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Topic 2 Revision
2.1 The sets of numbers:
You are expected to know the symbols and the definitions of each type of number that
come up in the course. They are:
N – natural numbers {0, 1, 2, 3…}
Z – integers {…-3, -2, -1, 0, 1, 2, 3…}
p
Q – rational numbers {
p, q  Z } i.e. rational numbers are numbers that can be
q
written as a fraction.
Notice here that it is all numbers that can be written as a fraction. They may be given to
you in decimal form. The following decimals are all rational as they can be written in
fraction form:
0.321
321
1000
-1.4
 14
10
0.333333…
1
3
-5
5
1
Copy and complete the following passage using the missing words below.
decimals fraction rational fractions terminate irrational recur fraction decimals recur
rational recur integers subset terminate
Rational numbers are numbers that can be written as ________________. They are often
also given as _______. Decimals which ______ or _____ can all be written as
___________ and so they are _____________. All _________ can be written as
_________ so are _____________. We say that the set of ______ is a _________ of
rational numbers. Some __________ (e.g. 2 , 3 7 , sin 28, …} neither _______ or
__________ and are therefore _________________.
R – real numbers – these are any numbers that can be written on a number line. This
includes every type of number you have come across in your maths so far.
Notice here that N  Z  Q  R
This means that if a number is a natural number then it must be an integer. If it is an
integer… (copy and complete).
1
The other types of number you are expected to be aware of include:
Multiples: This means ‘in the times table of’. E.g. multiplies of 5 are {5, 10, 15…}
Lowest common multiple: The smallest number which is a multiple of two numbers
e.g. To find the LCM of 6 and 8:
Multiples of 6: {6, 12, 18, 24, 30, 36, 42, 48…}
Multiples of 8: {8, 16, 24, 32, 40, 48, 56, 64…}
The LCM is therefore 24 as it is the smallest number that appears in both lists.
Factors: Numbers which divide exactly into a natural number
e.g. to find factors of 24:
1 x 24 = 24
2 x 12 = 24
3 x 8 = 24
4 x 6 =24
Notice that you do not need to continue the list as 5 does not divide into 24 and the next
pair would give you 6 x 4 (which we already have).
Factors of 24 are therefore {1, 2, 3, 4, 6, 8, 12, 24}
Highest common factor: The largest number which is a factor of 2 numbers
e.g. to find the HCF of 12 and 18
Factors of 12 {1, 2, 3, 4, 6, 12}
Factors of 18 {1, 2, 3, 6, 9, 18}
The HCF is therefore 6 as it is the largest number which appears in both lists.
Prime numbers: Prime numbers have 2 factors – 1 and themselves
e.g. to see if 8 is prime:
Factors of 8 {1,2 , 4, 8} – therefore 8 is not prime as it has 4 factors
e.g. to see if 11 is prime:
Factors of 11 {1, 11} – therefore 11 is prime as it has 2 factors
e.g. to see if 1 is prime:
Factors of 1 {1} – therefore 1 is not prime as it only has 1 factor
Prime factors: Every natural number can be written as a
product of prime numbers
e.g. to write 24 as a product of prime factors.
Divide the number by the smallest prime i.e. {2, 3, 5, 7..}
You are left with:
2 x 2 x 2 x 3 = 24
Or 24  23  3
2
Questions on 2.1:
1.
Tick the correct boxes:
N
4.2
-3
1.3
2.571571571…
13
19
4

7
8
2
Z
Q
R
4
3
16
2 8
2.
Draw a Venn diagram showing the sets N, Z, Q and R and place the numbers in
the correct position.
3.
How many times larger is 6300 x 120 than 6.3 x 1.2
4.
My bank account contained €12400. I withdraw €260 a month for 18 months.
What is the new balance? How many months do I need to carry on withdrawing for until
I am left with an overdraft?
5.
List the first 5 powers of 4.
6.
Write 2 3  32  5 as a natural number
7.
