Download Mathematics 3º ESO

Mathematics
3º ESO
María Isabel Muñoz Molina
Juan Francisco Antona Blázquez
Real Instituto de Jovellanos
1st TERM
Numbers
Polynomials
Equations
Numbers
E xpressing N um bers and
Operations in E nglish
1. Numbers
1
2
3
4
5
6
7
8
9
10
one
two
three
four
five
six
seven
eight
nine
ten
11
12
13
14
15
16
17
18
19
20
Cardinal numbers from 1 through 1 000 000
eleven
21 twenty-one
31
twelve
22 twenty-two
40
thirteen
23 twenty-three
50
fourteen
24 twenty-four
60
fifteen
25 twenty-five
70
sixteen
26 twenty-six
80
seventeen
27 twenty-seven
90
eighteen
28 twenty-eight
100
nineteen
29 twenty-nine
1 000
twenty
30 thirty
1 000 000
thirty-one
forty
fifty
sixty
seventy
eighty
ninety
one hundred
one thousand
one million
Examples:
305
three hundred and five
754
1 507
seven hundred and fifty-four
one thousand, five hundred and seven
2 328
1 800
two thousand, three hundred and twenty-eight
one thousand, eight hundred or eighteen hundred
1 436 209
one million, four hundred thirty-six thousand, two hundred and nine
2 700 400
two million, seven hundred thousand, four hundred
(NOTE: The British use and before the last two figures, but the Americans usually leave the and
out).
The number 0:
Nought (UK English)
Zero (US English)
Oh (like the letter O): used when reading out numbers figure by figure (e.g. telephone numbers,
credit card numbers, etc).
1
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2
Decimal numbers:
In English we use a full stop (.) instead of a comma (,) to separate the main part of a number from
the decimal part:
0.1
nought point one (UK) or
0.75
1.263
nought point seven five (UK)
one point two six three
zero point one (US)
or
zero point seven five (US)
You can use a comma (,) in large numbers to separate the hundreds, thousands, and millions. For
example, 5,074 means five thousand and seventy four. In UK English, spaces are sometimes used
instead of commas (5 074).
Place value:
In our decimal number system, the value of each digit depends on its place in the number. Each
place is 10 times the value of the next place to its right.
Millions
Hundred Thousands
Ten Thousands
Thousands
Hundreds
Tens
Units
Tenths
Hundredths
Thousandths
Ten Thousandths
1 000 000
100 000
10 000
1 000
100
10
1
0.1
0.01
0.001
0.0001
Ordinal numbers:
1st
2nd
3rd
4th
5th
6th
7th
8th
9th
10th
first
second
third
fourth
fifth
sixth
seventh
eighth
ninth
tenth
11th
12th
13th
14th
15th
16th
17th
18th
19th
20th
Ordinal numbers from 1 through 1 000 000
eleventh
21st twenty-first
31st
twelfth
22nd twenty-second
40th
thirteenth
23rd twenty-third
50th
fourteenth
24th twenty-fourth
60th
fifteenth
25th twenty-fifth
70th
sixteenth
26th twenty-sixth
80th
seventeenth
27th twenty-seventh
90th
eighteenth
28th twenty-eighth
100th
nineteenth
29th twenty-ninth
1 000th
twentieth
30th thirtieth
1 000 000th
D pto. M atemáticas. IES Jovellanos. 2011
thirty-first
fortieth
fiftieth
sixtieth
seventieth
eightieth
ninetieth
one hundredth
one thousandth
one millionth
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2. Operations
Operation symbols
+
plus
−
× or ·
minus
times
/ or :
divided by
=
6
=
equals or is equal to
is not equal to
<
>
is less than
is greater than
≤
≥
is less than or equal to
is greater than or equal to
Addition (to add):
2+3 =5
Two plus three equals five
The answer is called the sum or total.
Subtraction (to subtract):
7−3 =4
Seven minus three equals four
The answer is called the difference.
Multiplication (to multiply):
5 · 3 = 15
Five times three equals fifteen or Five multiplied by three. . .
The answer is called the product.
Division (to divide):
8:2=4
9
2
1
4
Eight divided by two equals four
9:
2:
4:
1:
dividend
divisor
quotient
remainder
✁✃✁✃✁✃✁✃✁✃✁✃✁✃
D pto. M atemáticas. IES Jovellanos. 2011
Rational N um bers
1. Natural Numbers
The natural numbers are the counting numbers from one to infinity. We use the letter N to refer
to the set of all natural numbers:
N = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, . . .}
The set of natural numbers is endless. As there is no largest counting number, we say that the set
of natural numbers is infinite.
The factors of a natural number are all the natural numbers which divide exactly into it, leaving no
remainder.
For example, the factors of 10 are: 1, 2, 5 and 10.
5 is a factor of 20 because 5 divides exactly into 20; we can also say that 20 is divisible by 5.
The even numbers are the natural numbers which are divisible by 2. The sequence of even numbers
is: 2, 4, 6, 8, 10, 12, 14 . . . and so on.
The odd numbers are the natural numbers which are not divisible by 2. The sequence of odd
numbers is: 1, 3, 5, 7, 9, 11, 13 . . . and so on.
A prime number is a natural number which has exactly two distinct factors, itself and 1.
For example, 17 is a prime number since it has only two factors, 1 and 17.
A composite number is a natural number which has more than two factors.
For example, 26 is a composite number since it has more than two factors: 1, 2, 13 and 26.
Notice that these definitions indicate that one (1) is neither prime nor composite.
Every composite number can be factorised as a product of prime factors in one and only one
way (apart from order).
For example,
72 = 2 · 2 · 2 · 3 · 3
or
72 = 23 · 32
(in exponent form).
A multiple of any natural number is obtained by multiplying it by another natural number.
For example, the multiples of 3 are: 3, 6, 9, 12, 15, 18,. . . and these are obtained by multiplying 3
by each of the natural numbers in turn:
3 · 1 = 3,
3 · 2 = 6,
3 · 3 = 9,
1
3 · 4 = 12 etc.
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Example 1
a) Find the largest multiple of 9 less than 500.
b) Find the smallest multiple of 11 greater than 1000.
a) We begin dividing 500 by 9:
500
5
9
55
We have 5 remainder, so 9 ·55 is smaller than 500. Hence,
as 9 · 55 = 495, the largest multiple of 9 less than 500 is
495.
b) We begin dividing 1000 by 11:
1000
11
10
90
We have 10 remainder, so 11 · 90 is smaller than 1000:
11 · 90 = 990. Hence, the smallest multiple of 11 greater
than 1000 is 11 · 91 =1001.
The HCF (highest common factor) of two or more natural numbers is the largest factor which is
common to both of them. For example:
The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24.
The factors of 40 are: 1, 2, 4, 5, 8, 10, 20, 40.
Hence, the HCF of 24 and 40 is 8.
In general, to find the HCF of two or more natural numbers, we write them as a product of prime
factors in exponent form, and we multiply the common factors raised to the smallest power.
For example: 24 = 23 · 3 and 40 = 23 · 5, so HCF is 23 = 8.
The LCM (lowest common multiple) of two or more natural numbers is the smallest multiple which
is common to both of them. For example:
The multiples of 6 are: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, . . .
The multiples of 8 are: 8, 16, 24, 32, 40, 48, 56,. . .
Hence, the LCM of 6 and 8 is 24.
In general, to find the LCM of two or more natural numbers, write them as a product of prime
factors in exponent form, and multiply common and non common factors raised to the highest
power.
For example: 6 = 2 · 3 and 8 = 23 , so LCM is 23 · 3 = 24.
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Exercises - Set A
1. Find the largest multiple of 7 which is less than 1000.
2. Find the smallest multiple of 13 which is greater than 1000.
3. Find the largest multiple of 17 which is less than 2000.
4. Find the smallest multiple of 15 which is greater than 10 000.
5. Find the HCF of 24, 72, 120.
6. Find the LCM of 12, 18, 27.
7. Three bells chime at intervals of 4, 5 and 6 seconds respectively. If they all chime at the same
instant, how long before they all chime together again?
2. Integers and order of operations
The negative whole numbers, zero and the positive whole numbers (not fractions or decimals) form
the set of all integers . We use the letter Z to refer to the set of integers:
Z = {· · · − 5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5 . . .}
We can show these numbers on a number line . Zero is neither positive nor negative.
b
−5
b
−4
b
−3
b
b
−2
−1
b
b
b
b
b
b
0
1
2
3
4
5
Remember the rules for handling addition and subtraction of integers:
+
−
+
−
(positive) = (positive)
(positive) = (negative)
(negative) = (negative)
(negative) = (positive)
The following rules apply to multiplication and division with integers:
(positive) · (positive) = (positive)
(positive) · (negative) = (negative)
(negative) · (positive) = (negative)
(negative) · (negative) = (positive)
(positive) : (positive) = (positive)
(positive) : (negative) = (negative)
(negative) : (positive) = (negative)
(negative) : (negative) = (positive)
Order of operations with integers:
• Perform the operations within brackets first.
• Calculate any part involving exponents (powers, roots).
• Starting from the left, perform all divisions and multiplications as you come to them.
• Finally, restart from the left and perform all additions and subtractions as you come to them.
The word BEDMAS may help you remember this order:
Brackets
Exponents
Division
Multiplication
Addition
Subtraction
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Rules for brackets:
• If an expression contains one set of grouping symbols, i.e., brackets, work that part first.
• If an expression contains two or more sets of grouping symbols one inside the other, work the
innermost first.
• The division line of fractions also behaves as a grouping symbol. This means that the numerator
and the denominator must be found separately before doing the division.
Exercises - Set B
1. Simplify:
a) −5 + 4 · (−2 + 1)3 − (−9 + 6)2
b) 12 − 2 · [25 : (−4 − 1) + (−2) − (−6 − 10)]
c) −7 − (−3) + (−8) · (−1) − (−12) : (−4)
d) −5 − 4 · [−8 : 2 − 2 · (−3)]
e) 10 − 10 · [−6 + 5 · (−4 + 7 − 3)]
f) 6 − 5 · [−4 − 1 + (−2)2 − 32 ]
2. Simplify:
a)
12 + (5 − 7)
18 : (6 + 3)
b)
57
7 − (2 · 3)
c)
3·8+6
6
d)
(3 + 8) − 5
3 + (8 − 5)
3. Insert grouping symbols if necessary, to make the following true:
a) 120 : 4 · 2 = 15
d) 5 · 7 − 3 − 1 = 19
b) 120 : 4 · 2 = 60
e) 5 · 7 − 3 − 1 = 33
c) 5 · 7 − 3 − 1 = 15
f) 3 + 2 · 8 − 4 = 36
g) 3 + 2 · 8 − 4 = 11
h) 3 + 2 · 8 − 4 = 15
i) 8 − 6 · 3 = 6
Using the calculator:
Most modern calculators have the order of operations rules built into them. They also have grouping
symbols keys (left hand bracket and right hand bracket).
12
For example, if we consider
and key in 12 : 4 + 2 = the calculator gives an answer
4+2
of 5, which is incorrect. However, if we type 12 : (4 + 2) = we obtain the answer 2, which is
correct.
Remember to use the sign change key before a negative number.
Exercises - Set C
1. Evaluate each of the following using your calculator:
a) 17 + 23 · 15
b) (17 + 23) · 15
d) 128 : (8 + 8)
e)
89 + (−5)
−7 · 3
D pto. M atemáticas. IES Jovellanos. 2011
c) 128 : 8 + 8
f)
−15 − 5
6−8:4
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3. Rational numbers (fractions)
A rational number is a number which can be written as a fraction, that is, in the form
and b are both integers, and b 6= 0.
4
−1
2
0
Examples:
,
,
,
,
7
10
3
1
a
, where a
b
−2
...
1
A fraction indicates a part of a unit or a part of a quantity; it consists of two whole numbers, a
numerator and a denominator , separated by a bar symbol:
• the denominator, b, shows how many equal parts the whole has been split into.
• the numerator, a, tells us how many of those equal parts are being described.
The set of rational numbers is represented by the letter Q:
o
na
, a, b ∈ Z, b =
6 0
Q=
b
Remember that in general:
−a
a
a
=
=−
b
−b
b
However:
−a
a
=
−b
b
How to say a fraction:
Here are some examples:
1
a half or one half
2
2
two fifths or two over five
5
8
eight over thirty-five
35
1
a third or one third
3
4
four sevenths or four over seven
7
54
fifty-four over seventy-two
72
1
4
a quarter
Proper and improper fractions:
A fraction which has numerator less than its denominator is called a proper fraction .
3
For example, is a proper fraction.
4
A fraction which has numerator greater than or equal to its denominator is called an improper fraction .
7
For example, is an improper fraction.
4
When an improper fraction is written as a whole number and a fraction, it is called a mixed number .
7
3
For example, can be written as the mixed number 1 43 , as it is really 1 + .
4
4
Example 2
Write the improper fraction
29
as a mixed number.
3
First, we divide 29 by 3:
29
2
3
9
Hence:
29
2
= 9 + = 9 32
3
3
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Simplifying fractions:
We can simplify a fraction by cancelling common factors in the numerator and denominator.
When a fraction is written as a rational number with the smallest possible denominator, we say
it is in lowest terms .
We can simplify a fraction to its lowest terms dividing both numerator and denominator by their
HCF.
Example 3
Simplify:
16
24
As HCF of 16 and 24 is 8, dividing both numerator and denominator by 8 we get:
16
2
= ,
24
3
which is the fraction in its lowest terms
Two fractions are equal or equivalent if they can be written in the same lowest terms.
We can convert a fraction to an equivalent fraction by multiplying or dividing both the numerator
3
6
9
12
and denominator by the same non-zero number:
= =
=
= ...
4
8
12
16
We can also test if two fractions are equivalent by cross-multiplying their numerators and denomi12
24
nators. For example,
and
are equivalent because 12 · 40 = 20 · 24 = 480.
20
40
Comparing fractions:
To compare two or more fractions:
• Find the common denominator , which is the LCM of the original denominators.
• Work out the equivalent fractions.
• Write the fractions in ascending or descending order. Remember that if two fractions have the
same denominator, the greater one is the fraction with the highest numerator.
D pto. M atemáticas. IES Jovellanos. 2011
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Example 4
Which fraction is bigger,
4
7
or ?
5
9
The common denominator of
4
36
=
5
45
7
35
=
9
45
4
7
and is 45, the LCM of 5 and 9.
5
9
(multiply numerator and denominator by 9).
(multiply numerator and denominator by 5).
The order is
36
35
4
7
>
, so > .
