sides over D = B 00 Put in numbers and evaluate D = 40 = = 60 and the units must be mm. D.J.Dunn 5

6 WORKED EXAMPLE No.5 Given the equation A R = + C make B the subject. Given A = 1, C = and R = 4 evaluate B B First take C to the other side and change sign. Multiply both sides by B A R - C = B BR BC = A Use brackets B(R-C) = A Divide both sides (R-C) A B = (R - C) Evaluate 1 B = = 6 (4 - ) SELF ASSESSMENT EXERCISE No.3 1. The volume of a solid cylinder of height H and cross sectional area A is V = A H Make H the subject and evaluate it when V = 100 cm 3 and A = 40 cm. Ohm s Law states V = I R Make R the subject and find the resistance when I = Amperes and V = 4 Volts. C 3. The law of a lifting jack is E = + D Make η the subject of the formula. η Given E = 0 N, C = 5 (constant no units) and D = N, calculate η (Greek letter ETA) E 4. The electric current I flowing from a battery is given by I = where E is the EMF, r the R + r internal resistance and R the external resistance. Measurements gave I = 4 Amps, E = 1.08 V and R = 3 Ohms. Determine the internal resistance. 5. Make y the subject in the following equations. i. yz = x ii. x = yz iii. x = 4z/y iv. z = 4x y v. xy = 4 + x vi. x/y = 3z vii. 1 3z + y 4x = 0 D.J.Dunn 6

7 1.6. DISTRIBUTIVE LAW If we have an expression such as: = 10 The equality is maintained if we multiply (or divide) everything by the same number. Suppose we multiply every term by 5 3x5 + 9x5 x5 = 10x = 50 The sum is still correct. We could have written this as (3 + 9 )x5 = 10x5 and get 10x5 = 10x5 This is also correct. If instead of the number 5 we could have used a symbol say y and the same would be true. We would have: (3 + 9 ) y = 10y or multiplying out 3y + 9y y = 10y This is the distributive law. Doing the reverse process is finding a common factor in this case y is the common factor. WORKED EXAMPLE No. 6 What do you get by multiplying (a + b + c) by z (a + b + c) x z = az + bz + cz WORKED EXAMPLE No. 7 What do you get by dividing (a + b + c) by z (a + b + c) z = a/z + b/z + c/z WORKED EXAMPLE No. 8 Change the following expression into a bracket multiplied by a common factor. 1a 3b 6c The common factor must be a number that common to each term and trying the lowest number 3 we can write the expression as 3x4a 3b 3 x c Now 3 is a common factor and we can remove it and form brackets. 3(4a b c) D.J.Dunn 7

8 SELF ASSESSMENT EXERCISE No.4 1. Multiply out the following expressions 5n(a+3b-4c) 3x(a b + c). Remove the common factor and create brackets for the following expressions. x + y + z 10x 5y PRODUCT OF TWO BINOMIAL EXPRESSIONS Suppose you wanted to evaluate the expression ( 3 + ) x (5 1) Clearly the answer 5 x 4 = 0 We could get the answer by first multiplying (5 1) by 3 and then by and adding the two results as follows. ( 3 + ) x (5 1) = 3 x (5 1) + x (5 1) = 3 x 4 + x 4 = = 0 If we used symbols the same rule applies. WORKED EXAMPLE No. 9 Multiply out (a + b) x (c + d) (a + b) x (c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd SELF ASSESSMENT EXERCISE No.5 1. Multiply out the following expressions ( + x) (y + z) (p q) (a b) (g + n) (a b) D.J.Dunn 8

9 ANSWERS TO SELF ASSESSMENT EXERCISES SAE 1 = 7 = 1 = 4 = 8 ½ SAE x = 40 y = 30 z = 196 α = 4 β = 5 γ = 3 SAE 3 1. H=V/A = ½ cm. R = V/I = 1 Ohms 3. C 5 η = = (no units) E - D Ohms 5. i. y = x/z ii. y = x/z iii. y = 4z/x iv. y = (4x z)/ v. y = (4 + x)/x vi. y = x/3z vii. y = (4x +3z -1)/ SAE na + 15nb 0nc and 3xa - 3xb + 6xc. (x+y+z) and 5(x y + 4) SAE 5 y + z + xy + xz pa pb ga + gb ga gb + na nb D.J.Dunn 9