Write 660 as the product of prime factors.
8.
List all the factors of 120.
9.
List the multiples of 7 between 80 and 120.
10.
Find the largest number which divides into 85 and 119.
11.
Evaluate 24  12  2 2
12.
Find the Lowest common multiple of 15 and 18.
13.
A bag of sweets has to be able to be shared equally between either 5, 6 or 9 kids.
What is the smallest number of sweets there could be in the bag?
3
2.2 Approximation:
Numbers can be rounded to any degree of accuracy. A number line often helps.
e.g. 2047 people attended a concert.
Round this to the nearest 20.
Notice that only multiples of 20 are
on the line.
2047 is closest to 2040 so we can say
that 2047 rounded to the nearest 20 is 2040.
2047 rounded to 2040 because it was less than 2050 (the half-way point). By convention,
if a number is exactly half way between the two rounded possibilities we would round up.
So 2090 rounded to the nearest 20 would be 2100.
Rounding to decimal places:
The first number after the decimal point is the 1st decimal
place
The second number after the decimal point is the 2nd
decimal place
e.g. the number given to the right is ∏ to 9 decimal places.
i)
Round ∏ to 1 decimal place
Firstly we underline the first decimal place. We then circle the next digit. If the circled
digit is 5 or larger then the underlined digit increases by 1 otherwise the underlined digit
remains the same.
4 is less than 5 therefore ∏=3.1 (to 1 d.p.)
ii)
Round ∏ to 3 d.p.
The circled 5 causes the underlined digit to increase by 1 therefore ∏=3.142 to 3 d.p.
If you are really lazy you can do these on a calculator:
1. Press mode
2. On the 2nd row change ‘float’ to 3
3. Type in 3.141592654 and press enter
4. Make sure you change your mode back to ‘float’
Note that when the last digit of a rounded number is 0 you must leave the 0 in.
4
e.g. 31.402 to 2 d.p. is 31.40 (not 31.4)
Rounding to significant figures:
When counting significant figures your count should begin at the first non-zero
number.
We can now round in exactly the same way as before.
e.g. round 0.0035478 to 3 s.f.
We underline the third s.f. and circle the next digit
The circled digit is 5 or larger so the underlined
increases by 1.
So 0.0035478=0.00355 (to 3 s.f.)
Accuracy of measurement:
Judy’s height is measured as 175 cm to the nearest cm. What is the possible range in
Judy’s height?
The number line above shows that if Judy’s height were less than 174.5 cm it would
have been rounded down to 174 cm.
If her height were more than 175.5 cm it would have been rounded up to 176 cm.
Remember our convention: If it had been exactly 174.5 cm then it would have been
rounded up to 175. If exactly 175.5 cm it would have been rounded to 176 cm.
Judy’s height is greater than or equal to 174.5 cm but it must be less than 175.5 cm
e.g. Find the bounds of the area and perimeter of a square whose side length x is
given as 2.4 cm to 1 d.p.
The boundaries for side length are 2.35  x  2.45
5
P  4x
 9.4  P  9.8cm
A  x2
 5.5225  A  6.0025
Percentage error:
Your formula book gives the following formula for percentage error.
e.g. John complains to his teacher saying that it will take him 1 hour 30 minutes to
complete his homework. Actually it takes him 55 minutes. What is the percentage
error in his calculation?
v A  90,
 
v E  55
90  55
100  63.6 % (3 s.f.)
55
e.g. Find the percentage error when ∏ is given to 3 decimal places
vA   ,
v E  3.142
 
  3.142
100  0.0130 % (3 s.f.)

Estimation:
Before performing a calculator operation you should have an idea of the order of
magnitude of your answer (how large you expect it to be).
You can do this by rounding all numbers to 1 s.f.
e.g. Johnty claims that  2  0.987 (to 3s.f.) why must he be wrong?
 3
 2  9
Doing these sorts of calculations in your head will help you avoid silly answers.
