Download Essential Mathematics

Essential Mathematics
by
Janine Upton
Integers and fractions
• Integer = Whole number
– >0
– <0
Positive integer
Negative integer
• Digit = one of the integers
–2
– 55
– 765
One digit
Two digits
Three digits
Mathematical Operators
Mathematical Operation
Multiplication
Division
Addition
Subtraction
Symbol
Equations - Multiplication
• Mathematical statement
e.g. 5 x 7 = 35
• Multiplication - Symmetrical (can be reversed
and still give same answer)
• Signs:
Division
Dividend/Numerator
Divisor/Denominator
3 (Quotient)
• No symmetry – Can’t swap numbers to get the same answer
- 𝑥
𝑦
- 𝑥
- 𝑦
= Negative answer
= Positive answer
Addition and Subtraction
• Addition:
– Has symmetry i.e. 2 + 3 is the same as 3 + 2
• Subtraction:
– No symmetry i.e. 2 – 3 is not the same as 3 - 2
Fractions
𝑎
𝑏
• Denominator can take any value, except 0
• Proper fraction:
2
5
Numerator < Denominator
• Improper fraction:
12
5
Numerator > Denominator
Fractions: Adding them
• Same denominators:
1 3
+
8 8
4
8
1
3
5
6
=
• Different denominators:
2
9
+ +
• Look for the Lowest Common Denominator (LCD)
– the smallest number into which the denominators will all divide
– The smallest number itself is called the lowest common multiple
(LCM)
• LCM = 18
6
18
+
4
18
+
15
18
=
25
18
Fractions: Subtracting them
• Same rules as addition
• Same denominators:
11
15
+
7
15
-
14
15
=
4
15
• Different denominators:
5
16
15
48
-
1
1
+
8 12
6
4
+
48 48
=
13
48
Fractions: Multiplying them
• Multiply the numerators to get the new
numerator
• Multiply the denominators to get the new
denominator
2
3
5
6
x =
10
18
Fractions: Dividing them
• Invert the second fraction
• Multiply it by the first
• Therefore changing it into a multiplication
statement
1
5
÷
2 12
1 12
x
2
5
=
12
10
=
6
5
Decimals
• Fraction where the numerator is divided by
the denominator to yield an equivalent
decimal expression
• A decimal consists of three components:
• an integer, followed by
• a decimal point, followed by
• another integer
Remember!
• If a number is expressed in decimal form, any
zeros on the right-hand end after the decimal
point and after the last digit do not change
the number’s value
• E.g. 0.5, 0.50, 0.500, 0.5000
• 0.60 can and often is written as .6
Decimals: Addition and subtraction
• Similar to those of integers
• Decimal points aligned under each other
2.3 + 0.34 + 1.672
2.300
Add 0’s to help align
0.340
1.672
4.312
Decimals: Multiplication and division
Multiplication:
When multiplying two decimals, follow these
steps:
1. Count the number of digits to the right of the decimal
point for each number.
2. Add the number of digits found in Step 1 to obtain a
number, say x.
3. Multiply the two original decimals, ignoring the
decimal points.
4. In the answer in Step 3, mark the decimal point so
that there are x digits to the right of the decimal point.
This is the answer to the original multiplication.
Decimals: Multiplication
3.12 x 2.7
1. 3.12 and 2.7 have 2 and 1 digits after their
decimal points respectively = 3
2. Perform the multiplication ignoring the
decimal points:
312
x 27
8424
3. 3 digits to the right of the point = 8.424
Decimals: Division
When dividing decimals, follow these steps:
1. Count the number of digits that are in the divisor to
the right of the decimal point.
Call this number x.
2. Move the decimal point in the dividend x places to the
right (adding zeros if
necessary). Do the same to the divisor.
3. Divide the transformed dividend (Step 2) by the
transformed divisor (which now
has no decimal point). The quotient of this division is the
answer.
Decimals: Division
• 3.248 ÷ 0.04
– Divisor has 2 decimal places after point
• Move the decimal point 2 places to the right in the
dividend (3.248) to obtain 324.8
• Do the same to 0.04 to obtain 4
Exponents
• An exponent (or power) is a number that is
written as a superscript to another number called
the base
• It tells us how many times the base is to be
multiplied by itself
• A number that is expressed as a base with an
exponent is said to be in exponential form.