45
45
5
9
Exercises - Set D
1. Express with denominator 12:
a)
2
3
b)
3
4
5
6
c)
d)
6
18
e)
15
45
d)
24
28
e)
18
42
e)
33
77
2. Express with numerator 12:
a)
3
7
b)
6
5
4
9
c)
3. Express in lowest terms:
a)
6
10
6
18
b)
c)
25
10
d)
14
35
f)
48
72
g)
78
117
h)
125
1000
4. Simplify:
a)
8−2·5
4·3
b)
−1 + 8 : 2
8−5
c)
32 − 2 · 5
8 − 32
d)
−5 · (−6)
−11 − 4
e)
6−3:3
2 + 10 : 2
f)
12 + 8 : 2
12 − 8 · 2
g)
6 + 22
6 − 23
h)
11 − 3
16 : 4
5. Place these fractions in ascending order:
a)
2 3 3 17
, , ,
5 8 4 40
b)
4 7 5
3
6
, , ,− ,−
3 5 7
4
11
c)
1
2 3
1
3
,− ,
,− ,−
8
3 11
6
4
6. Place these fractions in descending order:
a)
5 11 7 5
,
,
,
6 24 12 8
b)
2 3 4 5 6
, , ,
,
5 7 9 13 10
5
1
4
7
6
c) − , − , − , − , −
8
2
7
11
13
7. Write as a mixed number:
a)
45
4
b)
11
2
a)
37
5
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4. Operations with fractions
Addition and subtraction:
To add or subtract fractions:
• If necessary, convert the fractions so they have the lowest common denominator.
• Add or subtract the new numerators. The denominator stays the same.
For example:
3 2 1
9
8
6
9−8+6
7
− + =
−
+
=
=
4 3 2
12 12 12
12
12
When adding or subtracting mixed numbers, you can first convert them to improper fractions and
then perform the operation. However you can also add the whole numbers and fractions separately,
then combine the result.
Example 5
Find: 2 13 − 3 12 + 1 41
First, we convert the mixed numbers into improper fractions:
1
6 1
7
1
6 1
7
1
4 1
5
2 13 = 2 + = + = ; 3 12 = 3 + = + = ; 1 14 = 1 + = + =
3
3 3
3
2
2 2
2
4
4 4
4
Hence: 2 31 − 3 21 + 1 41 =
7 7 5
28 42 15
1
− + =
−
+
=
3 2 4
12 12 12
12
Multiplication:
To multiply two fractions, we multiply the two numerators to get the new numerator, and multiply
the two denominators to get the new denominator:
a c
a·c
· =
b d
b·d
To help make multiplication easier, we can cancel any common factors in the numerator and denominator before we multiply.
For example:
4 3
4·3
4·1
4
· =
=
=
9 5
9·5
3·5
15
Two numbers are reciprocals of each other if their product is one.
For any fraction
a
a b
a
b
, we notice that · = 1. So, the reciprocal of is .
b
b a
b
a
Division:
To divide two fractions, multiply the first by the reciprocal of the second:
a c
a d
a·d
: = · =
b d
b c
b·c
For example: 1 31 : 3 21 =
4 7
4·2
8
: =
=
3 2
3·7
21
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Exercises - Set E
1. Find:
a) 1 32 − 2
b) 3 43 − 1 21
c)
3
− 2 21
4
d) 1 23 + 3 14
e) 4 31 + 2 61
f) 2 32 − 5 56
g) −2 41 + 3 18
h) 4 15 − 2 16
2. Find:
a)
2 1 3
· ·
3 4 5
3 5 4
− ·
5 2 3
2
3
2
− · 1 32
g)
3
4
d)
b)
3
4
2
· −
· −
8
3
5
c)
2 3 2
+ ·
3 4 3
e)
3 1 2 1
· + ·
5 3 3 4
f)
4 1 1 2
· − ·
3 2 6 3
d)
4
:3
5
h) 4 · 1 13 − 5 ·
2
7
3. Find:
2 1
a) :
3 6
e)
1
−
2
3
g) 1 21 : −
4
5 1
b) :
7 3
1
: 1 23
4
f) 2 34 :
3
c) :
4
2
3
h) 3 15 : 1 31
4. Find:
2 3 4
a) − :
3 2 5
5 1 4
b) : +
3 2 3
1 2 3 6
c) · − :
2 5 4 5
2
d) :
5
1
3 2
−
+ ·
2
4 5
5. Find and give your answers in their simplest form:
a)
3 7 7
+ −
4 6 8
:
25
12
3 1
(−3) ·
−
5 3
d)
4 6
(−2) ·
−
3 5
b)
7
9
13
13
−
·
+ −
15 25
22
33
1
3
2
3− ·
−
4
5 15
e)
4
1 3
6+
·
−
25
2 4
c)
1
−
2
3
−1
4
3
+1
4
2 5
3 5
−
·
−
3 9
4 6
f) 7
5
4
−
· +1
12 6
3
5. Problem solving
To find a fraction of a quantity we multiply the quantity by the numerator, and divide the result by
the denominator.
3
3
3 · 85
For example:
of 85 e are:
· 85 =
= 51 e.
5
5
5
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Example 6
If
3
of a shipping container holds 2100 identical cartons, how many cartons will fit into:
8
a)
5
of the container
8
b) the whole container?
3
1
of the container holds 2100 cartons; hence of the container holds
8
8
2100 : 3 = 700 cartons.
5
holds 700 · 5 = 3500 cartons
8
8
b) The whole is which is 700 · 8 = 5600 cartons.
8
a) So,
Example 7
Rob eats
remains?
1
3
of a watermelon one day and of it the next day. What fraction of the watermelon
3
8
We let the whole watermelon be represented by 1. The fraction remaining will be:
24
8
9
7
1 3
1− − =
−
−
=
3 8
24 24 24
24
Exercises - Set F
1. Millie calculated that her bicycle cost
$, what did her bicycle cost?
1
of the cost of her father’s car. If the car cost 38 014
83
1 1
1
, and of the brickwork of his new garage. What
3 5
4
fraction must he complete on the fourth and final day?
2. Over three successive days Colin builds
3. 200 kg of sugar must be poured into packets so there is
packets will be filled?
2
kg of sugar per packet. How many
5
4. 2400 kg of icecream is put into plastic containers which hold
containers are needed?
3
kg each. How many plastic
4
5. John says that his income is now 3 21 times what it was 20 years ago. If his current annual
income is 63 000 e , what was his income 20 years ago?
1
1
1
1
of its weekly budget on rent, on food, on clothes,
on entertainment
3
4
8
12
and the remainder is banked. How much is banked if the weekly income is 864.72 e?
6. A family spends
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Exercises - Set G
1. Renee used
pipe is left?
2.
1
2
of a length of pipe and later used of what remained. What fraction of the
3
3
2
1
of an amount of money is 519 $. Find: a) of the money
3
3
b) the whole amount.
3
of a field was searched for truffles and 39 were found. How many truffles would we expect
13
to find in the remainder of the field?
3.
4. The seven tenths of a quantity are 210. What is the quantity?
5. Alfredo sent
2
2
of his potato crop to market last week. This week he sent of the remainder.
5
3
a) What fraction of his crop has now gone to market?
b) If he has 860 kg remaining, what was the original weight of the crop?
2
9
of the weight of a loaf of bread comes from the flour used in making the bread. If of
10
9
the weight of the flour is protein, what fraction of the weight of a loaf of bread is protein?
6.
7. A tree is losing its leaves. Two thirds fall off in the first week, two thirds of those remaining
fall off in the second week and two thirds of those remaining fall off in the third week. Now
there are 37 leaves. How many leaves did the tree have originally?
8. Out of a group of 40 people in a shop, 32 were aged thirty or over. What fraction of the
people were under thirty?
4
1
of the cars on the road are saloons. Of these saloons are red. What fraction of the cars
5
8
on the road are red saloons?
9.
13
of the distance between college and home. She wants to split
28
the remaining distance into 5 equal parts. What fraction of the whole journey is each part?
10. Sadie has already driven
1
of his time when awake playing on his
8
computer. What fraction of his day does Marc spend playing on the computer?
11. Marc spends a third of the day asleep. He spends
12. A 9 tonne load of top soil is divided into
can be filled?
3
tonne plant containers. How many containers
25
2
4
of a bar of chocolate. Linda eats of what remains. What fraction of the bar
5
9
of chocolate have they eaten between them?
13. John eats
14. In his will a man leaves half his estate to his wife and the rest is shared equally among his
five sons. What fraction of his estate does each son inherit?
15. Out of a deposit of oil you empty one half. Out of what remains, you empty one half again,
11
and then you empty
of what remains. Finally, there are 36 litres left. How many litres were
15
there at the beginning?
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D pto. M atemáticas. IES Jovellanos. 2011
D ecim al N um bers
1. What are decimal numbers?
6
3
+
, which can also be written as an improper
The number 4.63 is a quick way of writing 4 +
10 100
463
63
fraction
or as a mixed number 4 100
.
100
Numbers such as 4.63 are commonly called decimal numbers . 4 is the whole part and 63 is the
decimal part .
You say: four point six three, or four units six tenths and three hundredths, or four units and sixtythree hundredths.
2
6
+
(expanded fractional form). You
100
1000
say: fourteen point oh six two, or fourteen units six hundredths two thousandths, or fourteen units
sixty-two thousandths.
Likewise, 14.062 is the quick way of writing 14 +
Exercises - Set A
1. Write the following in expanded fractional form:
a) 2.5
b) 2.05
c) 2.0501
d) 4.0052
e) 0.0106
2. Write the following in decimal form:
a) 3 +
d)
2
10
7
9
+
100 1000
b)
7
8
+
10 100
e) 4 +
1
10 000
c)
6
3
+
10 1000
f) 5 +
3
2
+
100 10 000
2. Types of decimal numbers
There are three different types of decimal numbers: terminating, recurring and irrational numbers.
A terminating or exact decimal is one which does not go on forever, so you can write down all its
digits. For example: 0.125.
A recurring decimal is a decimal number which does go on forever, but where some of the digits
are repeated over and over again. For example: 0.12525252525 . . . is a recurring decimal, where 25
1
D ecim al N um bers - 3 o E SO
2
is repeated forever.
Sometimes recurring decimals are written with a bar over the digits which are repeated, or with dots
over the first and last digits that are repeated.
For example: 3.2014014014 · · · = 3.2014 = 3.20̇14̇
Irrational numbers are those which go on forever and don’t have digits which repeat. For example:
√
2 = 1.4142135 . . . , π = 3.14159265 . . .
All rational numbers (fractions) can be expressed as either terminating decimals or recurring
decimals, dividing the numerator by the denominator.
For example:
7
= 0, 175;
40
8
= 0.72727272 . . . ;
11
5
= 0.4166666 . . .
12
Conversely, all terminating and recurring decimals can be expressed as fractions.
However, irrational numbers cannot be expressed as fractions, they are not rational. All rational and
irrational numbers form the set of real numbers , which is represented by the letter R.
How to convert a terminating decimal into a fraction:
• Write the decimal as a fraction with denominator 10, 100, 1000,. . . , according to the number
of decimal places.
45
For example: 0.45 =
100
• Simplify the fraction to its lowest terms:
0.45 =
45
9
=
100
20
How to convert a recurring decimal into a fraction:
Look at these examples:
5.454545. . .
1. Let x = 5.454545 . . .
(A)
2. Multiply by 100 (because there are two recurring figures; if there were three recurring figures,
you would multiply by 1000):
(B)
100x = 545.454545 . . .
3. Subtract B−A:
100x =
x =
99x =
4. Divide by 99 and simplify: x =
545.454545 . . .
5.454545 . . .
540
540
60
=
99
11
2.5636363. . .
1. Let x = 2.5636363 . . .
2. Multiply by 10 (because there is one figure between the whole part and the recurring figures;
if there were two figures between the whole part and the recurring figures, you would multiply
by 100):
10x = 25.636363 . . .
(A)
D pto. M atemáticas. IES Jovellanos. 2011
D ecim al N um bers - 3 o E SO
3
3. Multiply (A) by 100 (because there are two recurring figures):
(B)
1000x = 2563.636363 . . .
4. Subtract B−A:
1000x =
10x =
2563.636363 . . .
25.636363 . . .
990x =
5. Divide by 990 and simplify: x =
2538
2538
141
=
990
55
Exercises - Set B
1. Write as decimals, and state in each case whether they are terminating or recurring decimals:
a)
7
50
b)
2
125
c)
5
9
d)
f)
87
60
g)
11
3
h)
33
22
i)
4
11
3
20
e)
44
7
j)
3
5
2. Write as a fraction in its lowest terms:
a) 0.15
b) 0.046
c) 0.7̇
d) 0.5̇4̇
e) 0.12̇
f) 3.407̇
g) 0.3̇05̇
h) 0.4̇8̇
i) 4.2̇
j) 1.2̇7̇
3. Calculate. Firstly you must express decimals as fractions:
2
3
1
3
a) 0.4̇6̇ − + 3.6
b) · 2.4̇ −
c) 5 − 2 ·
− 0.1̇4̇
5
3
5
5
1
to a decimal, and I got the answer 0.07692308.
13
There is no repeating pattern, so the decimal does not recur.” Explain why Shula is wrong.
4. Shula says, “I used my calculator to change
3. Approximation and rounding
Rounding a number is a way of writing it approximately. Sometimes we don’t need to write all the
figures in a number, as an approximate one will do.
For example: for a population of 27 653 the number is large and will change daily. It is better to
round up and say 28 000.
When rounding numbers to a given degree of accuracy, look at the next digit:
• If it is 5 or more then we have to round up , that is, increase the previous digit by one.
• Otherwise we round down , that is, leave the previous digit unchanged.
For example, to round 7.365 to 2 decimal places, look at the thousandths digit: 7.365.
The thousandths digit is 5, so we round up to 7.37: 7.365≈7.37 (to 2 decimal places).
Numbers can be rounded:
D pto. M atemáticas. IES Jovellanos. 2011
D ecim al N um bers - 3 o E SO
4
• to decimal places (d.p.) :
4.16 = 4.2 to 1 decimal place
• to the nearest unit, ten, hundred, thousand,. . .
32 559 = 33 000 to the nearest thousand
• to significant figures (s.f.) : the first non-zero digit in a number is the 1st significant figure;
it has the highest value in the number. When rounding to significant figures, count from the
first non-zero digit.
For example:
54.76 ≈ 55 (to 2 s.f.)
0.00405 ≈ 0.0041 (to 2 s.f.)
6.339 ≈ 6.34 (to 3 s.f.)
Exercises - Set C
1. Round these decimal numbers to the nearest whole number:
a) 5.8
b) 21.67
c) 39.175
2. Round these numbers to the nearest thousand:
a) 2239
b) 12 563
c) 155 669
3. Round these numbers to two decimal places:
a) 0.317
b) 15.304
c) 16.445
4. Use a calculator to work these out. Write your answers correct to 3 s.f. where necessary:
a) (16.8 + 12.4) · 17.1
d)
27.4 18.6
−
3.2
16.1
b) 37.4 − 16.1 : (4.2 − 2.7)
e)
27.9 − 17.3
+ 4.7
8.6
c)
16.84
7.9 + 11.2
f)
0.0768 + 7.1
18.69 − 3.824
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D pto. M atemáticas. IES Jovellanos. 2011
Percentage and Ratio
1. Percentage
Percentages are comparisons with the whole amount (which we call 100 %).
% reads percent which is short for per centum, which loosely translated from Latin, means out of
every hundred.
1
25
1
10
Thus 10 % means 10 out of every 100 or
=
. Likewise, 25 % means
= .
100
10
100
4
x
In general, x % =
(we can simplify the fraction).