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Questions on 2.2:
1. The profit gained by a large company last year was €1,342,987,645. Give this
number:
a) To the nearest thousand
b) To the nearest million
c) To the nearest billion
34.27  1.84
giving your answer to 2 decimal places
1.273.254  4.31
Round the following numbers to 3 decimal places:
2.41655
8.1997
3.9299
1
Write the fraction as a decimal to 3 significant figures
7
Write the following to 3 s.f.
15648
0.0045798
1.0006
1.99999
2. Evaluate
3.
a)
b)
c)
4.
5.
a)
b)
c)
d)
6. A cuboid shaped house has dimensions 6m by 3m by 4m (measurements given to
the nearest metre). Calculate the upper and lower bounds for the surface area and
volume of the house.
7. Find the percentage error when 72,346 is given to the nearest 1000.
8. Round each of the numbers in question 2 to 1 s.f. and use this to estimate the
answer. What is the percentage error in your estimation?
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2.3 Standard Index Form / Scientific Notation
A number in standard form is given as a 10 k ,
1  a  10,
k Z
None of the following are in standard form:
5 .6  9 .8 3
11.2  10 5
6.3  10 0.6
The first one is not standard form as the base is not 10. The second one is not in
standard form as a is too large and the third is not in standard form as the power is not
an integer.
Converting from standard form to decimal form:
If k is positive then the decimal point moves k units to the right
If k is negative then the decimal point moves k units to the left
Zeroes may need to be inserted as place-holders.
e.g. convert the following to standard form:
i) 6.3  10 5
We move the decimal point five spaces to the right
We now need to insert zeroes in the blank spaces and we are left with:
6.3  10 5  630000
ii) 3.54  10 2
We move the decimal point 2 spaces to the left
Again we insert zeroes and we are left with 3.54  10 2  0.0354
On a calculator the button EE means  10 x (note that only one E appears on screen)
If you type 6.3 EE 5 Enter or 3.54 EE -2 Enter you should see the correct answer.
Notice that with small numbers, the value of k is negative while for large numbers it
is positive.
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Converting from Decimal Form to Standard Form
We do exactly the same process in reverse ensuring that the value of a is between 1
and 10.
e.g. Write 0.000356 in standard form
This is a small number so k will be negative, the decimal point must be moved so it is
between the 3 and the 5. So we are moving it 4 spaces.
0.000356  3.56  10 4
e.g. Write 287.2 in standard form
This is a large number so k will be positive, the decimal point must be moved so it is
between the 2 and the 8. So we are moving it 2 spaces.
287.2  2.872  10 2
On the calculator:
1. Press MODE and change to SCI (scientific notation)
2. Type the number 0.000356 and press Enter
3. You get the answer 3.56 E -4. Remember this must be written as 3.56  10 4 .
Calculations with standard form can all be done on your calculator
e.g. 2.7  10 4  9  10 7  3  10 2
Questions on 2.3:
1.
a)
b)
c)
2.
a)
b)
3.
i)
ii)
Express the following in standard form:
0.00376
12500000
625.4  10 2
Write in normal form:
3.7  10 4
5.82  10 4
a  5.2  10 2 b  3.4  10 3 c  5.6  10 4
Arrange a, b and c in order of size
Evaluate the following:
ab
c
a)
b)
c)
a bc
b
c
a
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2.4 Conversion of units:
Prefixes: You should be familiar with the following prefixes:
Giga
A billion of…
Mega
A million of …
Kilo
A thousand of
Centi
One hundredth of
Milli
One thousandth of
Micro
One millionth of
Nano
One billionth of
10 9
10 6
10 3
10 2
10 3
10 6
10 9
e.g. A running track is 450m long. I use a number of 15cm rulers placed end to end
to measure it. How many rulers must I have?
1. Before calculating you must ensure that the units are both the same. You can
convert both to metres or both to cm.