• Written as: 𝑎𝑛
• a - base
• n - exponent or power
Positive exponents
• The product of two numbers will have the same
base with an exponent equal to the sum of the
two original exponents. That is:
𝑎𝑚 x 𝑎𝑛 = 𝑎𝑚+𝑛
23 x 24 = 23+4 = 27
• The quotient of those two numbers will have the
same base with an exponent equal to the
difference of the exponents. That is:
𝑎𝑚 ÷ 𝑎𝑛 = 𝑎𝑚−𝑛
36 ÷ 34 = 36−4 = 32
Positive exponents
• If a number already expressed in exponential
form is itself raised to another exponent, the
result is a number with the original base and
an exponent equal to the product of the two
exponents. That is:
(𝑎𝑚 )𝑛 = 𝑎𝑚𝑛
(53 )2 = 56
Negative exponents
• A number expressed in exponential with a
negative exponent is called the reciprocal (or
inverse) of the same number with the
negative sign removed from the exponent.
That is:
1
−𝑛
𝑎 = 𝑎𝑛
−3
2
=
1
23
=
1
8
Fractional exponents
• The exponent (𝑎𝑛 ) extended where n is a
fraction
𝑎
1/𝑘
= kth root of a
• The kth root of a number is one such that
when it is multiplied by itself k times you get
that number
𝑎
1/𝑘
𝑎
=
𝑚/𝑛
𝑘
𝑎
𝑛
= ( 𝑎)
𝑚
Fractional exponents
1/2
9
2
= 9
1/2 4
(5
3
) = 54/2 = 52 = 25
1/3
27 = 27
3 1/3
= (3 )
= 31 = 3
Zero exponent
𝑎0
= 𝑎𝑛 −𝑛
=
𝑎𝑛
𝑎𝑛
=1
Percentages
• Converting to a percentage:
– Multiply by 100
1
2
= 0.5
0.5 x 100 = 50%
0.872
0.872 x 100
87.2%
Percentages 2
• To convert to fraction from a percentage:
37.5% =
72% =
37.5
100
72
100
=
=
18
25
3
8
(divide by 12.5)
• To convert to percentage from a fraction:
80% = 0.80
45.78% = 0.4578
Algebra
• Constant
– Values that does not change
– Assign a letter of the alphabet to it
– Most times a, b, c, d
• Variable
– Can have different values
– Mostly letters at the end of the alphabet x, y, z
• Algebraic expression
– combination of constants and variables by the use of
arithmetic operations (such as addition, subtraction,
multiplication and division)
Algebra
• Term
– Part of an expression that is connected to another term by
a + or a – sign
– 4x + 6y – 9z
Terms
Coefficient
• Degree of an expression is the highest exponent (or
power) of any variable contained in the expression
– 3x + 7 – linear/first degree
– 9𝑥 2 - 4x + 1 – quadratic/second degree
– 10𝑦 3 + 4 𝑦 2 - y + 8 – cubic/third degree
Algebra
• In the expression 15x – 5y + 32 – 4x + 6xy +
𝑥 2 + 3y,
– Like terms:
• 15x and –4x
• –5y and +3y
• Equation
– statement that two expressions are equal
– E.g. 3x + 6 = 5x - 2
Algebra
• Solving the equation = find the value of the
variable
3x + 6 = 5x – 2
6 + 2 = 5x – 3x
8 = 2x
4=x
Solution: 4
X = 4, satisfies the equation
Algebraic expressions
• Rules
• Rule 1:
– If an expression contains like terms, these terms may
be combined into a single term. Like terms differ only
in their numerical coefficient. Constants may also be
combined into a single constant
5x – 2y + 4z + 7 – 3x + 7y – 6z + 4
5x and –3x are like terms
-2y and 7y are like terms
4z and –6z are like terms
7 and 4 are constants
Algebraic expressions
• Like terms can be combined to form a single
term:
5x and –3x may be combined to give a single term: 5x –
3x = 2x
-2y and 7y may be combined to give a single term: -2y +
7y = 5y
4z and –6z may be combined to give a single term: 4z –
6z = -2z
7 and 4 may be combined to form a single constant: 7 +
4 = 11
Hence, the expression may be simplified to: 2x + 5y – 2z
= 11
Algebraic expressions
• Rule 2:
• When an expression is contained in
parentheses (brackets), each term within the
parentheses is multiplied by any coefficient
written outside the parentheses.