100
We often need to calculate a percentage of a quantity. To do this, we multiply the quantity by the
percentage expressed either as a fraction or as its decimal equivalent. For example:
9 % of 24 =
9 · 24
216
9
· 24 =
=
= 2.16
100
100
100
37 % of 58 = 0.37 · 58 = 21.46 (0.37 is the decimal equivalent of 37 %).
To convert any fraction to a percentage, we write it as a decimal and multiply by 100. For example:
2
3
= 0.75 ; 0.75 · 100 % = 75 %
= 0.4 ; 0.4 · 100 % = 40 %
4
5
Exercises - Set A
1. Copy and complete these tables (simplify any fractions).
Percentage
12 %
Fraction
Decimal
Percentage
75 %
Fraction
0.36
Decimal
0.8
3/10
4/25
25 %
85 %
0.4
0.15
1
3/5
2. In a football squad of 24 players, 5 of the players are goalkeepers. What percentage of the
football squad are not goalkeepers?
3. Leon scores 68 % in his French exam and gets
he do the best?
37
in his German exam. In which subject did
54
4. A train journey is 395 km long. Lola is traveling on a train that has completed 23 % of the
journey. How many kilometres has Lola’s train traveled?
1
Percentage and R atio - 3 o E SO
2
2. Ratio
A ratio is an ordered comparison of quantities of the same kind.
For example: the ratio of cordial to water in a bottle is 1 : 5. This means that we have 1 part of
cordial and 5 parts of water (6 parts in total).
Example 1
Write as a ratio, without simplifying your answer:
a) Jack has 5 e and Jill has 50 cents.
b) Mix 200 ml of cordial with 1 l of water.
a) Jack : Jill = 5 e : 50 cents = 500 cents : 50 cents = 500 : 50
b) cordial : water = 200 ml : 1 l = 200 ml : 1000 ml = 200 : 1000
If we have a ratio and multiply or divide both parts by the same non-zero number, we obtain
an equal ratio .
For example:
45 : 15 = 3 : 1 (dividing both parts by 15)
0.4 : 1.4 = 4 : 14 = 2 : 7 (multiplying both parts by 10 and dividing by 2 later).
Ratios are equal if they can be expressed in the same simplest form.
For example: 3 : 5 = 6 : 10
15 : 20 = 3 : 4 = 12 : 16
Example 2
The ratio of walkers to guides on a demanding bushwalk is to be 9 : 2. How many guides are
required for 27 walkers?
walkers : guides = 9 : 2 = 27 : ?
If 9 parts is 27, then 1 part is 27 : 9 = 3, so 2 parts is 3 · 2 = 6.
Hence, 9 : 2 = 27 : 6, and 6 guides are needed.
Quantities can be divided in a particular ratio by considering the number of parts the whole is
to be divided into.
D pto. M atemáticas. IES Jovellanos. 2011
Percentage and R atio - 3 o E SO
3
Example 3
An inheritance of 60 000 $ is to be divided between Donny and Marie in the ratio 2 : 3. How
much does each receive?
There are 2 + 3 = 5 parts.
2
2
Donny gets of 60 000 $ = · 60 000 = 24 000 $
5
5
Marie gets
3
3
of 60 000 $ = · 60 000 = 36 000 $
5
5
Exercises - Set B
1. Write as a ratio, simplifying your answer if possible:
a) 10 $ is to 7 $
b) 2 l is to 5 l
c) 80 kg is to 50 kg
d) 2 $ is to 50 cents
e) 500 ml is to 2 l
f) 800 m is to 1.5 km
2. Express the following ratios in simplest form:
a)
3 1
:
4 4
b) 0.5 : 0.2
c) 18 : 24
d) 2 12 : 1 21
e) 1.5 : 0.3
3. A hospital employs nurses and doctors in the ratio 7 : 2. If 84 nurses are employed, how many
doctors are employed?
4. A farmer has pigs and chickens in the ratio 3 : 8. If she has 360 pigs, how many chickens
does she have?
5. The price of a TV is reduced from 500 e to 400 e. A DVD player costing 1250 e is reduced
in the same ratio as the TV. What does the DVD player sell for?
6. Divide: a) 50 $ in the ratio 1 : 4
b) 35 $ in the ratio 3 : 4
7. A fortune of 400 000 $ is to be divided in the ratio 5 : 3. What is the larger share?
8. The ratio of girls to boys in a school is 5 : 4. If there are 918 students at the school, how
many are girls?
9. A glass contains alcohol and water in the ratio 1 : 4. A second glass has the same quantity
of liquid but this time the ratio of alcohol to water is 2 : 3. each glass is emptied into a third
glass. What is the ratio of alcohol to water for the final mixture?
10. One full glass contains vinegar and water in the ratio of 1 : 3. Another glass of twice the
capacity of the first has vinegar and water in the ratio 1 : 4. If the contents of both glasses were
mixed together what is the ratio of vinegar to water?
11. A glass contains alcohol and water in the ratio 1 : 3. Another glass of the same capacity
contains alcohol and water in the ratio 3 : 5. What is the ratio of alcohol to water if we mix
both glasses?
12. The same as before, but the ratios are 1 : 4 and 2 : 5.
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D pto. M atemáticas. IES Jovellanos. 2011
E rrors and accuracy of
m easurem ent
1. Measurement
The measurement of length, area, volume and capacity is of great importance. Builders, architects,
engineers and manufacturers need to measure the sizes of objects to considerably accuracy.
The most common system of measurement is the Systeme International (SI) .
Important units that you should be familiar with include:
Measurement of
Length
Mass
Capacity
Time
Temperature
Speed
Standard unit
metre
gram
litre
hours, minutes, seconds
degrees Celsius
metres per second (ms−1 )
What it means
How long or how far
How heavy an object is
How much liquid or gas is contained
How long it takes
How hot or cold
How fast it is travelling
The SI uses prefixes to indicate an increase or decrease in the size of a unit:
Prefix
tera
giga
mega
kilo
hecto
Symbol
T
G
M
k
h
Meaning
1 000 000 000 000
1 000 000 000
1 000 000
1 000
100
Prefix
centi
milli
micro
nano
pico
Symbol
c
m
µ
n
p
Meaning
0.01
0.001
0.000 001
0.000 000 001
0.000 000 000 001
2. Errors
Whenever we take a measurement, there is always the possibility of error. Errors are caused by
inaccuracies in the measuring device we use, and in rounding off the measurement we take. They
can also be caused by human error, so we need to be careful when we take measurements.
There are two types of error:
• The absolute error due to rounding or approximation is the difference between the actual
or true value and the measured value, regardless the sign:
εa = |true value − rounded value|
• The percentage or relative error is the absolute error compared with the true value,
expressed as a percentage:
εr =
εa
· 100
true value
1
E rrors - 3 o E SO
2
Example 1
The crowd at a tennis tournament was 14 869, but in the newspaper it was reported as 15 000.
Find the absolute and percentage errors in this approximation.
εa = |15 000 − 14 869| = 131
εr =
131
· 100 ≈ 0.881 %
14 869
3. Accuracy
When a measurement is written, it is always written to a given degree of accuracy. The real
measurement can be anywhere within ± half a unit.
For example: A man walks 23 km (to the nearest km). Because the real measurement has been
rounded, it can lie anywhere between 22.5 km (minimum) and 23.5 (maximum).
When we take measurements, we are usually reading from some sort of scale. The scale of a ruler
may have millimetres marked on it, but when we measure an object’s length it is likely to lie between
two marks.
So, when we round or estimate to the nearest millimetre, our answer may be inaccurate by up to half
a millimetre. We say the ruler is accurate to the nearest half a millimetre. In general:
A measurement is accurate to ±
1
of the smallest division on the scale.
2
Example 2
Rod’s height was measured using a tape measure with centimetre graduations. It was recorded
as 188 cm. For this measurement state the absolute and the percentage error.
The tape measure is accurate to ±
1
cm.
2
The absolute error is 0.5 cm.
0.5
The percentage error is:
· 100 ≈ 0.266 %.
188
D pto. M atemáticas. IES Jovellanos. 2011
E rrors - 3 o E SO
3
Exercises - Set A
1. Find the absolute error and percentage error in saying that:
a) there were 300 people at the conference when there were actually 291.
b) 2.95 can be approximated by 3.
c) 31 823 $ can be rounded to 32 000 $.
2. State the accuracy possible when using:
a) a ruler marked in mm.
b) a set of scales marked in kg.
c) a tape measure marked in cm.
d) a jug marked with 100 ml increments.
3. Su-Lin’s height was measured using a tape measure with centimetre markings. Her height
was recorded as 154 cm.
a) State the range of possible heights in which her true height lies.
b) Find the absolute error in the measurement.
c) Find the percentage error.
4. When we measured the height of a building we got 14,48 m, but the true height actually is
14,39. When measuring the height of a tree, we got 7,85 m, whereas the true height is 7,92 m.
a) Find the absolute and percentage errors in each measurement.
b) Which measurement is more accurate? Why?
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D pto. M atemáticas. IES Jovellanos. 2011
N um ber sets and interval
notation
1. Number sets
A set is a collection of objects or things.
For example:
V = {vowels} = {a,e,i,o,u}
E = {even numbers} = {2, 4, 6, 8, 10 . . . }
V is a finite set as it has a finite number of elements.
E is an infinite set as it has infinitely many elements.
are both sets.
We use the symbol ∈ to mean “is a member of ” or “is in”. So, a ∈ V and 28 ∈ E, but h 6∈ V or
119 6∈ E.
We use:
• N to represent the set of all natural numbers {0, 1, 2, 3, 4, 5, 6 . . . }
• Z to represent the set of all integers {0, ±1, ±2, ±3, ±4, ±5 . . .}
p
• Q to represent the set of all rational numbers
, p, q ∈ Z, q 6= 0
q
• R to represent the set of all real numbers; these are the numbers which can be placed on the
number line, that is, all decimal numbers (exact, recurring or irrational).
Exercises - Set A
1. True or false?
a) −3 ∈ N
b) 6 ∈ Z
e) −
f) 2 31 ∈ Z
1
6∈ Q
4
2. Which of these are rational?
3
a) 8
b) −7
c)
4
e) π
f) 2 15
g) 9, 176
3
∈Q
4
g) 0, 3684 ∈ R
c)
√
3
√
h) 400
d)
1
d)
h)
√
2 6∈ Q
1
∈Z
0, 1
N um ber sets and interval notation - 3 o E SO
2
2. Interval notation
Interval or set notation allows us to quickly describe sets of numbers using mathematical symbols
only.
For example:
{x | − 3 < x ≤ 2, x ∈ R}
including 2”.
reads “the set of all real x such that x lies between minus 3 and 2,
Unless stated otherwise, we assume we are dealing with real numbers, thus the set can also be written
as {x | − 3 < x ≤ 2}.
not included
included
b
We can represent the set on a number line as
|
|
−3
2
Sometimes we want to restrict a set to include only integers or rationals.
For example: {x | − 5 < x < 5, x ∈ Z}
minus 5 and 5”.
reads “the set of all integers x such that x lies between
We can represent the set on a number line as
b
|
b
b
b
b
−5
b
b
0
1. Write verbal statements for the meaning of:
b) {x | x ≤ 5}
d) {x | 1 ≤ x ≤ 4}
c) {y | 0 < y < 8}
e) {t | 2 < t < 7}
f) {n | n ≤ 3 or n > 6}
2. Sketch the following number sets:
a) {x | 4 ≤ x < 8, x ∈ N}
b) {x | − 5 < x ≤ 4, x ∈ Z}
c) {x | − 3 < x ≤ 5, x ∈ R}
d) {x | x > −5, x ∈ Z}
e) {x | x ≤ 6}
f) {x | − 5 ≤ x ≤ 0}
3. Write in set notation:
a)
c)
e)
|b
0
b
|
|
0
3
b
b
|
|
−1
2
b
b
3
b
b
b)
d)
b
|
|
2
5
|b
b
b
b
b
0
|b
6
f)
b
|
5
|
0
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D pto. M atemáticas. IES Jovellanos. 2012
b
|
5
Exercises - Set B
a) {x | x > 4}
b
Index Law s
1. Index notation
Rather than write 2 · 2 · 2 · 2 · 2, we write such a product as 25 .
25 reads “two to the power of five”, “two raised to five” or “two with index five”.
If n is a positive integer, then an is the product of n factors of a
i.e., an = |a · a · a{z· a . . . a}
n factors
a is the base.
n is the power, index or exponent.
Other examples:
52 reads “five to the power of two”, “five raised to two” or “five squared”.
73 reads “seven to the power of three”, “seven raised to three” or “seven cubed”.
Exercises - Set A
1. What is the last digit of 3100 ?
(Hint: Consider 31 , 32 , 33 , 34 , 35 , 36 . . . and look for a pattern.)
2. What is the last digit of 7200 ?
2. Negative bases
So far we have only considered positive bases raised to a power.
We will now look at negative bases. Consider the statements below:
(−1)1 = −1
2
(−1) = (−1) · (−1) = 1
(−1)3 = (−1) · (−1) · (−1) = −1
(−1)4 = (−1) · (−1) · (−1) · (−1) = 1
(−2)1 = −2
(−2)2 = (−2) · (−2) = 4
(−2)3 = (−2) · (−2) · (−2) = −8
(−2)4 = (−2) · (−2) · (−2) · (−2) = 16
In he pattern above it can be seen that:
A negative base raised to an odd power is negative; whereas
a negative base raised to an even power is positive.
1
Index law s - 3 o E SO
2
Be careful:
to a power.
(−2)4 = 16, but −24 = −16, so always use brackets when raising a negative base
Exercises - Set B
1. Simplify:
a) (−1)17
b) −54
c) (−2)5
a) 2, 86
2. Find using your calculator:
b) (−5)5
d) −(−3)2
c) −94
d) (−1, 14)23
3. Find using your calculator, correct to 3 decimal places:
a) (2, 6 + 3, 7)4
3
3, 2 + 1, 92
d)
1, 47
b) 8, 63 − 4, 23
4
0, 52
e)
0, 09·, 14
c) 12, 4 · 10.74
a)
648
3, 624
3. Index laws
Recall the following index laws where the indices m and n are positive integers:
am · an = am+n
To multiply numbers with the same base, keep the base and add
the indices.
am
= am−n
an
To divide numbers with the same base, keep the base and subtract the indices.
(am )n = am·n
When raising a power to a power, keep the base and multiply
the indices.
(a · b)n = an · bn
a an
= n
b
b
The power of a product is the product of the powers.
The power of a quotient is the quotient of the powers.
Exercises - Set C
1. Simplify using the index laws (leave your answer in index form):
a) 22 · 24
b) 114 · 11
c) 1315 · 136
d) 172 · 175
e)
f)
g) 149 : 144
h) 98 : 92
25
22
56
5
i) (22 )4
j) (104 )2
k) (93 )7
l) (74 )5
m) 63 · 53
n) 84 : 24
o) 57 · 37
p) (63 )2 · 96
q) (83 )3 · 82
r) (102 )5 : 210
s) (62 )3 · 6
t) (73 )4 : (72 )4
4. Zero and negative indices
Look at this example:
23
8
= =1
23
8
D pto. M atemáticas. IES Jovellanos. 2012
Index law s - 3 o E SO
3
23
= 23−3 = 20 .