450m = 450 x 100 cm =45000 cm
2. Now calculate
45000 15  3000
Therefore I would need 3000 rulers
e.g. A rectangle has side lengths 1.2m and 3.6m
i) Find its area in m 2
A  1.2  3.6  7.32m 2
ii) Find its area in cm 2
A common error here is to multiply the above answer by 100. To be safe you must
instead change the original units to cm:
Length=120cm Width=360cm
A  120  360  43200cm 2
Other units: You may be asked to convert other units but you will be given the
conversion factors.
e.g. 1760 yards is 1 mile, 1 mile is 1.62 km. How many yards are there in 5 km?
These are easiest to solve as ratio problems
So 5km is 3.086 x 1760 yards which is 5432 to the nearest yard
10
Questions on 2.4:
1. A bicycle wheel has radius 15cm. How many complete turns must it make on a
journey of 3km?
2. How long would it take (to the nearest minute) to walk 10 km at 3km/h?
3. Which is faster 10km/h or 3m/s?
5
4. If C  F  32  convert 84°F to Centigrade. Convert 10°C to Fahrenheit.
9
5. A cube has side length 2m. What is its volume in cm 3 ?
6. 1 pound is 16 ounces. An ounce is 28 grams. Jodie weighs 85kg. What is his
weight in pounds?
7. A teaspoon contains 5ml. How many spoonfuls of medicine can be taken from a
2 litre bottle?
11
2.5A Arithmetic sequences:
A sequence is a number pattern – it can be generated by a rule
e.g. write the first 5 terms of the sequence given by:
u n  3  2n (notice that u n is the symbol given to the nth term)
u1  3  2  1  5
u2  3  2  2  7
u3  3  2  3  9
u 4  3  2  4  11
u 5  3  2  5  13
So u1 means the 1st term and u 27 would mean the 27th term
The sequence 5, 7, 9, 11, 13… is an arithmetic sequence. It has a common difference
between terms of 2. We can always find the common difference by taking a term and
subtracting the term before.
We require two things to define an arithmetic sequence: u1 - the first term and d – the
common difference.
Finding a formula for an arithmetic sequence:
In your formula book you
are given:
e.g. Find a formula for the general term (this means ‘express un as a function of n) of
6, 2, -2… and calculate the 20th term.
1. We check that the sequence is arithmetic.
The sequence is arithmetic as the common difference is -4. We subtract 4 every step
we take. The first term is 6.
2. Substitute into the formula.
un  6  n  1 4
3. Simplify.
u n  6  4n  4
u n  10  4n
4. Check by substituting n=1, 2, 3
10  4  1  6
10  4  2  2
10  4  3  2
5. Substitute n=20 to find the 20th term
12
u 20  10  40  20  70
If you are really stuck in the exam:
1. Write down the values of u1 and d as you may get marks for this.
2. On your calculator press ‘6 Enter’
3. Now type ‘-4 Enter’
4. Repeatedly press Enter until you arrive at the 20th term.
Finding other information using the formula:
In the above formula there are 4 unknowns: u n , u1 , n and d . In many questions you
will be given three of them and asked to work out the other one.
e.g. The first term of an arithmetic sequence is 120. The 10th term is 57. Find:
1) The value of the common difference.
2) Which term is the first negative term.
1) a)
Write down the known information and what is unknown:
u1  120,
n  10,
u10  57
d ?
b) Substitute:
57  120  10  1d
c) Solve:
57  120  9d
 63  9d
d  7
2) a)
Rewrite the general formula for the sequence:
u n  120  n  1  7
u n  127  7n
b) Set up an inequality
127  7n  0
c) Solve
127  7 n
127
n
7
n  18.14
d) Give the first whole number solution
The 19th term is the first negative one
13
Using the GDC:
1. Set up a list with the position (n) in L1 and the term
( u n ) in L2
2. Perform a linear regression on the data (STAT, CALC,
LinRegL1, L2)
3. This shows that u n  7n  127 as shown before
4. Enter
5. Use TBLSET to start your table at 1 and go in steps of 1
6. Go to TABLE and scroll down to see the numbers
Which shows that the 19th term is the first negative one.