2(3x + 4y – 1)
2(3x) + 2(4y) +2(-1)
6x + 8y – 2
Algebraic expressions
• Rule 3:
• To multiply one expression by another, multiply
each term of one expression by each term of the
other expression. The resulting expression (which
should be simplified by collecting any like terms)
is said to be the product of the two expressions.
5x(2x – y + 4)
5x(2x) +5x(-y) + 5x(4)
10x2 – 5xy + 20x
Example
Find the product of the two expressions 3x + 2
and 2x – 1
(3x + 2)(2x – 1)
3x(2x – 1) + 2(2x – 1)
3x(2x) + 3x(-1) + 2(2x) + 2(-1)
6𝑥 2 – 3x + 4x – 2
6𝑥 2 + 1x – 2
6𝑥 2 + x – 2
Solving linear equations
1. Place like terms on the same side of the
equation
2. Any arithmetic operation that you perform on
one side of the equation you must also perform
on the other side
When terms are moved from one side of an
equation, they are said to be transposed (or
transferred)
This process is called transposition.
Solving an equation that involves only one
variable follow these steps:
1. Place all the terms involving the variable on the
left-hand side of the equation and
the constant terms on the right-hand side.
2. Collect the like terms involving the variable, and
collect the constant terms.
3. Divide both sides of the equation by the
coefficient of the variable.
The left-hand side of the equation should now
consist of the variable only.
The right-hand side of the equation is the solution.
Example
Solve the following equation for x:
9x – 27 = 4x + 3
9x – 27 – 4x = 4x + 3 – 4x
5x – 27 = 3
5x – 27 +27 = 3 + 27
5x = 30
x=6
Check by substituting x = 6 into the original
equation:
9(6) – 27 = 4(6) + 3
54 – 27 = 24 + 3
27 = 27
Example 2
• Solve the following equation for m:
12 – 4m = 6m + 37
12 – 4m – 6m = 6m + 37 – 6m
12 – 10m = 37
12 – 10m + 12 = 37 - 12
-10m = 25
m = -2.5
Example 3
5(3x – 4) = 2(6 – 2x)
15x – 20 = 12 – 4x
15x + 4x = 12 +20
19x = 32
x=
32
19
= 1.685m
Example 4
• Solve the following equation for z:
3𝑧
3(2z + 5) = + 36
4
3𝑧
6z + 15 = + 36
4
24z + 60 = 3z + 144
24z – 3z = 144 – 60
21z = 84
z=4
Solving simultaneous linear equations
• Simultaneous equations
– Equation that must be solved at the same time to
find the values of the variables that will solve both
equations
An example of simultaneous equations in the
variables x and y is:
4x + 3y = 11
5x – 2y = 8
Solving simultaneous linear equations
The steps involved in the solution of these equations are as
follows:
1. Make the coefficient of either of the variables in one equation equal to its
coefficient in the other equation. It does not matter which variable is used for this
purpose. This must be achieved by multiplying one equation (or possibly both
equations) by a constant.
2. Eliminate the variable that now has the same coefficients in both equations.
How you do this depends on the signs (+,-) before the coefficients:
(a) If the signs are the same, subtract one equation from the other. (It does not matter which
equation is subtracted).
(b) If the signs are different, add the equations. Write the equations under one another for
this step, making sure that like terms are directly under each other.
3. After the subtraction or addition is made, you should be left with an equation
in only one variable, namely the variable that was not eliminated in Step 2. Solve
this equation using the technique discussed in the previous section.
4. Find the value of the eliminated variable by substituting the value of the
variable found in Step 3 into either of the two equations.
5. Verify your solution by substituting the values of the two variables into each of
the two original equations and checking that they are true statements.
Example
• Solve the following simultaneous equations
for x and y :
3x + 4y = 33
(1)
2x – 3y = 5
(2)
First eliminate the variable x from both
equations (Step 1)
6x + 8y = 66
(3)
6x – 9y = 15
(4)
Example cont…
• Now because the signs are the same, subtract
one equation from the other (Step 2).