23
But, applying the second index law:
20 = 1.
So, we deduce that:
a0 = 1, for all a 6= 0.
In general:
72
7·7
1
1
=
=
= 3
75
7·7·7·7·7
7·7·7
7
72
But, applying the second index law:
= 72−5 = 7−3 .
75
1
So, we deduce that:
7−3 = 3 .
7
Now, look at this example:
In general, if a is any non-zero number, and n is an integer, then:
• a−n =
1
an
(i.e., an and a−n are reciprocals of one another).
• In particular, a−1 =
1
a
and
a −n
b
=
n
b
.
a
Example 1
Simplify, giving answers in simplest rational form:
−2
3
−2
a) 5
b)
c) 80 − 8−1
5
a) 5−2 =
1
1
=
52
25
b)
−2 2
3
5
52
25
=
= 2 =
5
3
3
9
c) 80 − 8−1 = 1 −
1
7
=
8
8
Exercises - Set D
1. Simplify, giving answers in simplest rational form:
a) 5 − 70
b) 60 − 20
c) (6 − 2)0
d) 4−1
e) 3−2
−1
1
i)
3
f) 3−3
−1
2
j)
5
h) 2−4
m) 20 + 21 + 2−1
n) 1 21
g) 10−5
−2
3
k)
4
−2
1
o)
+ 2−1
3
−3
l) 50 − 5−1
p)
−2 4
4
1
−
3
2
2. Write as powers of 2, 3 or 5:
a) 8
f)
1
125
b)
1
8
g) 32
c) 9
h)
1
32
d)
1
9
i) 81
e) 125
j)
1
81
D pto. M atemáticas. IES Jovellanos. 2012
Index law s - 3 o E SO
4
Exercises - Set E
1. Simplify:
a) 73 · 72
b) 54 · 53
c) a7 · a2
d) a4 · a
e) b8 · b5
f) a3 · an
g) b7 · bm
h) m4 · m2 · m3
k) 77 : 74
l)
59
52
b10
m) 7
b
1113
119
p5
n) m
p
i)
j)
o)
a6
a2
ya
y5
p) b2x : b
q) (32 )4
r) (53 )5
s) (24 )7
t) (a5 )2
u) (p4 )5
v) (b5 )n
w) (xy )3
x) (a2x )5
2. Express in simplest form with a prime number base:
a) 8
b) 25
c) 27
d) 43
e) 92
f) 3a · 9
g) 5t : 5
h) 3n · 9n
k) (54 )x−1
l) 2x · 22−x
3x+1
3x−1
4y
n) x
8
16
2x
2y
m) x
4
i)
j)
o)
3x+1
31−x
p)
2t · 4t
8t−1
3. Remove the brackets of:
a) (a · b)3
b) (a · c)4
c) (b · c)5
d) (a · b · c)3
e) (2a)4
f) (5b)2
g) (3n)4
i) (4abc)3
j)
h) (2bc)3
5
2c
l)
d
a 3
k)
b
m 4
n
4. Express the following in simplest form, without brackets:
a) (2b4 )3
3 4
m
e)
2n2
b) (5a4 b)2
4 2
4a
f)
b2
c) (−6b2 )2
3
−2a2
g)
b2
d) (−2a)3
2
−3p2
h)
q3
5. Rational indices
Look at this example:
1
1
1
1
Since 3 2 · 3 2 = 3 2 + 2 = 31 = 3, and
√
1
3 2 = 3.
√ √
3 · 3 = 3 also, then, by direct comparison:
√ √ √
1
1
1
Likewise, 2 3 · 2 3 · 2 3 = 21 = 2, compared with 3 2 · 3 2 · 3 2 = 2, suggests:
√
1
2 3 = 3 2.
√
√
1
In general:
an = n a
( n a reads “the nth root of a”)
m
1
Also: a n = (am ) n =
√
n
am
D pto. M atemáticas. IES Jovellanos. 2012
Index law s - 3 o E SO
5
Example 2
Write as a single power of 2:
√
1
a) 3 2
b) √
2
a)
√
3
c)
√
5
4
1
1
1
b) √ = 1 = 2− 2
2
22
1
2 = 23
c)
√
√
2
5
5
4 = 22 = 2 5
Exercises - Set F
1. Write as a single power of 2:
√
√
1
a) 5 2
b) √
c) 2 2
5
2
√
√
4
f) 2 · 3 2
g) √
h) ( 2)3
2
2. Write as a single power of 3:
√
√
1
a) 3 3
b) √
c) 4 3
3
3
√
d) 4 2
1
i) √
3
16
√
d) 3 3
1
e) √
3
2
1
j) √
8
1
e) √
9 3
3. Write the following in the form ax where a is a prime number and x is rational:
√
√
√
√
√
a) 3 7
b) 4 27
c) 5 16
d) 3 32
e) 7 49
1
f) √
3
7
1
g) √
4
27
1
h) √
5
16
1
i) √
3
32
1
j) √
7
49
✁✃✁✃✁✃✁✃✁✃✁✃✁✃
D pto. M atemáticas. IES Jovellanos. 2012
Standard form (Scientific
notation)
10 000
1 000
100
10
1
Observe the pattern.
As we divide by 10, the exponent (power) of 10 decreases by one.
1
10
1
100
1
1000
=
=
=
=
=
=
104
103
102
101
100
10−1
= 10−2
= 10−3
etc.
We can use this pattern to simplify the writing of very large and very small numbers.
For example, 5 000 000 = 5 · 1 000 000 = 5 · 106 , and
1
3
=3·
= 3 · 10−6
0, 000 003 =
1 000 000
1 000 000
Standard form (or scientific notation) involves writing any given number as a number between
1 and 10, multiplied by an integer power of 10,
i.e., a · 10n
where a lies between 1 and 10.
If the original number is > 10, the power of 10 is positive.
If the original number is < 1, the power of 10 is negative.
If the original number is between 1 and 10, leave it as it is.
More examples:
37 600 = 3, 76 · 104
3, 2 · 102 = 320
0, 00086 = 8, 6 · 10−4
5, 76 · 10−5 = 0, 0000576
Note about the use of calculators:
If you perform on your calculator 2 300 000·400 000 your calculator will display 9.211 or 9.2E11 ,
depending on the model, which actually represents 9, 2 · 1011 .
If you get something like 2.4−10 or 2.4E − 10 , this actually represents 2, 4 · 10−10 .
Numbers which are already represented in standard form can be entered into the calculator using the
EXP key.
1
Scientific notation - 3 o E SO
2
Exercises - Set A
1. Express the following in scientific notation:
a) 259
b) 259 000
c) 25, 9
d) 0, 259
e) 0, 000 259
f) 40, 7
g) 4070
h) 0, 0407
i) 407 000 000
j) 0, 000 040 7
2. Express the following in scientific notation:
a) The distance from the Earth to the Sun is 149 500 000 000 m.
b) Bacteria are single cell organisms, some of which have a diameter of 0, 0003 mm.
c) A speck of dust is smaller than 0, 001 mm.
d) The probability that your six numbers will be selected for Lotto on Friday night is
0, 000 000 141 62.
e) The central temperature of the Sun is 15 million degrees Celsius.
f) A single red blood cell lives for about four months and during this time it will circulate
around the body 300 000 times.
3. Write as an ordinary decimal number:
a) 4 · 103
b) 2, 1 · 10−3
c) 4, 33 · 107
d) 8, 6 · 10−1
e) 3, 8 · 105
f) 2, 9 · 10−5
g) 5, 86 · 109
h) 5, 86 · 10−9
i) 5 · 10−2
j) 6, 3 · 108
4. Write as an ordinary decimal number:
a) The wave length of light is 9 · 10−7 m.
b) The world population in the year 2000 was 6, 130 · 109 .
c) The diameter of our galaxy, the Milky Way, is 1 · 105 light years.
d) The smallest visuses are 1 · 10−5 mm in size.
e) The mass of a bee’s wing is 10−7 kg.
5. Use your calculator to find correct to 2 decimal places, and giving your answer in scientific
notation:
a) 0, 06 · 0, 002 : 4000
d) 320 · 600 · 51 400
b) 426 · 760 · 42 000
e) 0, 004 28 : 120 000
c) 627 000 · 74 000
f) 0, 026 · 0, 0042 · 0, 08
6. If a missile travels at 5400 km/h how far will it travel in:
a) 1 day
b) 1 week
c) 2 years?
(Give your answers in standard form with decimal part correct to 2 places and assume that 1
year=365,25 days.)
7. Light travels at a speed of 3 · 108 metres per second. How far will light travel in:
a) 1 minute
b) 1 day
c) 1 year?
(Give your answers in standard form with decimal part correct to 2 places and assume that 1
year=365,25 days.)
8. Find, with decimal part correct to 2 places:
a) (3, 42 · 105 ) · (4, 8 · 104 )
b) (6, 42 · 10−2 )2
d) (9, 8 · 10−4 ) : (7, 2 · 10−6 )
e)
D pto. M atemáticas. IES Jovellanos. 2012
1
3, 8 · 105
c)
3, 16 · 10−10
6 · 107
f) (1, 2 · 103 )3
Scientific notation - 3 o E SO
3
Example 1
Simplify the following, giving answers in standard form:
a) (5 · 104 ) · (4 · 105 )
b) (8 · 105 ) : (2 · 103 )
a) (5 · 104 ) · (4 · 105 ) = (5 · 4) · (104 · 105 ) = 20 · 109 = 2 · 1010
b) (8 · 105 ) : (2 · 103 ) =
8 105
·
= 4 · 102
2 103
Exercises - Set B
1. Simplify the following, giving your answer in standard form:
a) (3 · 103 ) · (2 · 102 )
b) (5 · 102 ) · (7 · 105 )
c) (5 · 104 ) · (6 · 103 )
e) (5 · 104 )2
d) (3 · 103 )2
f) (8 · 10−2 )2
g) (8 · 104 ) : (4 · 102 )
h) (8 · 105 ) : (2 · 103 )
✁✃✁✃✁✃✁✃✁✃✁✃✁✃
D pto. M atemáticas. IES Jovellanos. 2012
Polynomials
Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona
Algebraic notation and Polynomials
Algebraic Notation
The ability to convert worded sentences and problems into algebraic symbols and to
understand algebraic notation is essential in the problem solving process.
Notice that:
•
•
•
2 + 3
2 + 3 = 8
2 + 3 > 28
Is an algebraic expression, whereas
Is an equation, and
Is an inequality or inequation.
When we simplify repeated sums, we use product notation:
For example:
+
= 2 ‘lots’ of =2 × =2
and
++
= 3 ‘lots’ of =3 × =3 When we simplify repeated products, we use index notation:
For example:
× = and
× × = EXAMPLE 1
Write, in words, the meaning of:
a) − 5
b) + c) 3 + 7
a) Is “5 less than x”
b) Is “the sum of a and b” or “b more than a”
c) Is “7 more than three times the square of x”
EXAMPLE 2
Write the following as algebraic expressions:
a) The sum of p and the b) The square of the sum of p
square of q
a) + b) ( + )
c) 2 − and q
c) b less than double a
1
Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona
Algebraic notation and Polynomials
2
EXAMPLE 3
Write, in sentence form, the meaning of:
a) = b)
a) D is equal to the product of c and t
=
b) A is equal to a half of the sum of b and c,
or, A is the average of b and c.
EXAMPLE 4
Write “S is the sum of a and the product of g and t” as an equation.
The product of g and t is The sum of a and is + So, the equation is
= + TO PRACTICE
EXERCISE 1
Write in words, the meaning of:
a. 2
b. e. − 3
f.
b+c
g. 2 + h. (2)
j.
− k. + l.
i.
2
c. √
d. (a + b)
EXERCISE 2
Write the following as algebraic expressions:
a. The sum of " and #
i.
The difference between p and q if p>q
b. The sum of p, q and r
j.
a less than the square of b
c. The product of a and b
k. Half the sum of a and b
d. The sum of r and the square of s
l.
e. The square of the sum of r and s
m. A quarter of the sum of a and b
f.
n. The square root of the sum of m and n
The sum of the squares of r and s
g. The sum of twice a and b
h. The sum of x and its reciprocal
The sum of “a” and a quarter of “b”
o. The square root of the sum of the
squares of x and y
Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona
Algebraic notation and Polynomials
3
EXERCISE 3
Write in sentence form:
a. $ = + b. % =
d. ) = e. * = f.
+ = h. = √ + i.
=
-
g. % = , .
c. ' = 3(
&
&
EXERCISE 4
Write the following as algebraic equations:
a. S is the sum of p and r
b. D is the difference between a and b where b>a
The difference
between two
numbers is the
larger one minus
the smaller one
c. A is the average of k and m
d. M is the sum of a and its reciprocal
e. K is the sum of t and the square of s
f.
N is the product of g and h
g. Y is the sum of x and the square of d and e
Algebraic Substitution
To evaluate an algebraic expression, we substitute numerical values for the unknown, then
calculate the result.
Consider the number
crunching machine
alongside:
Input x
5x - 7
calculator
Output
If we place any number / into the machine, it calculates 0/ − 1. So, / is multiplied by 5, and
then 7 is subtracted:
For example: if = 2
5 − 7
=5×2−7
= 10−7
=3
and
if = −2 ;
5 − 7
= 5 × (−2) − 7
= −10 − 7
=−17
Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona
Algebraic notation and Polynomials
4
Notice that when we substitute a negative number such as −2, we place it in brackets. This
helps us to get the sign of each term correct.
TO PRACTICE
EXERCISE 5
If 4 = 3, = 1 and 5 = −2, evaluate:
7
b.
78
c.
7:98
f.
978
g.
a.
8
e.
(7:8)
9
7:9
9:7
8
9:87
78:9
d.
;8:9
h.
7
7:9
9
−4
EXERCISE 6
If = −3, = −4 and = −1, evaluate:
a. b. e. + f.
( + )
c. + d. ( + )
g. 2
h. (2)
EXERCISE 7
If = 4, = −1 and = = 2, evaluate:
a. > + b. > + e. >= − f.
> + c. >= − d. > − g. > + = + 2
h. >2 − 5=
Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona
Algebraic notation and Polynomials
DEFINITIONS: product, factors, sum, terms
A product is an expression where the last operation is multiplication.
In a product, the things being multiplied are called the factors.
A sum is an expression where the last operation is addition.
In a sum, the things being added are called the terms.
As an example, consider
the expression
then here is the order that computations would be done:
1. Add b and c .
2. Multiply this sum by a .
( + )
Notice that the last operation done is multiplication.
Thus, the expression ( + ) is a product.
The factors are and + .
As a second example,
consider the expression
here is the order that computations would be done:
1. Multiply a and b .
2. Add this result to c .
+ .
Notice that the last operation done is addition.
Thus, + is a sum.
The terms are and .
EXAMPLES
The expression 35 is a product.
The factors are @, /, A
The expression −4( + 2) is a product.
The factors are −B, /, / + C
The expression 5 − 5 + 1 is a sum.