e.g. The common difference of an arithmetic sequence is 8. The 16th term is 192.
a) Find the first term
b) Determine whether 270 is a member of the sequence
1) a)
Write down the known information and what is unknown:
u1  ?, n  16,
u16  192
d 8
b) Substitute:
192  u1  16 1 8
c) Solve:
192  u1  15  8
192  u1  120
u1  72
2) a)
Rewrite the formula for the sequence:
u n  72  n  1 8
u n  64  8n
b) Set up an equation
64  8n  270
c) Solve
8n  206
206
8
n  25.75
As n is not a whole number 208 is not in the sequence
n
14
Using the formula for simultaneous equations:
If you are given two terms of the sequence (not including the first term) you will have
two unknowns ( u1 and d). This requires you to use simultaneous equations.
e.g. u3  8 u8  17 , find u1 and d.
1. Subsitute the known information into two separate equations:
8  u1  3  1d
 17  u1  8  1d
2. Simplify:
8  u1  2d
 17  u1  7d
3. Solve either algebraically or using Polysmlt
This shows that the first term is -1 and the difference is -5.
Algebraic questions:
Most algebraic questions rely on you to understand that the common differences must
be equal.
3k  1
e.g. three terms of an arithmetic sequence are given by 2k  4, 4k  6,
Find the value of k and of the three terms:
a) Form an equation showing that the differences are equal. Remember that the
difference between terms is found by subtracting the previous term.
4k  6  2k  4  3k  1  4k  6
b) Simplify and solve
2k  2   k  7
3k  9
k 3
c) Substitute back to find the terms and check that they form an arithmetic sequence
23  4  2
43  6  6
3  3  1  10
15
Questions on 2.5A:
1. Find a formula for the general term and evaluate the 10th term for the following
sequences:
a) 3, 7, 11…
b) -2, -4, -6…
1 1 2
, , …
c)
3 2 3
2. The common difference of a sequence is 2, the fifth term is 15. Find the first and
twelfth terms.
3. The first term of a sequence is -5, the twelfth term is 0.5. When does the
sequence first exceed 180?
4. The first term of a sequence is 13, the 14th term is 52. Is 312 a member of the
sequence?
5. The fifth term of a sequence is -2 and the twelfth term is -12.5. What is the first
term which is less than -20?
6. The eleventh term of a sequence is -16, the eighth term is -11.5. Find the 100th
term.
7. For each of the following questions find the value of k and the three terms:
a) k  1, 2k  1, 13
7k
b) k  1, 2k  3,
c) k ,
k2,
k 2  6 (assume k is positive)
16
2.5B Arithmetic Series:
An arithmetic series is the sum of an arithmetic sequence
e.g. 3, 5, 7, 9… is a sequence
3+5+7+9… is a series
You are given two formulae for the sum of terms of a sequence. The first one is used
if you know d (the difference), the second one is used if you know the value of u n (the
last term that you are adding.
e.g. Find the sum of 3, 7, 11, 15 and 19
i) By adding
3+7+11+15+19=55
ii) By using the first formula
u1  3, n  5, d  4
5
2  3  5  1 4
2
5
S 5  6  4  4 
2
5
S 5   22  55
2
iii) By using the second formula
n  5, u1  3, u n  19
S5 
5
3  19  5  22  55
2
2
You will probably notice that the second formula is a lot easier than the first one but
you often will not have the choice of which to use.
S5 
e.g. Find the sum of -3, 1, 5…45
This one is harder than it looks. We have not been given the value of n so we will
have to work it out.
1. Set up an equation to find n.
Using the formula for the general term:
u n  u1  n  1d
45  3  n  1  4
48
 n 1
4
n  13
So we know that 45 must be the 13th term.