• Suppose that we subtract equation (4) from
equation (3):
6x + 8y = 66
6x - 9y = 15_
17y = 51
y=
51
=
17
3
•
Example cont… 2
To obtain the solution for x (Step 4), we substitute the value y = 3
3x + 4y = 33
2x – 3y = 5
(1)
(2)
3x +4(3) = 33
2x – 3(3) = 5
3x = 21
2x = 14
x=7
Therefore, the solution to the simultaneous equations is x =
7 and y = 3.
Step 5 involves verification of the solution.
Substituting these values into Equations (1) and (2)
Practical example
• A customer left his car for service and received a
combined bill (for parts and labour) of $ 228. If the
labour cost twice as much as the parts, find the
amount that parts and labour each cost.
• Parts (in dollars) = x
• Labour = 2(x)
• Parts + labour = $228
Therefore:
x + 2x = 228
3x = 228
x = 76
Parts = $76
Labour = $152
Practical example 2
• A family consisting of 2 adults and a number of
children went to the cinema. Each adult paid $ 9
and each child paid $ 3.50 for the tickets. If the
total bill was $ 39, how many children went?
• Adult = $9 – there were two ($9 x 2 = $18)
• Child (x) = $3.50
18 + 3.5x = $39
3.5x = 21
x=6
There were six children
Inequalities
• An inequality is expressed using the following
symbols:
> “greater than”
< “less than”
≥ “greater than or equal to”
≤ “less than or equal to”
Solving linear inequalities
x+2>5
x>3
2x – 5 < x + 3
2x – x < 3 + 5
x<8
Note: Inequality sign did not change
Solving linear inequalities
• If one of the inequalities contains a negative:
-x>2
Need to multiply by “-”
x < -2
Example 2:
x + 4 > 2x – 8
x – 2x > -8 -4
-x > -12
x < 12
Functions
• f(x)
• x = argument of the function
Linear function: f(x)= 3x + 2
f(5)
3(5) + 2 = 17
Quadratic function: f(x)= 2𝑥 2 + x – 6
f(5)=
2(5)2 + 5 – 6 = 49
Roots of equations
• Finding the root(s) of equations involves
finding value(s) that make the function zero.
– Linear function:
a + bx = 0
bx = -a
x=-
𝑎
𝑏
Quadratic equation:
Completing the square
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
𝑏
𝑎(𝑥 2 + 𝑎) + 𝑐 = 0
𝑏 2
1 𝑏
𝑏2
( 2(𝑎) = ( 2𝑎 ) = 4𝑎2 )
𝑏
𝑎
𝑏2
)+
4𝑎2
𝑐=0
𝑏
𝑎
𝑏2
)+
4𝑎2
𝑐=
𝑎(𝑥 2 + 𝑥 +
If we multiply the a out:
𝑎(𝑥 2 + 𝑥 +
𝑏 2 𝑏2
) =
2𝑎
4𝑎
𝑎(𝑥 +
Divide by a
(𝑥 +
𝑏 2 𝑏2
) = 2
2𝑎
4𝑎
-
-c
𝑐
𝑎
Get common denominators
𝑏 2
𝑏2
𝑐
4𝑎
(𝑥 + 2𝑎 ) = 4𝑎2 - 𝑎 x 4𝑎
𝑏 2
To get rid of the square:
(𝑥 + 2𝑎 ) =
(𝑥 +
𝑏2 −4𝑎𝑐
4𝑎2
𝑏 2
) =
2𝑎
𝑏2 −4𝑎𝑐
4𝑎2
x + 2a = ±
𝑏2 −4𝑎𝑐
2𝑎
b
±
2a
𝑏2 −4𝑎𝑐
2𝑎
b
x =−
𝑥=
−𝑏± 𝑏2 −4𝑎𝑐
2𝑎
𝑏2
4𝑎
Examples
f(x) = 2𝑥 2 + 𝑥 − 6 = 0
−1 ± 12 − 4(2)(−6)
2(2)
−1± 49
4
=
−1±7
4
= 1.5 and -2
Practical example: Linear cost
function
• Cost is defined in terms of two components:
Total variable cost
Total fixed cost
𝑐 𝑥 = 𝐹 + 𝑉𝑥
𝑐 = Total cost
𝑉 = Variable cost per unit
𝐹 = fixed cost per period
Practical example
• A company that produces a single product wants
to determine the function that expresses total
cost as a function of the number of units
produced. The fixed expenditure each year is R 50
000. The estimated raw material and labour costs
for each unit produced are R 5.50. What will be
the total cost to produce 120 items?