The terms are0/, −A, D
The expression 2 + 253 − 7 is a sum.
The terms are /2, CA3, −1
EXERCISE 8
In the following expressions, how many terms are there? And each term has how many
factors?
a) 2 + 4 + 5( + )
b) E + 5FG + 2
5
Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona
Algebraic notation and Polynomials
6
Algebraic expressions
An algebraic expression is an expression that contains one or more numbers, one or more
variables, and one or more arithmetic operations.
It doesn't include an equal sign.
Algebraic expressions can be or many forms, for example:
3x 2 − 2 x − 1
3x 2 + 2 x +
1
x
− 6x 2 y − 2
3x 2 + 2 x
x2 − 3
A term consist of products of numbers and letters, so 3x2, -2x, -1, -6x2y, etc. are terms
The number multiplying the letters is the coefficient of the term.
3x2 (x2 term. Coefficient is 3)
-2x (x term. Coefficient is -2)
-1 (constant term is -1)
Polynomials
Polynomials are algebraic expresions. A polynomial in is a sum of terms, each of the form
k
,
where:
is a real number,
H is a nonnegative integer. That is, H = {0, 1,2, 3, . . . }.
DEFINITION: standard form of polynomials; degree;
leading coefficient
The standard form of a polynomial is:
"n/n + "n-1/n-1+. . . +"1/ + "0
Here, n denotes the highest power to which is raised; this highest power is
called the degree of the polynomial.
Thus, in standard form, the highest power term is listed first, and subsequent
powers are listed in decreasing order.
Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona
Algebraic notation and Polynomials
Notice that in the notation i i (read as " "KLMNONPQK/OROSQN "), the
number i denotes the coefficient of the i term.
The number n , which is the coefficient of the highest power term, is called
the leading coefficient of the polynomial.
Note that a constant (like 5 ) can be written as 5x0 .
This is why the power is allowed to equal zero in the definition of polynomial—to allow for
constant terms.
EXAMPLE 5
The expression 5x4-x3-3x2+7x-5
coefficient and degree.
is a polynomial. Find its terms, coefficients, leading
The terms are: 5x4, -x3, -3x2, 7x, and -5 .
Comparing each term with the required form ax k , we have:
term
writing in the
form ax k
a
k
5x4
5x4
a=5
k=4
-x3
(-1)x3
a=-1
k=3
-3x2
-3x2
a=-3
k=2
7x
7x
a=7
k=1
-5
-5x0
a=-5
k=0
Notice that every value of a is a real number, and every value of k is a
nonnegative integer.
The standard form of this polynomial is: 54−3−3 2+7 − 5
Here, the highest power term is written first, and subsequent terms decrease in
power.
The degree is 4 , since this is the highest power.
The leading coefficient is 5 , since this is the coefficient of the highest power term.
Notice that the leading coefficient actually leads (comes at the beginning of) the
polynomial, WHEN the polynomial is written in standard form.
EXAMPLE 6
The following expressions are NOT polynomials. Why?
a)
1x + x- 1
b) x - x1/2
c) 7x2 - 7x + 7x1/ 2
7
Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona
Algebraic notation and Polynomials
8
a) 1x + x- 1 is not a polynomial; no negative powers are allowed.
b) x - x1/2 is not a polynomial; the number 1/2 is not an allowable power.
c) 7x2 - 7x + 7x1/ 2 is not a polynomial; the number 1/2 is not an allowable power.
DEFINITION: monomial, binomial, trinomial
A polynomial with exactly one term is called a monomial.
A polynomial with exactly two terms is called a binomial.
A polynomial with exactly three terms is called a trinomial.
DEFINITION: quadratic, cubic, quartic
A polynomial of degree
A polynomial of degree
A polynomial of degree
A polynomial of degree
•
•
•
•
1
2
3
4
is called linear
is called quadratic.
is called a cubic.
is called a quartic.
Polynomials have beautiful smooth graphs—no breaks and no kinks.
The higher the degree of a polynomial, the more it is allowed to "turn" (change
direction).
Indeed, it can be shown easily (using calculus) that a polynomial of degree n can have
at most n-1 turning points.
The graph below is the polynomial T() = 3 − .
Notice that this polynomial has degree 3 and has 2 turning points.
Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona
Algebraic notation and Polynomials
EXERCISE 9
1) What is the degree of these polynomials?
a) 2x16+80x8
b) 7769x-97x7-56x9+31x19
c) 21x4-12x11+6x2-1710x10+4171x14
d) 60x3+65x5-3425x15
e) 8+27x11+234x18
2) What name is given to a polynomial with exactly one term?
3) What is the leading coefficient of these polynomials?
a)
b)
c)
d)
85-54x20
100x19-57x20+10029x6+2x4
61x4-6x3-7726x12+1425x11+45x19
9643x3-45x16-97x19
4) A polynomial is a sum of terms, each of a particular form. What is this form?
5) Is −28 an allowable term in a polynomial?
6) Is 2954 an allowable term in a polynomial?
7) Is −7x70 an allowable term in a polynomial?
8) Is the term x−9 an allowable term in a polynomial?
9) What is a quartic function?
10) What name is given to a polynomial with exactly two terms?
11) What is a trinomial?
12) What name is given to a polynomial with exactly one term?
13) Suppose that a polynomial has degree 8 . What (if anything) can be said about the
number of turning points for this polynomial?
14) Suppose that a polynomial has 5 turning points. What (if anything) can be said
about the degree of this polynomial?
15) Write the following polynomial in standard form:
a)
b)
c)
d)
e)
-3x14+8x27+6x5+4x6
4x2-7x4+5x6
-6x26+3x9-7x6
5x-2x25-2x2
2x3+8x6+8x7+4x-3x2
16) Can the graph of a polynomial have a break in it?
17) Can the graph of a polynomial have a kink in it?
9
Departamento de Matemáticas. Real Instituto de Jovellanos. J. F. Antona
Adding and subtracting polynomials
1
Adding and subtracting polynomials
Only like terms (those with identical letters and powers) can be added or subtracted.
Collecting like terms
In algebra,
Like terms are terms which contain the same variables (or letters) to the same indices.
For example:
• and −2 are like terms
• 2 and 3 are unlike terms because the power of x are not the same
Algebraic expressions can often be simplified by adding or subtracting like terms. We call this
collecting like terms.
Consider
2 + 3 = + + + + = 5
2
lots
of
a
3 lots
of a
5
lots
of
a
EXAMPLE 1
Where possible, simplify by collecting like terms:
a) 4 + 3
b) 5 − 2
c) 2 − 1 + d) − 2
e) 2−4
a) 4 + 3 = 7
b) 5 − 2 = 3
c) 2 − 1 + = 3 − 1
(since 2 and are like terms)
d) − 2 = −
(since and −2 are like terms)
2
e) −4
cannot be simplified since 2 and −4 are unlike terms.
EXAMPLE 2
Simplify:
a) 3 2 − 7 2
b) 3 2−7
c) 3 3−7 3−2 + 26−3 6−4 2
a) 3 2 − 72 = −42
b) 3 2−7 cannot be expressed as a single term.
c) 3 3−7 3−2 + 2 6−36−42= −4 3−2 − 6−4 2
Departamento de Matemáticas. Real Instituto de Jovellanos. J. F. Antona
Adding and subtracting polynomials
2
EXERCISE 1
Simplify, where possible, by collecting like terms:
a)
5++4
b)
e)
+−3
i) + 2
m)
2 + 3 − 5
c)
−2+5
d)
+1+
f) 5 + g)
5 − h)
− 5
j) + + k)
5 + 5
l) − 5 + 5
o)
4 + p)
n)
6+3+
2 + 3 − 3 − EXERCISE 2
Simplify, where possible:
a)
7 − 7
b)
7 − c)
7 − 7
d)
+ 2
e)
− 2
f)
4 − g)
+ 3 + 2 + 4
h)
2 + + 3 − 4
i)
2 − + 3 + 3
j)
3 + 2 − − k)
+ 4 − 3 + 2
l)
+ 2 − − 5
n)
+ + + 4
o)
2 − 3 − − 7
m) + 5 + 2 − 3
EXERCISE 3
Simplify, where possible:
a)
4 + 6 − − 2
b)
2 + − 2
c)
3 − 2 + d)
+ 2 + 2 − 5
e)
− 6 + 2 − 1
f)
3 + 7 − 2 − 10
g)
−3 + 2 − − h)
+ 2 − i)
2 − − + 3
j)
4 − − k)
+ + l)
4 − 2 − − Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona
Multiplication of polynomials
1
Multiplication and expansion of polynomials
A. PRODUCT NOTATION
In algebra we agree:
To leave out the “×” signs between any multiplied quantities provided that at least
one of them is an unknown (letter)
• To write numerals (numbers) first in any product
• Where products contain two or more letters, we write them in alphabetical order
For example:
• 3 is used rather than 3 × or 3
• 2 is used rather than 2
•
ALGEBRAIC PRODUCTS
The product of two or more factors is the result obtained by multiplying them together
Consider the factors −3
22 . Their product −3 × 22 can be simplified by
following the steps below:
•
•
•
Step 1: Find the product of the signs.
Step 2: Find the product of the numerals or numbers.
Step 3: Find the product of the variables or letters.
−2 × 3 = −6 So.
× = − × + = −
3 × 2 = 6
EXAMPLE 1
Simplify the following products:
a) −3 × 4
b) 2 × − c) −4 × −2 a) −3 × 4
b) 2 × − c) −4 × −2 = −12
= −2 = 8 For−2, the
sign is−, the
numeral is 2,
and the variable
is Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona
Multiplication of polynomials
2
EXERCISE 1
Write the following algebraic products in simplest form:
a) × b) × 2 × c) × d) × 2
e) 2 × 3
f) 4 × 5
g) −2 × 7
h) 3 × −2
i) 2 × j) 3 × 2
k) −2 × l) −3 × 4
m) −2 × (−)
n) −3 × o) − × (−2)
p) 3
× (−2
)
q) (−)
r) (−2)
s) 2 × t) × (−3)
EXERCISE 2
Simplify the following:
a)
2 × 5 + 3 × 4
b)
5 × 3 − 2 × c)
3 × + 2 × 4
d)
× 2 + × 3
e)
4 × − 3 × f)
3 × − 2 × 2
g)
3 × + 2 × 2
h)
4 × − 3 × 2
i)
3 × − 2 × •
Multiplication is often written using the multiplication sign or cross symbol "×" between the
factors:
×
•
In algebra multiplication is sometimes denoted by a middle dot
∙
•
In algebra, multiplication involving variables is often written as a juxtaposition
•
This notation can also be used for quantities that are surrounded by parentheses
()()
EXERCISE 3
Simplify the following:
(−5 ) ∙ (4 ) ∙ ( )
a)
(2) ∙ (3)
b)
d)
(12) ∙ (6 ) ∙ ( 23)
e) (−3 3) ∙ (−26 5)
c)
(3 ) ∙ (−5 )
f)
(−4 736) ∙ (332 5)
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Multiplication of polynomials
3
B. THE DISTRIBUTIVE LAW
Consider the expression 2( + 3). We say that 2 is the coefficient of the expression in the
brackets. We can expand the brackets using the distributive law:
( + ) = + The distributive law says that we must multiply the coefficient by each term within the
brackets, and add the results.
EXAMPLE 2
Expand the following:
a) 3(4 + 1)
a)
b) 2(5 − 2)
c) −2( − 3)
b)
c)
3(4 + 1)
2(5 − 2)
= 3 × 4 + 3 × 1
= 2 × 5 + 2 × (−2)
= 12 + 3
= 10 − 4
2
−2( − 3)
= (−2) × + (−2) × (−3)
2
= −2 +6
With practice, we do not need to write all of these steps.
EXAMPLE 3
Expand and simplify:
a) 2(3 − 1) + 3(5 − )
a)
b) (2 − 1) − 2(5 − )
b)
2(3 − 1) + 3(5 − )
(2 − 1) − 2(5 − )
= 6 − 2 + 15 − 3
= 2 − − 10 + 2 = 3 + 13
= 4 − 11
Notice in b) that
the minus sign in
front of 2 affects
both terms inside
the following
brackect
EXERCISE 4
Expand and simplify:
a) 3( + 1)
b) 2(5 − )
c) – ( + 2)
d) ( + 3)
e) −3( + 2)
f) −2( − )
g) (2 − 1)
h) 2( − − 2)
i) 1 + 2( + 2)
j) 13 − (4( + 3)
k) 3( − 6) + 2
l) ( − 1) + m) 2(3 − ) + n) 4 − 5(2 − 3)
o) 7 − 4( + 2)
p) 3( − ) + 5
Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona
Multiplication of polynomials
4
EXERCISE 5
Expand and simplify:
a) 3( − 5) + 2(5 + )
b) 2 + (5 − 7)
c) 2 − (3 − 6)
d) 3( + 2) + 5(4 − )
e) 6( − 2) − 4(3 + 5)
f) 4 − 3(2 − 3)
g) 2 − (3 − 2)
h) 2 (4 − 3) − 3(5 + 3) i) −3 ( − 4)
j) (5 − )
k) 3 (2 − 4 )
C. THE PRODUCT
(! + ")(# + $)
l) −4 (2 − 5) − 2(3 − )
( + )( + ) = + + + To remember
FOIL rule
= is the product of First terms of each bracket
= is the product of Outer terms of each bracket
= is the product of Inner terms of each bracket
= is the product of Last terms of each bracket
EXAMPLE 4
Expand and simplify:
a) ( + 2)( + 3)
a)
b) (2 + 3)(3 − 5)
b)
( + 2)( + 3)
(2 + 3)(3 − 5)
= + 3 + 2 + 6
= 6 − 10 + 9 − 15
= + 5 + 6
= 6 − − 15
Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona
Multiplication of polynomials
5
EXERCISE 6
Expand and simplify:
a) ( − 6)( + 7)
b) ( + 3)( − 2)
c) ( + 5)( − 5)
d) (2 + 2)(4 − 6)
e) ( − 2)(1 − 3)
f) (6 − 1)(6 − 1)
g) (4 − 6)(3 − 2)
h) (4 − 3)(5 + 3)
i) ( − 3)( − 4)
j) (2 − 5)(3 + 4)
k) (3 + 4 )(2 − 5)
l) (5 − 4)(7 + 5)
EXAMPLE 5
Expand and simplify:
a) ( + 5)( − 5)
a)
( + 5)( − 5)
What do you
notice about
the two
middle terms?
b) (4 − 3)(4 + 3)
b)
(4 − 3)(4 + 3)
= − 5 + 5 − 25
= 16 + 12 − 12 − 9
= − 25
= 16 − 9
EXERCISE 7
Expand and simplify:
a) ( − 6)( + 6)
b) ( + 3)( − 3)
c) (2 + 5)(2 − 5)
d) (4 + 6)(4 − 6)
e) (3 − 2)(2 + 3)
f) (6 + 1)(6 − 1)
EXAMPLE 6
Expand and simplify:
a) (2 + 5)
a)
(2 + 5)
b) (4 − 3)
What do you
notice about
the two
middle terms?
b) (4 − 3)
= (2 + 5)(2 + 5)
= (4 − 3)(4 − 3)
= 4 + 10 + 10 + 25
= 16 − 12 − 12 + 9
= + 20 + 25
= 16 − 24 + 9
EXERCISE 8
Expand and simplify doing the product:
a) ( + 6)
b) ( − 3)
c) (2 − 5)
d) (4 − 3)
e) (3 − 2)
f) (6 + )
Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona
D.
