Download edexcel national certificate unit 4 – mathematics for

EDEXCEL NATIONAL CERTIFICATE
UNIT 4 – MATHEMATICS FOR TECHNICIANS
OUTCOME 1
TUTORIAL 1 – ALGEBRAIC LAWS
Determine the fundamental algebraic laws and apply algebraic manipulation techniques to the
solution of problems involving algebraic functions, formulae and graphs
CONTENTS
Algebraic laws and algebraic manipulation techniques
Algebraic laws and manipulation: introduction to algebraic expressions and equations, algebraic
operations, rules for manipulation and transposition of formulae; manipulation and algebraic
solution of algebraic expressions; direct and inverse proportion and constants of proportionality;
linear, simultaneous and quadratic equations; graphical solution of simple equations; use of
standard formulae to solve surface areas and volumes of regular solids; use of calculator for
algebraic problems
Indices, logarithms and functions: the index as a power, numerical and literal numbers in index
form, laws of indices and logarithms, common and Naperian logarithms, functional notation and
manipulation, logarithmic and exponential functions, other simple algebraic functions.
Algebra is essential for any student studying Engineering and Technology. Students following
the Edexcel National should have a GCSE pass grade in Maths. This probably means that
many will not have done any Algebra in their GCSE studies. For this reason the level in this
tutorial is set at very basic.
It is assumed that the student is familiar with arithmetical operations, the use of brackets and
fractions.
The answers to all the exercises are given at the end.
© D.J.Dunn www.freestudy.co.uk
1
1. BASIC ALGEBRAIC MANIPULATION
1.1 SYMBOLIC REPRESENTATION
You should already be familiar with mathematical sums such as:
12 + 5 = 17
and
18 ÷ 2 = 9
18
=9
2
In engineering it is common to use the solidus instead of the dividing line and write 18/2 = 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.
The division sign ÷ can be replaced by a dividing line so we can write 18 ÷ 2 = 9 as
For example in the sum
12 +
= 17 The apple has replaced the number 5
SELF ASSESSMENT EXERCISE No.1
1. What does the strawberry represent in this sum?
+ 15 = 22
2. What does the cherry represent in this sum? 16 -
=4
3. What does the banana represent in this sum?
4. What does the pineapple represent in this sum?
÷4=6
x 2 = 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.2
1.
2.
3.
4.
5.
6.
x + 140 = 180
y – 27 = 3
z ÷ 7 = 28
20/α = 5
7–β=2
9 x γ = 27
Note that when using symbols we often leave out the multiplication sign so in the last question we
could write 9 γ = 27 instead of 9 x γ = 27
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 2 = 14 becomes 7 . 2 = 14 but take care as this is easily confused with the decimal point.
© D.J.Dunn www.freestudy.co.uk
2
1.2.
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:
2 apples + 3 plums = 5 pears
It would make sense if the equation was
2 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
2[m] x 3[m] = 6 [m2]
This is because the product of metres (m) and metres is square metres [m2]
We can mix the units so long as the product is correct for example
2 [N] x 3 [m] = 6 [Nm]
The product of Newton [N] and metre [m] is Newton Metre [Nm]
Here is another example
6[m]÷ 2[s] = 3 [m/s] or 6[m]/2[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.
1.3.
SWAPPING SIDES
Consider the sum 12 = 6 x 2
It is just as correct to write 6 x 2 = 12
In the same way an equation such as x = y z
is just as correct written as
yz=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 www.freestudy.co.uk
3
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 bc
=
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
a
We can swap sides and write b = and b is now the subject. Similarly we could have divided both
c
a
sides by b to make c the subject. 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 20 m2 and B is 5 m, what is the depth?
SOLUTION
A = B x D divide both sides by B
A
A BD
=
= D swap sides and D =
B
B
B
20
Put in the known values D =
and so D = 4 and further the units must be metres
5
B
C
If we multiply both sides by C the equality is maintained and A C = B
Now consider the equation A =
Swap both sides and B = A C so B is now the subject of the equation.
WORKED EXAMPLE No.2
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 20 Volts is applied to 5 Ohm resistor.
SOLUTION
V
R=
I
Multiply both sides by I and R I = V
V
Evaluate and I = 20/5 = 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.
Divide both sides by R and:
© D.J.Dunn www.freestudy.co.uk
I=
4
1.5 SUBTRACTING OR ADDING TO BOTH SIDES
Consider the sum 20 = 15 + 5
Now add 2 to both sides. 20 + 2 = 15 + 5 + 2 or 22 = 15 + 5 + 2 = 22
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.
20 - 3 = 15 + 5 – 3
or 17 = 15 + 5 – 3 = 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
20 = 15 + 5
Subtract 5 from both sides.
20 – 5 = 15 + 5 - 5
20 – 5 = 15
Here is another example.
16 – 4 = 12
Add 4 to both sides.