𝑐 𝑥 = 𝐹 + 𝑉𝑥
𝑐 120 = 50000 + 5.50(120)
𝑐 120 = 𝑅50660
Practical example: Linear revenue
function
• Revenue = money that flows into a business from
either selling products or providing services
𝑅 𝑥 =𝑝 ×𝑥
𝑝 = price
𝑥 = quantity
• Profit = difference between total revenue and
total cost
P 𝑥 = 𝑅 𝑥 − 𝐶(𝑥)
When Total Revenue > Total Costs = Net gain
When Total Revenue < Total Costs = Net loss/deficit
Practical example
•
The price of a single product is R 65. Variable costs per unit are R 20 for materials
and R 27.50 for labour. Annual fixed costs are R 100 000. Construct the profit
function and determine the profit if annual sales are 20 000 units.
Price = R65
Variable = R20 + R27.50
Fixed = R100 000
x = 20 000
P 𝑥 =𝑅 𝑥 −𝐶 𝑥
𝑐 𝑥 = 𝐹 + 𝑉𝑥 = 100 000 + R47.50x
𝑅 𝑥 = 𝑝 × 𝑥 = 65x
P 𝑥 =𝑅 𝑥 −𝐶 𝑥
65x – (100 000+47.5x)
(65x – 47.5x) - 100000
17.5x – 100 000
P 20 000 = 17.5(20 000) – 100 000
P(20 000) = R250 000
Break-even analysis
• Break-even analysis is used to determine the number of units
that must be sold (either in rands or units of output) for the
business to break even; that is, neither to earn profits nor
incur losses
Example:
A company makes plastic containers and sells them at R 45 each. The costs of
production consist of a fixed cost of R 28 050 and a variable cost of R 12 per
container. What is the breakeven production level for this company?
𝑐 𝑥 = 𝐹 + 𝑉𝑥 = R28050 + R12x
𝑅 𝑥 = 𝑝 × 𝑥 = 45x
P 𝑥 = 𝑅 𝑥 − 𝐶 𝑥 = 45x – (R28 050 +R12x)
P 𝑥 = 33x – R28 050
For break-even P 𝑥 = 0
Therefore, P 𝑥
33x – R28 050 = 0
33x = R28 050
x = 850 containers to break-even
Quadratic Profit Functions
• Sometimes the profit made by a business is
described by a quadratic function (nonlinear)
• This occurs when either (both) the cost or
(and) revenue functions are quadratic
Example
• The profit function for an organisation is given by
𝑝 𝑥 = 𝑥 2 + 200𝑥 − 30 000
𝑥 = number of units sold
1. Determine the profit if 600 units are sold.
𝑝 600 = 6002 + 200(600) − 30 000
= R450 000
2. What is the break-even level for this
organisation?
−200± 2002 −4(1)(−300000)
2(1)
=
−200± 160000
2
= -300 or
100
Cant be negative – therefore the solution = 100 units
The value of money
Simple interest
• Interest earned in equal amounts every year
(or month) and which is given in proportion of
the original investment (capital)
𝑺 = 𝑿 + 𝒏𝒓𝑿
𝑿 = Capital
𝒏 = Number of periods
𝒓 = interest rate
Simple Interest example
• How much will an investor have after four
years if he invests R 5000.00 at 10% simple
interest per annum?
𝑺 = 𝑿 + 𝒏𝒓𝑿
𝑺 = 𝟓𝟎𝟎𝟎 + 𝟒(𝟎. 𝟏)5000
𝑺 = 𝑹𝟕𝟎𝟎𝟎
Compound Interest
• This type of interest is calculated by means of
compounding
𝑆 = 𝑋(1 + 𝑟)𝑛
Example:
• How much will an investor have after four years if he invests R
5000.00 at 10% compound interest per annum?
𝑆 = 𝑋(1 + 𝑟)𝑛
𝑆 = 5000(1 + 0.1)4
𝑆 = 𝑅7320.50
Present Value
• The amount of money which must be invested now for
years at an interest rate of %, to earn a given future
sum of money at the time the money is due
1
𝑋=𝑆 ×
(1 + 𝑟)𝑛
• You consider buying a business for R 900 000. The
business is expected to run for 4 years and the following
net returns are expected.