Multiplication of polynomials
6
DIFFERENCE OF TWO SQUARES
If and are perfect squares then − is called the difference of two squares
Notice that
( + )( − ) = − + − = − The middle two terms add to zero
Thus,
( + )( − ) = − EXAMPLE 7
Expand and simplify:
a) (3 + 5)(3 − 5)
a)
b) (4 − 3)(4 + 3)
(3 + 5)(3 − 5)
b)
(4 − 3)(4 + 3)
= (3) − (5) = (4) − (3)
= 9 − 25
= 16 − 9 EXERCISE 9
Expand and simplify using the rule(! + ")(! − ") = !& − "& :
a) ( − 6)( + 6)
b) ( + 3)( − 3)
c) (2 + 5)(2 − 5)
d) (4 + 6)(4 − 6)
e) (3 − 2)(2 + 3)
f) (6 + 1)(6 − 1)
g) (2 − 1)(2 + 1)
h) (2 + 3)(2 − 3)
i) (2 + 5)(2 − 5)
j) (4 + 6)(4 − 6)
k) (3 − 2)(2 + 3)
l) (6 + )(6 − )
m) (4 + 5)(4 − 5)
n) (3 + )(3 − )
o) (2 + 4)(2 − 4)
E. PERFECT SQUARES EXPANSION
( + ) and ( − ) are called perfect squares.
Notice that
( + ) = ( + )( + )
= + + + using FOIL
= + 2 + Notice that
the middle
two terms
are identical
Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona
Multiplication of polynomials
7
Thus, we can state the perfect square expansion rule:
( + ) = + 2 + We can remember the rule as follows:
Step 1: Square the first term.
Step 2: Add twice the products of the first and last terms.
Step 3: Add on the square of the last term.
( − ) = ( − )( − )
Notice that
= − − + using FOIL
= − 2 + So, we can state:
( − ) = − 2 + EXAMPLE 8
Expand and simplify:
a) (5 + 3)
a)
b) (6 − 4)
(5) + 2 ∙ 5 ∙ 3 + 3
= 25 + 30 + 9
b) 6 − 2 ∙ 6 ∙ 4 + (4)
= 36 − 48 + 16 EXERCISE 10
Expand and simplify using the perfect square expansion rule:
a) ( + 6)
b) ( − 3)
c) (2 − 5)
d) (4 − 3)
e) (3 − 2)
f) (6 + )
g) (3 + 4)
h) (2 − )
i) (7 + 2)
j) (12 − 3)
k) (3 − 4)
l) (25 − 2)
EXAMPLE 9
Expand and simplify:
a) (3 + 5)
a) (3 + 5)
b) 6 − (4 + 3)
b) 6 − (4 + 3)
= (3 ) + 2 ∙ 3 ∙ 5 + 5
= 6 − (4 + 2 ∙ 4 ∙ 3 + (3))
= 9 + 30 + 25
= 6 − (16 + 24 + 9 )
= 6 − 16 − 24 − 9 = −10 − 24 − 9 Notice the use of
square brackets in
the second line.
These remind us to
change the signs
inside them when
they are removed
Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona
Multiplication of polynomials
8
EXERCISE 11
Expand and simplify:
a) ( + 4)
b) (2 − )
c) (7 + 2 )
d) (5 − 3 )
e) (2 − 3)
f) (2 − 5 )
g) ( − 2)
h) (2 − 4 )
i) (2 − 6)
EXERCISE 12
Expand and simplify:
a) 3 + 1 − ( + 4)
b) 5 − 3 + ( − 2)
c) ( + 1)( − 1) − ( + 4)
d) ( + 3)( − 3) − ( − 2)
e) (2 − 3) + ( + 5)( − 5)
f) (3 + 2) − ( + 4)( − 4)
g) (3 + 2)(3 − 2) − (5 − )
h) (2 + 9)(2 − 9) − (4 − 6)
i) ( − 7) + ( + 8)
j) (4 + 2) − ( − 9)
F. FURTHER EXPANSION
In this section we will expand more complicated expressions by repeated use of the expansion
laws.
( + )( + + ')
Consider the expansion of
( + )( + + ') = + + ' + + + '
Notice that there are 6 terms in this expansion and that each term within the first bracket
is multiplied by each term in the second
2 terms in the first bracket × 3 terms in the second bracket = 6 terms in the expansion
Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona
Multiplication of polynomials
9
EXAMPLE 9
Expand and simplify:
(2 + 4)( + 3 + 5)
(2 + 4)( + 3 + 5) = 2 + 6 + 10 + 4 + 12 + 20
= 2 + 10 + 22 + 20
{Collecting like terms}
EXAMPLE 10
Expand and simplify:
( + 2)
( + 2) = ( + 2)( + 2)
= ( + 2)( + 4 + 4)
= ( + 2)( + 4 + 4)
{each term within the first bracket is multiplied by
each term in the second bracket}
= + 4 + 4 + 2 + 8 + 8
= + 6 + 12 + 8
{Collecting like terms}
EXAMPLE 11
Expand and simplify:
a)
( + 1)(2 − 3 + 4)
a)
( + 1)(2 − 3 + 4)
b) ( + 2)( − 3)( + 3)
b) ( + 2)( − 3)( + 3)
= ( + 2)( − 3)( + 3)
= ( + 2)( − 9){difference of two squares}
= ( + )(2 − 3 + 4)
= − 9 + 2 − 18
= 2 − 3 + 4 + 2 − 3 + 4
= 2 − + + 4
{Collecting like terms}
Always look for
ways to make
your expansions
simpler. In b) we
can use the
difference of two
squares
= + 2 − 9 − 18
Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona
Multiplication of polynomials
10
EXERCISE 13
Expand and simplify:
a) ( + 3)( + + 5)
b) (2 − 4)(3 − 5 + 2)
c) (3 − 1)(5 + 4 − 6)
d) (4 + 3)(6 − 2 − 8)
e) ( − 2)
f) (2 + 3)
g) (2 + 1)
h) (3 − 2)
i) ( + 3)( + 3)
j) ( − 3)( + 1)
k) 2( − 4)( + 4)
l) ( + 2)( − 2)(2 + 3)
m) −3(2 − 1)( + 2)
n) (1 − 3)(2 + 5)
o) 2 ( − 1)
p) (3 − 2)( − 2)( + 1)
q) ( − 3)(1 − )(2 − 2)
r) ( − 1)( − 2)( + 2)
HARDER EXTENSION
EXERCISE 14
Expand and simplify:
a) ( + 3 )(4 ) + 2 + 5 )
b) (3 − 2 )( − 3 − 5 − 1)
c) (2 + + 3 − 2)( − 3 + 1)
d) ( − 7 + − 1)( + 4 − 6 + 2)
G.
THE BINOMIAL EXPANSION
Consider
( + )* .
•
•
•
We note that:
+ is called a binomial as it contains two terms
Any expression of the form ( + )* is called a power of a binomial
The binomial expansion of ( + )* is obtained by writing the expression without
brackets.
Now (! + ")+ = (! + ")(! + ")&
= (! + "),!& + &!" + "& = !+ + &!& " + !"& + "!& + &!"& + "+ = !+ + +!& " + +!"& + "+ Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona
Multiplication of polynomials
11
So, the binomial expansion of
(! + ")+ = !+ + +!& " + +!"& + "+
Doing the same process with
(! − ")+ = (! − ")(! − ")&
= (! − "),!& − &!" + "& -
= !+ − &!& " + !"& − "!& + &!"& − "+ = !+ − +!& " + +!"& − "+ So, the binomial expansion of
(! − ")+ = !+ − +!& " + +!"& − "+
EXAMPLE 12
Expand and simplify using the binomial expansion:
a)
( + 2)
b) (2 − 4)
a) {Using ( − ) = − 3 + 3 − }
b)
We substitute = and = 2
{Using ( − ) = − 3 + 3 − }
We substitute = (2) and = 4
∴ ( + 2) = + 3 2 + 32 + 2
∴ (2 − 4) = (2) − 3(2) 4 + 3(2)4 − 4
= + 6 + 12 + 8
= 8 − 48 + 96 − 64
We use brackets
to assist our
substitution
b)
{Using ( + ) = + 3 + 3 + }
We substitute = (2) and = (−4)
∴ (2 − 4) = (2) + 3(2) (−4) + 3(2)(−4) + (−4)
= 8 − 48 + 96 − 64
EXERCISE 15
Use the binomial expansion for (! + ")+ or (! − ")+ to expand and simplify:
a) ( + 1)
b) (3 − 2)
c) ( − 5)
d) (2 + 3)
e) (4 − 6)
f) (2 + 1)
g) ( − 2)
h) ( − 4 )
i) (3 + )
Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona
Multiplication of polynomials
EXERCISE 15 (Extension)
Find, in terms of x, the surface area of the following figures:
a)
c)
b)
d)
EXERCISE 16 (Extension)
Find, in terms of x, the surface area of the orange zone of the following figures:
a)
EXERCISE 17 (Extension)
Find, in terms of x, the volume of the following figure:
a)
b)
12
Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona
Multiplication of polynomials
EXERCISE 18 (Extension)
a) Find, in terms of/, the surface area and the volume, of the following
cylindrical water tank trailer, with hemispherical endings.
b) Find the surface and volume if / = + m.
10 m
EXERCISE 19 (Review)
a) Expand and simplify:
1.
4 ∙ 8
2.
6 ∙ 2 3.
2 ∙ 5 ∙ (−3 ) ∙ 4.
(−4)(−7)
5.
5 2 ∙ (−32 3)
6.
(5 − 6)
7.
4 − 2(2 + 3)
8.
2( − 6) + 3(2 − )
9.
5(2 − 3) − 8(4 − 7)
10.
3 (4 ) − 7 )
11.
−6 (3 − 4) − 3(5 − 2)
12.
(3 + 2)( − 4)
13.
( − 5)
14.
(3 + 4)
15.
(2 − 6)
16.
(2 − 6)(2 + 6)
17.
( − 5)( + 5)
18.
( − 3)( + 3)
13
Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona
19.
– 5( + 2)( − 2)
20.
−( + 3)
21.
(2 − 5)( − 1)( + 1)
22.
( − 5)( − 1)( − 1)
23.
( − 1)
24.
5 + 2 − ( + 2)
25.
(3 − 2)( − 2 + 7)
26.
( − 1)( + 2)( − 3)
27.
( + 2)
28.
( + 1)( + 1)( − 1)
29.
( + 5)( − 5)(2 + 3)
30.
(2 − )(3 − + 5 − 2)
31.
(3 − 4)
Multiplication of polynomials
b) Find, in terms of x, the surface area of the orange zone of the following figures:
1.
2.
14
Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona
Multiplication of polynomials
KEY POINTS
Useful products:
(In Spanish: “productos notables”)
( + ) = + ( + )( + ) = + + + ( + )( − ) = − ( + ) = + 2 + ( − ) = − 2 + ( + ) = + 3 + 3 + ( − ) = − 3 + 3 − Expansion and factorisation relation:
Expansion
( + ) = + 2 + Factorisation
Note: !& + "& cannot be factorised.
!& − "& is the difference of squares
MORE EXERCISES TO PRACTICE AT HOME:
EXERCISE 20
1)
2)
15
Departamento de Matematicas. Real Instituto de Jovellanos. J. F. Antona
3)
4)
5)
6)
7)
8)
9)
Multiplication of polynomials
16
Equations
Linear equations
Remember:
• x2 + 3x
• x2 + 3x = 8
• x2 + 3x > 28
is an algebraic expression, whereas
is an equation, and
is an inequality or inequation.
Many mathematical situations can be described using an equation. This is a formal statement that
one expression is equal to another. The expressions are connected by the equals sign =.
An algebraic equation is an equation which involves at least one variable.
For example, 3x + 5 = 8 is an algebraic equation.
We read the symbol = as “equals” or “is equal to”.
By following a formal procedure, we can solve the equation to find the value of the variable which
makes it true. In this case we see that x must be 1, since 3 + 5 = 8.
1. Linear equations
A linear equation contains at least a variable which can only be raised to the power 1.
For example, 3x + 5 = 8,
whereas
1
2x
x2 − 3x + 2 = 0,
−3=2
and
1 √
= x, and
x
1+x
=x
3
x3 = 8
are linear equations
are not linear equations.
The left hand side (LHS) of an equation is on the left of the = sign.
The right hand side (RHS) of an equation is on the right of the = sign.
For example,
3x + 5 = |{z}
8
| {z }
RHS
LHS
The solution of a linear equation is the value of the variable which makes the equation true. In
other words, it makes the left hand side (LHS) equal to the right hand side (RHS).
In the example
true is x = 1.
3x + 5 = 8
above, the only value of the variable x which makes the equation
1
Linear equations - 3 o E SO
2
2. Maintaining balance
For any equation the LHS must always equal the RHS. We can therefore think of an equation as a
set of scales that must always be in balance.
The balance of an equation is maintained provided we perform the same operation on both sides of
the equal sign.
Any mathematical operation we perform on one side of an equation we must also perform on
the other side.
Example 1
What equation results when:
3x − 2
a) both sides of
= −1 are multiplied by 4
4
b) both sides of 5x = −15 are divided by 5?
3x − 2
3x − 2
= −1 ⇒ 4 ·
= 4 · (−1) ⇒ 3x − 2 = −4
4
4
5x
−15
b) 5x = −15 ⇒
=
⇒ x = −3
5
5
a)
Exercises - Set A
1. What equation results from adding:
a) 6 to both sides of x − 6 = 4
b) 7 to both sides of x − 7 = 6
c) 11 to both sides of 2x − 11 = 3
d) 13 to both sides of 3x − 13 = −2?
2. What equations results from subtracting:
a) 2 from both sides of x + 2 = 3
c) 5 from both sides of 2x + 5 = 03
b) 4 from both sides of x + 4 = −2
d) 9 from both sides of 3x + 9 = −1?
3. What equation results from multiplying both sides of:
a) x = −3 by 3
d)
x
= −1 by 4
4
b) 3x = 1
e)
by 5
x
= −2 by −3
−3
4. What equation results from dividing both sides of:
x
= 4 by 2
2
x
f)
= 4 by −10?
−10
c)
a) 3x = 12 by 3
b) −5x = 30 by −5
c) 2x + 8 = 0 by 2
d) 3x − 9 = 24 by 3
e) 4(x + 2) = −12 by 4
f) −6(x − 1) = −24 by −6?
D pto. M atemáticas. IES Jovellanos. 2012
Linear equations - 3 o E SO
3
3. Formal solution of linear equations
When we use the = sign between two algebraic expressions we have an equation which is in balance.
Whatever we do to one side of the equation, we must do the same to the other side to maintain the
balance.