2. Find the sum of the sequence.
17
We have enough information to use both formulae so we’ll use the second one as it is
easier:
13
13
S13   3  45   42  273
2
2
Finding other information using the formula:
e.g. The first term of a sequence is 6. The sum of the first 8 terms is 160. Find the
12th term.
1. Write out the information that has been given:
u1  6 S8  160
n8
2. As we will need to find d we should use the first formula. Substitute in and solve:
8
160  2  6  8  1d 
2
160  412  7 d 
40  12  7 d
28  7 d
d 4
3. We now need to find the twelfth term (note it does not ask for the sum of 12
terms).
u n  u1  n  1d
u12  6  12  1  4
u12  6  44  50
e.g. The sum of the first n terms of a sequence is 24. The first term is -1, the common
difference is 2. Find n.
1. Write out the information that has been given:
u1  1
S n  24
d 2
2. Substitute into the formula (note that you do not know the last term)
n
24  2  1  n  1  2 
2
n
24   2  2n  2 
2
n
24  2n  4   nn  2 
2
24  n 2  2n
3. Notice we are left with a quadratic – rearrange them into decreasing powers and
solve manually or using polysmlt.
n 2  2n  24  0
And n must be positive so it is 6.
18
Questions on 2.5B
1. Find the sum of the first 50 terms of the sequence 17, 15, 13, 11…
2. An arithmetic sequence has 12 terms. The first term is 10, the last term is -16.
Find the sum of terms.
3. The first term of an arithmetic sequence is 14, the sum of 12 terms is 264. What
is the twelfth term?
4. Find the sum of 3, 10, 17…157
5. In a room there is 1 chair in the first row, 2 in the second, 3 in the third…How
many rows are there if 153 chairs are used in total?
19
2.6 Geometric Sequences and Series:
2, 6, 18, 54… is an example of a geometric sequence.
The next term in the sequence can always be found by multiplying by a common ratio
(r).
r can be calculated by dividing a term by the previous one:
6 18 54
r  
 3 so whichever two terms we use to find r it is always the same.
2 6 18
Notice that r can be a fraction:
27 1

e.g. 81, 27, 9, 3… - r 
81 3
40 2

e.g. 100, 40, 16, 6.4… - r 
100 5
r can also be negative:
24
2
  Notice that when r is negative the sequence
e.g. 36, -24, 16… - r 
36
3
oscillates between positive and negative values.
Using formulae for geometric sequences:
In your formula book you are given:
The nth term of a geometric sequence is un  u1r n 1
The sum of n terms of a geometric sequence is
u1  r n  1
u1 1  r n 
Sn 
(use this when r  1) 
(use this when r  1)
r 1
1 r
e.g. Consider the sequence 7, 21, 63, 189…
a) Find the common ratio r.
21
r
3
7
b) Find the 10th term.
u1  7 r  3 n  10
 u10  7  3101
 u10  7  39
 u10  7  19683  137781
c) Find the sum of the first ten terms.
Since r>1 we use the first formula for the sum
7  310  1 7  59049  1 7  59048
S10 


 206668
3 1
2
2
20
Questions requiring simultaneous equations:
As with arithmetic sequences – if we are given two terms of the sequence (apart from
the first term) then we will have two unknowns ( u1 and r). To find these we will need
to use simultaneous equations.
e.g. the 3rd term of a sequence is 4 and the 6th term is 0.256. Find u1 and r.