•
•
•
•
YEAR 1 : R 300 000
YEAR 2 : R 250 000
YEAR 3 : R 400 000
YEAR 4 : R 300 000
• Should you buy the business? A project of this type is
expected to return at least 15% per annum.
Example continued:
Index Numbers
• An index number is a relative figure, expressed as a percentage, which is
used to measure how much an economic variable changes over time or
differs between two locations
• Price index ( is the one most frequently used. It compares changes in price from one
period to another.
• Quantity or volume index (𝐼𝑞 ) measures how much the quantity of a variable changes
over time.
• Value index measures changes in total monetary worth by combining price and quantity.
• Base Period
– Point in time to which the comparison is made,
– Choice of a base period will determine the usefulness of any data. In choosing
a base period, the following should be considered:
• the period should be recent enough for comparisons with the base to be meaningful;
• it should be a typical period with respect to the activity of interest, and should be the
same period used by other series with which you are likely to compare your data;
• the period should be one of relative economic stability, without any abnormal influences
• Base year is always 100 –
– Each subsequent year will be above or below 100, depending on whether
there has been an increase or decrease in the data compared with the base
year
Steps:
Index numbers example
1. Select the period to be used as base
2. Divide the current value of the commodity by the
base value
3. Multiply this ratio by 100
𝐼𝑞 =
𝑃𝑖
𝑃𝑏
x 100
𝐼𝑞 = Price Index
𝑃𝑖 = Current period price
𝑃𝑏 = base period price
Example
• If milk cost R3.50 a litre in 1999 and R3.85 in
2000, the simple price index for 2000 would
be:
𝑃𝑖
𝐼𝑞 = x 100
𝑃𝑏
3.85
𝐼𝑞 =
x 100
3.5
𝐼𝑞 = 110
This means that milk increased in price by 10%
over the period under consideration.
Aggregate/Composite Index Numbers
• These indexes are used to measure the
relative change of a whole group (or basket) of
related items
• If each item in the overall index is of equal
importance, the index is said to be
unweighted
• If each item in the overall index is not of equal
importance, the index is said to be a weighted
index
Unweighted
• 𝐼𝑝 =
Σ𝑃𝑖
Σ𝑃𝑏
x 100
Σ𝑃𝑖 = Sum of the prices of all the
items in the given period
Σ𝑃𝑏 = Sum of the prices of all the items in
the base period
Example
• The following table shows the materials needed
for a course in statistics, and the price (in Rands)
per unit in 1999 and 2000
=
Σ𝑃𝑖
Σ𝑃𝑏
x 100 =
209
193
x 100 = 108.29
This means that the prices increased by 8.29% over
the period under consideration
Weighted Composite
• These methods assume that the quantity of a
commodity in a given period is a valid measure of
its comparative importance
Laspeyres’ Index
• This quantity uses the quantities consumed
during the base period as a weight factor, and
assumes that whatever the changes in price, the
quantities purchased will remain the same
• It does not take into consideration changes in
consumption patterns. If consumption patterns
change, the base period should also be changed.
Laspeyres’ Index
𝐿𝑝 =
Σ(𝑝𝑖 × 𝑞𝑏)
Σ(𝑝𝑏 × 𝑞𝑏 )
x 100
Example:
• Consider the price of three items for years 2008 and 2011.
Calculate the Laspeyres’ price index and Laspeyres’ quantity
index for 2011 using 2008 as the base year
Laspeyres’ Index cont…
Paashe’s Index
• This method uses the quantities consumed
during the current period as weights
• It measures the change in the total cost of
goods that represent a consumption pattern
typical of the current year
• It avoids the problem of changing
consumption patterns. This index tends to
underestimate the change in total cost
Paashe’s Index
𝑃𝑝 =
Σ(𝑝𝑖 ×𝑞𝑖 )
Σ(𝑝𝑏 ×𝑞𝑖 )
x 100
𝑃𝑝 =
Σ(𝑝𝑖 ×𝑞𝑖 )
Σ(𝑝𝑏 ×𝑞𝑖 )
x 100
82700
57600
x 100 = 143.6