Technique for solving equations:
Do the same to
both sides
Since both sides of an equation are equal, any operation performed on both sides will not affect the
equality. For example, if one side is multiplied by 2, so must the other side, to maintain equality.
Compare the balance of weights:
As we perform the inverse operations necessary to isolate the unknown, we must perform the same
operations on the other side to maintain the balance.
Consider how the expression has been built up and then isolate the unknown by using inverse
operations in reverse order.
Example 2
Solve for x:
3x + 7 = 22
3x + 7 − 7 = 22 − 7
3x = 15
3x
15
=
3
3
x= 5
Check:
(subtract 7 from both sides)
(simplify)
(divide both sides by 3)
(simplify)
LHS= 3 · 5 + 7 = 22 ⇒ LHS=RHS
Example 3
Solve for x:
11 − 5x = 26
11 − 5x − 11 = 26 − 11
−5x = 15
−5x
15
=
−5
−5
x = −3
Check:
(subtract 11 from both sides)
(simplify)
(divide both sides by −5)
(simplify)
LHS= 11 − 5 · (−3) = 26 ⇒ LHS=RHS
D pto. M atemáticas. IES Jovellanos. 2012
Linear equations - 3 o E SO
4
Example 4
Solve for x:
x
+2−2
3
x
3
x
·3
3
x
x
+ 2 = −2
3
= −2 − 2
(subtract 2 from both sides)
= −4
(simplify)
= −4 · 3
(multiply both sides by 3)
= −12
(simplify)
Example 5
Solve for x:
5·
4x + 3
= −2
5
4x + 3
= −2 · 5
5
4x + 3 = −10
(multiply both sides by 5)
(simplify)
4x + 3 − 3 = −10 − 3
4x = −13
4x
13
= −
4
4
x = −3 41
(subtract 3 from both sides)
(simplify)
(divide both sides by 4)
(simplify)
Exercises - Set B
1. Solve for x:
a) 6 − x = −5
b) −4x = 15
d) 5 − 4x = −7
2. Solve for x:
x
a) = 5
2
x
d) − 2 = 1
3
3. Solve for x:
2x + 7
a)
=0
5
3x − 1
d)
=4
2
c) 3 − 2x = 7
e) 3 − 7x = −2
2x
= −4
3
x−1
e)
=1
2
b)
1
(1 − 3x) = −1
2
1
e) (4 − 3x) = −1
5
b)
D pto. M atemáticas. IES Jovellanos. 2012
f) 17 − 2x = −1
x
+ 1 = −3
4
x+5
f)
= −1
3
c)
1 + 4x
=3
7
1
f) (1 + 2x) = −2
4
c)
Linear equations - 3 o E SO
5
4. Equations with a repeated unknown
Equations in which the unknown appears more than once need to be solved systematically. We use
the following procedure:
Step 1: If necessary expand any brackets and collect like terms.
Step 2: If necessary, remove the unknown from one side. Aim to be left with a positive unknown
on one side.
Step 3: Use inverse operations to isolate the unknown and maintain balance.
Step 4: Check that your solution satisfies the equation, i.e. LHS=RHS.
Example 6
Solve for x:
5(x + 1) − 2x = −7
5x + 5 − 2x = −7
3x + 5 = −7
3x + 5 − 5 = −7 − 5
3x = −12
3x
12
=−
3
3
x = −4
(expand brackets)
(collect like terms)
(subtract 5 from both sides)
(simplify)
(divide both sides by 3)
(simplify)
Example 7
Solve for x:
15 − 2x = 11 + x
15 − 2x + 2x = 11 + x + 2x
15 = 11 + 3x
15 − 11 = 11 + 3x − 11
4 = 3x
4
3x
=
3
3
x = 1 31
(add 2x to both sides)
(simplify)
(subtract 11 from both sides)
(simplify)
(divide both sides by 3)
(simplify)
D pto. M atemáticas. IES Jovellanos. 2012
Linear equations - 3 o E SO
6
Exercises - Set C
1. Solve for x:
a) 4(x − 1) − 2x = 2
c) 4(x − 2) − (x + 1) = 6
b) 2(x − 3) + 3x = 9
d) 5(3 + 2x) + 2(x + 1) = −7
2. Solve for x:
a) 3x + 2 = x + 10
d) 3x − 1 = 6x − 4
b) 7x + 2 = 4x − 8
c) 8x + 3 = 4x + 4
e) 4x − 2 = 8x + 10
f) 5 + 2x = 18 + 4x
b) 6 + 2x = 15 − x
c) 6 + x = 13 − 2x
3. Solve for x:
a) 3x + 5 = 1 − x
d) 9 − 2x = 12 − 7x
e) 4 − 3x = 8 − x
f) 9 − 3x = x + 2
4. Solve for x:
a) 3(x + 1) − x = 15
b) 5(x + 2) − 2x = 11
e) 4(3x + 1) + 18 = x
f) 7x − 2(x + 1) = 13
c) 6(1 − 2x) = −4 − 7x
g) 14x − 5(2x + 5) = 7
i) 2x − 4(4 − 3x) = x + 10
k) 2(x − 6) + 7x = 3(3x − 4)
d) 2(x + 4) + 8(x − 2) = 12
h) 19 − (3 − x) = 5x
j) 2(x + 4) = 7 − (3 + x)
l) 5(2x − 4) = 4x − 2(1 − 3x)
5. Fractional equations
More complicated fractional equations can be solved by:
• writing all fractions with the same lowest common denominator (LCD), and then
• equating numerators.
Example 8
Solve for x:
2
x
=
3
5
x·5
2·3
=
3·5
5·3
5x = 6
x =
1 51
x
2
=
3
5
has LCD of 15
(to achieve a common denominator)
(equating numerators)
(divide both sides by 5)
In this case, as we have only one fraction on the LHS and only one fraction on the RHS, we get
the same result cross-multiplying the fractions:
x
2
=
⇒ 5x = 2 · 3 ⇒ 5x = 6 ⇒ x = 1 51
3
5
D pto. M atemáticas. IES Jovellanos. 2012
Linear equations - 3 o E SO
7
Example 9
Solve for x:
2x + 3
x−2
=
4
3
(cross-multiply)
3(2x + 3) = 4(x − 2)
(expand brackets)
6x + 9 = 4x − 8
(use inverse operations)
6x − 4x = −9 − 8
(simplify)
2x = −17
17
x= −
2
(divide both sides by 2)
Example 10
Solve for x:
2x + 1 x − 2
−
=5
3
2
2x + 1 x − 2
−
= 5
3
2
2 · (2x + 1) 3 · (x − 2)
6·5
−
=
2·3
3·2
6
has LCD of 6
(to achieve a common denominator)
2(2x + 1) − 3(x − 2) = 30
(equating numerators)
x + 8 = 30
(collecting like terms)
4x + 2 − 3x + 6 = 30
(expanding brackets)
x = 22
Note the use of brackets in the original fractions.
Exercises - Set D
1. Solve for x:
6x
3
a)
=
4
2
b)
2
x
=
3
5
c)
4x
1
=
5
2
d)
1
5x
=
5
3
e)
2x + 3
2x
=
5
3
f)
x+1
2x + 5
=
3
4
g)
4−x
x+1
=
2
3
h)
3x + 8
=x−5
3
i)
x+2
5−x
=
3
4
2. Solve for x:
x x
a) + = 2
2
5
d)
3x 2x
1
+
=−
4
3
2
b)
x 2x
5
−
=
2
3
6
c)
3x x
− = 11
2
8
e)
5x x
3
− =
2
6
8
f)
x 3x
−
=4
7
2
D pto. M atemáticas. IES Jovellanos. 2012
Linear equations - 3 o E SO
8
Exercises - Set E
1. Solve for x:
x 2x + 1
=0
a) +
3
6
b)
2x + 3 x − 1
3
+
=
5
2
4
c)
2x − 3 x + 4
+
= −2
5
2
d)
x − 2 2x + 1
+
=2
3
12
e)
4x − 3 x + 2
1
+
=−
5
3
4
f)
x − 4 2x − 3
5
+
=
3
8
6
2. Solve for x:
x + 1 2x
−
=4
a)
2
5
b)
3x − 2 x − 1
1
−
=
3
2
2
c)
4x x − 2
−
= −3
5
2
d)
2x − 1 x + 2
1
−
=−
4
3
3
e)
4x + 1 2x − 3
−
=1
2
6
f)
x−5 x+1
−
= −2
6
4
Exercises - Set F
1. Solve for x:
30 − x
5
a)
=
20 + x
4
c) 3(2x − 5) + 8x − 6 =
x
− (5x + 3)
2
b)
x x
x
− = − 11
5
9
3
d)
x−3
x+5 x−1
=
−
4
6
9
3x + 1
= −5x − 3
2
e)
2x 3x − 5
x
−
= −3
15
20
5
f) −4(2x − 1) +
g)
8−x
4(3 − x) 4x + 10
−
=2−x−
5
2
10
h)
20 − x 3x − 5
x + 1 1 − 2x
−
=
+
3
4
12
4
j)
3x − 1 1 − 4x
1 − x 14 − x
−
=
−
4
5
4
6
i)
2x − 1 5x + 2
2x − 3
−
=
+1
3
12
4
k)
x−4
(−4x + 2)
5x + 6
− 4(−2x + 1) −
= 2(x − 3) +
5
10
2
✁✃✁✃✁✃✁✃✁✃✁✃✁✃
D pto. M atemáticas. IES Jovellanos. 2012
W ord problem s
1. Problem solving
Many problems can be translated into algebraic equations. When problems are solved using algebra,
we follow these steps:
Step 1: Read the problem.
Step 2: Decide on the unknown quantity and allocate a variable.
Step 3: Decide which operations are involved.
Step 4: Translate the problem into an equation.
Step 5: Solve the equation by isolating the variable.
Step 6: Check that your solution does satisfy the original problem.
Step 7: Write your answer in sentence form. Remember, there is usually no variable in the
original problem.
Example 1
When a number is trebled and subtracted from 7, the result is −11. Find the number.
Let x be the number, so 3x is the number trebled.
7 − 3x is this number subtracted from 7.
So, 7 − 3x = −11 ⇒ −3x = −18 ⇒ x = 6
So, the number is 6.
Check:
7 − 3 · 6 = 7 − 18 = −11
X
Example 2
Sarah’s age is one third her father’s age. In 13 years’ time her age will be a half of her father’s
age. How old is Sarah now?
Let Sarah’s present age be x years, so her father’s present age is 3x years.
So, 3x + 13 = 2(x + 13)
Table of ages:
Sarah
Father
Now
x
3x
3x + 13 = 2x + 26
3x − 2x = 26 − 13
13 years time
x + 13
3x + 13
x = 13
Sarah’s present age is 13 years.
1
W ord problem s - 3 o E SO
2
Exercises - Set A
1. When three times a certain number is subtracted from 15 the result is −6. Find the number.
2. Five times a certain number, minus 5, is equal to 7 more than three times the number. What
is the number?
3. Three times the result of subtracting a certain number from 7 gives the same answer as
adding eleven to the number. Find the number.
4. I think of a number. If I divide the sum of 6 and the number by 3, the result is 4 more than
one quarter of the number. Find the number.
5. The sum of two numbers is 15. When one of these numbers is added to three times the other,
the result is 27. What are the numbers?
6. What number must be added to both the numerator and denominator of the fraction
get the fraction 87 ?
2
5
to
7. What number must be subtracted from both the numerator and denominator of the fraction
3
1
4 to get the fraction 3 ?
8. Eli is now one quarter of his father’s age. In 5 years’ time his age will be one third of his
father’s age. How old is Eli now?
9. When Maria was born, her mother was 24 years old. At present, Maria’s age is 20 % of her
mother’s age. How old is Maria now?
10. Five years ago, Jacob was one sixth of the age of his brother. In three years’ time his age
doubled will match his brother’s age. How old is Jacob now?
2. Money and investment problems
Problems involving money can be made easier to understand by constructing a table and placing
the given information into it.
Example 3
Britney has only 2-cent and 5-cent stamps. Their total value is $1,78, and there are two more
5-cent stamps than there are 2-cent stamps. How many 2-cent stamps are there?
If there are x 2-cent stamps then there are (x + 2) 5-cent stamps.
2x + 5(x + 2) = 178
Type
2-cent
5-cent
Number
x
x+2
Value
2x cents
5(x + 2) cents
(equating values in cents)
2x + 5x + 10 = 178
7x = 168
x = 24
So, there are 24 2-cents stamps.
D pto. M atemáticas. IES Jovellanos. 2012
W ord problem s - 3 o E SO
3
Exercises - Set B
1. Michaela has 5-cent and 10-cent stamps with a total value of 5,75 e. If she has 5 more
10-cent stamps than 5-cent stamps, how many of each stamp does she have?
2. The school tuck-shop has milk in 600 ml and 1 litre cartons. If there are 54 cartons and 40
ml of milk in total, how many 600 ml cartons are there?
3. Aaron has a collection of American coins. He has three times as many 10 cent coins as 25
cent coins, and he has some 5 cent coins as well. If he has 88 coins with a total value $11,40,
how many of each type does he have?
4. Tickets to a football match cost 8 e, 15 e or 20 e each. The number of 15 e tickets sold
was double the number of 8 e tickets sold. 6000 more 20 e tickets were sold than 15 e tickets.
If the gate receipts totalled 783 000 e, how many of each type of ticket were sold?
5. Kelly blends coffee. She mixes brand A costing $6 per kilogram with brand B costing $8 per
kilogram. How many kilograms of each brand does she have to mix to make 50 kg of coffee
costing her $7,20 per kg?
6. Su Li has 13 kg of almonds costing $5 per kilogram. How many kilograms of cashews costing
$12 per kg should be added to get a mixture of the two nut types costing $7,45 per kg?
3. Motion problems
Motion problems are problems concerned with speed, distance travelled, and time taken. These
variables are related by the formulae:
speed =
distance
time
distance = speed × time
time =
distance
speed
Speed is usually measured in either kilometres per hour (denoted km/h or km h−1 ) or metres per
second (denoted m/s or m s−1 ).
Example 4
A car travels for 2 hours at a certain speed and then 3 hours more at a speed 10 km h−1 faster
than this. If the entire distance travelled is 455 km, find the car’s speed in the first two hours
of travel.
Let the speed in the first 2 hours be s km h−1 .
First section
Second section
Speed (km h−1 )
s
(s + 10)
Time (h)
2
3
Total
Distance (km)
2s
3(s + 10)
455
So, 2s + 3(s + 10) = 455
2s + 3s + 30 = 455
5s = 425
s = 85
The car’s speed in the first two hours was 85 km h−1 .
D pto. M atemáticas. IES Jovellanos. 2012
W ord problem s - 3 o E SO
4
Exercises - Set C
1. Joe can run twice as fast as Pete. They start at the same point and run in opposite directions
for 40 minutes. The distance between them is now 16 km. How fast does Joe run?
2. A car leaves a country town at 60 km per hour. Two hours later, a second car leaves the
town; it catches the first car after 5 more hours. Find the speed of the second car.
3. A boy cycles from his house to a friend’s house at 20 km h−1 and home again at 25 km h−1 .
9
If his round trip takes 10
of an hour, how far it is to his friend’s house?