16 = 12 + 4
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.
SOLUTION
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 = 2D + 2B.
Make B the subject.
Given that P is 200 mm and B is 40 mm, what is D?
SOLUTION
First divide both sides by 2
P 2D 2B
=
+
=D+B
2
2
2
Now subtract B from both sides to get rid of B on the right
P
−B= D
2
Now swap both sides over D =
Put in numbers and evaluate
© D.J.Dunn www.freestudy.co.uk
D=
P
−B = D + B-B
2
P
−B
2
200
− 40 = 100 − 40 = 60 and the units must be mm.
2
5
WORKED EXAMPLE No.5
Given the equation R =
A
+ C make B the subject. Given A = 12, C = 2 and R = 4 evaluate B
B
SOLUTION
First take C to the other side and change sign.
Multiply both sides by B
Use brackets
A
B
BR – BC = A
R -C =
B(R-C) = A
A
B=
(R - C)
12
B=
=6
(4 - 2)
Divide both sides (R-C)
Evaluate
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 cm3 and A = 40 cm2
2. Ohm’s Law states V = I R
Make R the subject and find the resistance when I = 2 Amperes and V = 24 Volts.
C
+ D Make η the subject of the formula.
η
Given E = 20 N, C = 5 (constant no units) and D = 2 N, calculate η (Greek letter ETA)
3. The law of a lifting jack is E =
E
where E is the EMF, r the
R+r
internal resistance and R the external resistance. Measurements gave I = 4 Amps, E = 12.08 V
and R = 3 Ohms. Determine the internal resistance.
4. The electric current I flowing from a battery is given by I =
5. Make y the subject in the following equations.
i.
ii.
iii.
iv.
v.
vi.
vii.
yz = x
x = 2yz
x = 4z/y
z = 4x – 2y
2xy = 4 + x
2x/y = 3z
12 – 3z + 2y – 4x = 0
© D.J.Dunn www.freestudy.co.uk
6
1.6.
DISTRIBUTIVE LAW
If we have an expression such as:
3 + 9 – 2 = 10
The equality is maintained if we multiply (or divide) everything by the same number. Suppose we
multiply every term by 5
3x5 + 9x5 – 2x5 = 10x5
15 + 45 – 10 = 50
The sum is still correct.
We could have written this as
correct.
(3 + 9 – 2)x5 = 10x5
and get
10x5 = 10x5 This is also
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 – 2) y = 10y
or multiplying out
3y + 9y – 2y = 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
SOLUTION
(a + b + c) x z = az + bz + cz
WORKED EXAMPLE No. 7
What do you get by dividing (a + b + c) by z
SOLUTION
(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.
12a – 3b – 6c
SOLUTION
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 2c
Now 3 is a common factor and we can remove it and form brackets.
3(4a – b –2c)
© D.J.Dunn www.freestudy.co.uk
7
SELF ASSESSMENT EXERCISE No.4
1. Multiply out the following expressions
5n(2a+3b-4c)
3x(a – b + 2c)
2. Remove the common factor and create brackets for the following expressions.
2x + 2y + 2z
10x – 5y + 20
1.7.
PRODUCT OF TWO BINOMIAL EXPRESSIONS
Suppose you wanted to evaluate the expression ( 3 + 2) x (5 – 1)
Clearly the answer 5 x 4 = 20
We could get the answer by first multiplying (5 – 1) by 3 and then by 2 and adding the two results
as follows.
( 3 + 2) x (5 – 1) = 3 x (5 – 1) + 2 x (5 – 1) = 3 x 4 + 2 x 4 = 12 + 8 = 20
If we used symbols the same rule applies.
WORKED EXAMPLE No. 9
Multiply out (a + b) x (c + d)
SOLUTION
(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
(2 + x) (y + z)
(p – q) (a – b)
© D.J.Dunn www.freestudy.co.uk
(2g + n) (a – b)
8
ANSWERS TO SELF ASSESSMENT EXERCISES
SAE 1
=7
SAE 2
x = 40y = 30
= 12
z = 196
α=4
= 24
β=5
SAE 3
1.
H=V/A = 2 ½ cm
2.
R = V/I = 12 Ohms
C
5
3.
η=
= (no units)
E - D 18
4.
0.02 Ohms
5.
i. y = x/z
ii. y = x/2z
v. y = (4 + x)/2x
vi. y = 2x/3z
γ=3
iii. y = 4z/x iv. y = (4x – z)/2
vii. y = (4x +3z -12)/2
SAE 4
1.
10na + 15nb – 20nc and 3xa - 3xb + 6xc
2.
2(x+y+z)
and
5(2x – y + 4)
SAE 5
2y + 2z + xy + xz pa – pb –ga + gb
© D.J.Dunn www.freestudy.co.uk
2ga – 2gb + na – nb
9
=8½