1. Substitute both terms into the general formula:
4  u1r 31
0.256  u1r 61
2. Simplify the expressions:
4  u1r 2 ________________(eq1)
0.256  u1r 5 _____________  eq 2 
3. To solve we need to divide one equation by the other e.g. eq2/eq1
0.256 u1r 5

4
u1r 2
4. Simplify (notice that u1 cancels from top and bottom)
0.064  r 3
r  3 0.064  0.4
5. Substitute r into either eq1 or eq2 to find u1
4  u1  0.42
4  u1  0.16
u1  25
Questions involving percentages:
Geometric sequences will often be disguised as percentage growth or decay.
e.g. An increase in value of 3% per year corresponds to r=1.03
Next year I will have 100% of what I had this year plus another 3%. In total 103% or
1.03
A reduction in value of 5% per year corresponds to r=0.95
e.g. An employee starts work on a salary of $20,000 per year with annual increase of
4%.
1. What will her annual salary be in ten years time?
u1  20000
r  1.04
n  10
u10  20000 1.04101  28466.24
So after ten years she will earn $28,466.24 (we leave money answers to 2 d.p.)
21
2. How much will she have earned in total over the ten years?
Note here that we are asked to add together her yearly salaries. We should use the
sum formula.
20000 1.0410  1
S10 
 240122.14
1.04  1
So in total she will have earned $240,122.14


3. How many years would she have to work for her total earnings to exceed
$500,000.
We substitute the information we know into the sum formula:
20000 1.04 n  1
500000 
1.04  1
As we cannot solve this algebraically (you never can when the power is unknown) we
need to solve it graphically.
Set Y1 to be theft side of the equation and Y2 to be the right side.


We now guess how long it will take – ten years got almost halfway there so lets say
that n must be somewhere between 0 and 25 years. The amount of money must go
beyond 500000. So let’s set our window and show the graph. We now find where the
curve and line intersect.
This shows that it will take 17.7 years. However the first full year in which her
earnings exceed $500,000 will be the 18th.
22
Questions on 2.6:
1. Find a formula for the general term of 4, 8, 16, 32… Use this to find the 15th term.
2. Find the value of k if k,
k+8,
9k are in geometric sequence
3. The third term of a geometric sequence is 8 and the sixth term is -1. Find the 10th
term.
4. What is the first term of 36, 18, 9… to be below 0.001
5. A colony of bears had a population of 1000 in 1990. What is the population now
if it declines at 2% per annum?
6. Find the sum of the first ten terms of 12, -6, 3, -1.5…
7. Tracy’s father gives her $100 this year, he promises to give her $85 next year and
$72.25 the year after.
a) Show that this is a geometric sequence and find the annual percentage reduction.
b) How much will Tracy’s dad give her in twenty years time?
c) How much will he have given her in total in twenty years time?
d) In how many years will he first give Tracy less than $2?
23
2.7: Simultaneous and Quadratic Equations:
Any question in which there are 2 unknowns will require you to solve simultaneous
equations. You should be able to set up a pair of equations and solve them both
algebraically and using your calculator application Polysmlt.
Setting up and solving simultaneous equations:
e.g. 7 CDs and 4 DVDs come to a total of €69.09. 3CDs and 5 DVDs cost €53.76.
Find the cost of 5CDs and 8 DVDs.
Our two unknowns are the cost of a CD (x) and the cost of a DVD (y). We can now
write our equations and label them:
7 x  4 y  69.09 ____________(1)
3x  5 y  53.76 _____________(2)
Solving simultaneous equations:
You are expected to be able to use four methods:
1. Polysmlt (try to solve these using the example on page 15)
2. Elimination:
a) Multiply your equations so that there is a matching coefficient of either x or y.
We could either multiply (1) by 3 and (2) by 7 so that both equations have an xcoefficient of 21 or we could:
(1)  5 35 x  20 y  345.45 ___(3)
to make the y-coefficient be 20 in both cases.
(2)  4 12 x  20 y  215.04 ___(4)
b) We now either add or subtract our equations using the rule SSS (same signs
subtract). The matching coefficient (20) is positive in both equations so we subtract.
(3)  (4)
23x  130.41
notice that we could have done (4) – (3) but we would be left with a negative in both
sides. The answer would be the same.
c) Solve:
130.41
x
 5.67
23
d) Substitute into one of the original equations to find y.
(1)
7  5.67  4 y  69.09
39.69  4 y  69.09
4 y  29.4
y  7.35
e) Substitute into the other original equation to check your answer.
3 5.67  5 7.35  53.76 this works so a CD cost €5.67 and a DVD €7.35.
24
3. Substitution:
a) Make either x or y the subject of either (1) or (2). Notice that this gives us 4
possibilities – we should choose whichever makes our life easiest.
e.g. Making x the subject of equation (2) gives us:
53.76  5 y
x
3
b) Substitute for x into equation (1)
 53.76  5 y 
7
  4 y  69.09
3


c) Simplify and solve
 376.32  35 y 

  4 y  69.09
3


35
125.44 
y  4 y  69.09
3
23
125.44 
y  69.09
3
23
56.35 
y
3
3  56.35
y
23
y  7.35
We now continue as step d) above.
4. Solving simultaneous equations graphically. This will be covered in the linear
functions section.
Setting up and solving quadratic equations:
In a quadratic equation there is only one unknown.
10m
e.g. A grass area length 10m, width 5m is enclosed by
a path with width x m. If the area of the path is
126m2 find the value of x.
5m
xm
1. Find an expression for the area of the path.
The area of the path is equal to the area of the whole
shape – the area of the grass.
The length of the whole shape is given by 10+2x
The width of the whole shape is given by 5+2x.
A  10  2 x  5  2 x   (10  5)
2. Substitute the value of A and simplify
126  50  20 x  10 x  4 x 2  50
126  30 x  4 x 2
3. Divide by any common factors and arrange in decreasing powers of x.
25
xm
63  15 x  2 x 2
0  2 x 2  15 x  63
4. To factorise we now look for two numbers which:
a) Multiply to give the product of the outside terms (2 x -63) = -126
b) Add together to give the inside term 15
The pairs of factors of 126 are {1,126}, {2,63}, {3,42}, {6,21}, {7,18}, {9,14}
The pair {6,21} has a difference of 15 so this looks promising
We find that -6 x 21 = -126 and -6 + 21 = 15
5. We now split the middle term:
0  2 x 2  6 x  21x  63
6. We can now factorise the first two terms and the second two terms:
0  2 x( x  3)  21 x  3
7. We bring the factors together:
0   2 x  21 x  3
8. The zero-product rule states that if two numbers multiply to give 0 then one of
them must be 0.
2 x  21  0
Either
21
x 
 10.5
2
x 3  0
Or
x  3
9. Decide which of your answers makes sense:
In this case x must be positive so the path is 3m wide.
Other forms of the quadratic equation:
The general form of the quadratic equation is y  ax 2  bx  c and can be factorised as
above. In some special cases there are easier ways to factorise:
1. If a=0 then the equation is no longer quadratic and should be solved as alinear
equation.
2. If b=0 then we are normally left with the difference of two squares:
a2  b2   a  b  a  b 
e.g. Solve 9 x 2  25  0
Using DOTS  3x  5 3x  5  0
5
5
or 
3
3
3. If c=0 then we can solve using common factors:
e.g. solve 5 x 2  4 x  0
x is a common factor
Therefore x 
26
x 5x  4  0
x  0 or
4
5
Questions on 2.7:
1. A soft drink comes in 500 ml and 300 ml bottles. A shopkeeper receives 96
bottles containing 39.2 litres in total. How many of each bottle did he receive?
2. John has €8000 which he splits between bank A and bank B. Bank A gives 4%
interest while bank B gives 6% interest (both are simple interest rates). After 1 year
John has earned €410 interest. How much did he put in each account?
3. A number squared plus 5 times the number gives 36. What could the number be?
4. The hypoteneuse of a right angled triangle is given by 5x-7. The other two sides
are 2x-3 and 3x. Calculate x.
5. The point (4,y) is 2.5 units away from the point (2,2). Calculate the possible
values of y.
27