4. A motor cyclist makes a trip of 500 km. If he had increased his speed by 10 km h−1 ,he could
have covered 600 km in the same time. What was his original speed?
5. Normally I drive to work at 60 km h−1 . If I drive at 72 km h−1 I cut 8 minutes off my time
for the trip. What distance do I travel?
Revision problems (1)
1. One-half of Heather’s age two years from now plus one-third of her age three years ago is
twenty years. How old is she now?
2. A piece of 16-gauge copper wire 42 cm long is bent into the shape of a rectangle whose
width is twice its length. Find the dimensions of the rectangle.
3. A collection of 33 coins, consisting of nickels, dimes, and quarters, has a value of $3.30. If
there are three times as many nickels as quarters, and one-half as many dimes as nickels, how
many coins of each kind are there?
4. A wallet contains the same number of pennies, nickels, and dimes. The coins total $1.44.
How many of each type of coin does the wallet contain?
5. An aircraft carrier made a trip to Guam and back. The trip there took three hours and the
trip back took four hours. It averaged 6 km/h on the return trip. Find the average speed of the
trip there.
6. A passenger plane made a trip to Las Vegas and back. On the trip there it flew 432 mph and
on the return trip it went 480 mph. How long did the trip there take if the return trip took nine
hours?
7. A cattle train left Miami and traveled toward New York. 14 hours later a diesel train left
traveling at 45 km/h in a effort to catch up to the cattle train. After traveling for four hours
the diesel train finally caught up. What was the cattle train’s average speed?
8. Jose left the White House and drove toward the recycling plant at an average speed of 40
km/h. Rob left some time later driving in the same direction at an average speed of 48 km/h.
After driving five hours Rob caught up with Jose. How long did Jose drive before Rob caught
up?
9. A passenger train leaves the train depot 2 hours after a freight train left the same depot. The
freight train is traveling 20 mph slower than the passenger train. Find the rate of each train, if
the passenger train overtakes the freight train in three hours.
10. Two cyclists start at the same time from opposite ends of a course that is 45 miles long.
One cyclist is riding at 14 mph and the second cyclist is riding at 16 mph. How long after they
begin will they meet?
D pto. M atemáticas. IES Jovellanos. 2012
W ord problem s - 3 o E SO
5
Revision problems (2)
1. A boat travels for three hours with a current of 3 mph and then returns the same distance
against the current in four hours. What is the boat’s speed in calm water? How far did the boat
travel one way?
2. A spike is hammered into a train rail. You are standing at the other end of the rail. You hear
the sound of the hammer strike both through the air and through the rail itself. These sounds
arrive at your point six seconds apart. You know that sounds travels through air at 1100 feet
per second and through steel at 16500 feet per second. How far away is that spike?
3. The sum of two consecutive integers is 15. Find the numbers.
4. A golf shop pays its wholesaler $40 for a certain club, and then sells it to a golfer for $75.
What is the markup rate?
5. A shoe store uses a 40 % markup on cost. Find the cost of a pair of shoes that sells for $63.
6. An item is marked down 15 %; the sale price is $127,46. What was the original price?
7. 2 m3 of soil containing 35 % sand was mixed into 6 m3 of soil containing 15 % sand. What
is the sand content of the mixture?
8. 5 fl. oz. of a 2 % alcohol solution was mixed with 11 fl. oz. of a 66 % alcohol solution. Find
the concentration of the new mixture.
9. 9 lbs. of mixed nuts containing 55 % peanuts were mixed with 6 lbs. of another kind of mixed
nuts that contain 40 % peanuts. What percent of the new mixture is peanuts?
10. Working alone, Ryan can dig a 10 ft by 10 ft hole in five hours. Castel can dig the same
hole in six hours. How long would it take them if they worked together?
11. It takes Trevon ten hours to clean an attic. Cody can clean the same attic in seven hours.
Find how long it would take them if they worked together.
12. Working together, Paul and Daniel can pick forty bushels of apples in 4,95 hours. Had he
done it alone it would have taken Daniel 9 hours. Find how long it would take Paul to do it
alone.
✁✃✁✃✁✃✁✃✁✃✁✃✁✃
D pto. M atemáticas. IES Jovellanos. 2012
More exercises to practice with answers:
J.F. Antona. Maths Dep.
3º ESO
Word Problems
(involving linear equations)
"Age" Word Problems
1) One-half of Heather's age two years from now plus one-third of her age three years ago is
twenty years. How old is she now?
Geometry Word Problems
1) Suppose a water tank in the shape of a right circular cylinder is thirty feet long and eight feet
in diameter. How much sheet metal was used in its construction?
2) A piece of 16-gauge copper wire 42 cm long is bent into the shape of a rectangle whose
width is twice its length. Find the dimensions of the rectangle.
"Coin" Word Problems
1) A collection of 33 coins, consisting of nickels, dimes, and quarters, has a value of $3.30. If
there are three times as many nickels as quarters, and one-half as many dimes as nickels, how
many coins of each kind are there?
2) A wallet contains the same number of pennies, nickels, and dimes. The coins total$1.44.
How many of each type of coin does the wallet contain?
"Distance" Word Problems
5) A passenger train leaves the train depot 2 hours after a freight train left the same depot. The
freight train is traveling 20 mph slower than the passenger train. Find the rate of each train, if
the passenger train overtakes the freight train in three hours.
6) Two cyclists start at the same time from opposite ends of a course that is 45 miles long. One
cyclist is riding at 14 mph and the second cyclist is riding at 16 mph. How long after they begin
will they meet?
= Problems to try for homework
J.F. Antona. Maths Dep.
3º ESO
7) A boat travels for three hours with a current of 3 mph and then returns the same distance
against the current in four hours. What is the boat's speed in calm water? How far did the boat
travel one way?
8) With the wind, an airplane travels 1120 miles in seven hours. Against the wind, it takes eight
hours. Find the rate of the plane in still air and the velocity of the wind.
9) A spike is hammered into a train rail. You are standing at the other end of the rail. You hear
the sound of the hammer strike both through the air and through the rail itself. These sounds
arrive at your point six seconds apart. You know that sound travels through air at 1100 feet per
second and through steel at 16,500 feet per second. How far away is that spike?
"Investment" Word Problems (simple interest)
1) You put $1000 into an investment yielding 6% annual interest; you left the money in for two
years. How much interest do you get at the end of those two years?
2) You invested $500 and received $650 after three years. What had been the interest rate?
3) You have $50,000 to invest, and two funds that you'd like to invest in. The You-Risk-It Fund
(Fund Y) yields 14% interest. The Extra-Dull Fund (Fund X) yields 6% interest. Because of
college financial-aid implications, you don't think you can afford to earn more than $4,500 in
interest income this year. How much should you put in each fund?"
4) An investment of $3,000 is made at an annual simple interest rate of 5%. How much
additional money must be invested at an annual simple interest rate of 9% so that the total
annual interest earned is 7.5% of the total investment?
5) A total of $6,000 is invested into two simple interest accounts. The annual simple interest
rate on one account is 9%; on the second account, the annual simple interest rate is 6%. How
much should be invested in each account so that both accounts earn the same amount of
annual interest?
6) An investor deposited an amount of money into a high-yield mutual fund that returns a
9% annual simple interest rate. A second deposit, $2,500 more than the first, was placed in a
certificate of deposit that returns a 5% annual simple interest rate. The total interest earned on
both investments for one year was $475. How much money was deposited in the mutual fund?
7) The manager of a mutual fund placed 30% of the fund's available cash in a 6% simple
interest account, 25% in 8% corporate bonds, and the remainder in a money market fund that
earns 7.5% annual simple interest. The total annual interest from the investments
was $35,875. What was the total amount invested?
"Number" Word Problems
1) The sum of two consecutive integers is 15. Find the numbers.
2) The product of two consecutive negative even integers is 24. Find the numbers.
= Problems to try for homework
J.F. Antona. Maths Dep.
3º ESO
3) Twice the larger of two numbers is three more than five times the smaller, and the sum of
four times the larger and three times the smaller is 71. What are the numbers?
"Percent of" Word Problems
1) A golf shop pays its wholesaler $40 for a certain club, and then sells it to a golfer for$75.
What is the markup rate?
2) A shoe store uses a 40% markup on cost. Find the cost of a pair of shoes that sells for$63.
3) An item originally priced at $55 is marked 25% off. What is the sale price?
4) An item that regularly sells for $425 is marked down to $318.75. What is the discount rate?
5) An item is marked down 15%; the sale price is $127.46. What was the original price?
Mixture Word Problems
Work Word Problems
= Problems to try for homework
Solutions to Word Problems
(involving linear equations)
_____________________________________________________________________________________
"Age" Word Problems
____________________________________________________________________________________
1) One-half of Heather's age two years from now plus one-third of her age three years ago is twenty
years. How old is she now?
This problem refers to Heather's age two years in the future and three years in the past. So I'll
pick a variable and label everything clearly:
age now: H
age two years from now: H + 2
age three years ago: H – 3
Now I need certain fractions of these ages:
one-half of age two years from now: ( 1/2 )(H + 2) = H/2 + 1
one-third of age three years ago: ( 1/3 )(H – 3) = H/3 – 1
The sum of these two numbers is twenty, so I'll add them and set this equal to 20:
H
/2 + 1 + H/3 – 1 = 20
/2 + H/3 = 20
3H + 2H = 120
5H = 120
H = 24
H
___________________________________________________________________________________
Geometry Word Problems
___________________________________________________________________________________
1) Suppose a water tank in the shape of a right circular cylinder is thirty feet long and eight feet in
diameter. How much sheet metal was used in its construction?
What they are asking for here is the surface area of the water tank. The total surface area of the
tank will be the sum of the surface areas of the side (the cylindrical part) and of the ends. If the
diameter is eight feet, then the radius is four feet. The surface area of each end is given by the
area formula for a circle with radius r: A = (pi)r2. (There are two end pieces, so I will be
multiplying this area by 2 when I find my total-surface-area formula.) The surface area of the
cylinder is the circumference of the circle, multiplied by the height: A = 2(pi)rh.
Side view of the
cylindrical tank, showing
the radius "r".
An "exploded" view of
the tank, showing the
three separate surfaces
whose areas I need to
find.
Then the total surface area of this tank is given by:
2 ×( (pi)r2 ) + 2(pi)rh (the two ends, plus the cylinder)
= 2( (pi) (42) ) + 2(pi) (4)(30)
= 2( (pi) × 16 ) + 240(pi)
= 32(pi) + 240(pi)
= 272(pi)
Since the original dimensions were given in terms of feet, then my area must be in terms of
square feet:
the surface area is 272(pi) square feet.
___________________________________________________________________________________
2) A piece of 16-gauge copper wire 42 cm long is bent into the shape of a rectangle whose width is
twice its length. Find the dimensions of the rectangle.
Do I care that the wire is made of copper, or that the wire is a length of sixteen-gauge? No; all I
care is that the length is forty-two units, that the units are centimeters, that the rectangle is twice
as long in one direction as the other, and that I'm supposed to find the values of each of these
directions. I can ignore the other information.
Since the wire is 42 centimeters long, then the perimeter of the rectangle is 42 centimeters. That
is:
2L + 2W = 42
I also know that the width is twice the length, so:
W = 2L
Then: Copyright © Elizabeth Stapel 2000-2008 All Rights Reserved
2L + 2(2L) = 42 (by substitution for W from the above equation)
2L + 4L = 42
6L = 42
L=7
Since the width is related to the length by W = 2L, then W = 14, and the rectangle
is 7centimeters long and 14 centimeters wide.
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"Coin" Word Problems
____________________________________________________________________________________
1) A collection of 33 coins, consisting of nickels, dimes, and quarters, has a value of $3.30. If there
are three times as many nickels as quarters, and one-half as many dimes as nickels, how many coins
of each kind are there?
I'll start by picking and defining a variable, and then I'll use translation to convert this exercise
into mathematical expressions.
Nickels are defined in terms of quarters, and dimes are defined in terms of nickels, so I'll pick a
variable to stand for the number of quarters, and then work from there:
number of quarters: q
number of nickels: 3q
number of dimes: (½)(3q) = (3/2)q
There is a total of 33 coins, so:
q + 3q + (3/2)q = 33
4q + (3/2)q = 33
8q + 3q = 66
11q = 66
q=6
Then there are six quarters, and I can work backwards to figure out that there are 9 dimes
and 18 nickels.
_________________________________________________________________________________
2) A wallet contains the same number of pennies, nickels, and dimes. The coins total$1.44. How
many of each type of coin does the wallet contain?
Since there is the same number of each type of coin, I can use one variable to stand for each:
number of pennies: p
number of nickels: p
number of dimes: p
The value of the coins is the number of cents for each coin times the number of that type of coin,
so:
value of pennies: 1p
value of nickels: 5p
value of dimes: 10p
The total value is $1.44, so I'll add the above, set equal to 144 cents, and solve:
1p + 5p + 10p = 144
16p = 144
p=9
There are nine of each type of coin in the wallet.
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"Distance" Word Problems
"Distance" word problems, often also called "uniform rate" problems, involve something travelling at
some fixed and steady ("uniform") pace ("rate" or "speed"), or else moving at some average speed.
Whenever you read a problem that involves "how fast", "how far", or "for how long", you should think of
the distance equation, d = rt, where d stands for distance, r stands for the (constant or average) rate of
speed, and t stands for time.
Warning: Make sure that the units for time and distance agree with the units for the rate. For instance, if
they give you a rate of feet per second, then your time must be in seconds and your distance must be in
feet. Sometimes they try to trick you by using the wrong units, and you have to catch this and convert to
the correct units.
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_____________________________________________________________________________________
5) A passenger train leaves the train depot 2 hours after a freight train left the same depot. The
freight train is traveling 20 mph slower than the passenger train. Find the rate of each train, if the
passenger train overtakes the freight train in three hours.
passenger train
freight train
total
d
d
d
---
r
r
r – 20
---
t
3
3+2=5
---
(As it turns out, I won't need the "total" row this time.) Why is the distance just "d" for both
trains? Partly, that's because the problem doesn't say how far the trains actually went. But mostly
it's because they went the same distance as far as I'm concerned, because I'm only counting from
the depot to wherever they met. After that meet, I don't care what happens. And how did I get
those times? I know that the passenger train drove for three hours to catch up to the freight train;
that's how I got the "3". But note that the freight train had a two-hour head start. That means that
the freight train was going for five hours.
passenger train
freight train
total
d
d = 3r
d = 5(r – 20)
---
r
r
r – 20
---
t
3
3+2=5
---
Now that I have this information, I can try to find my equation. Using the fact that d = rt, the
first row gives me d = 3r (note the revised table above). The second row gives me:
d = 5(r – 20)
Since the distances are equal, I will set the equations equal:
3r = 5(r – 20)
Solve for r; interpret the value within the context of the exercise, and state the final answer.
_____________________________________________________________________________________
6) Two cyclists start at the same time from opposite ends of a course that is 45 miles long. One
cyclist is riding at 14 mph and the second cyclist is riding at 16 mph. How long after they begin will
they meet?
slow guy
fast guy
total
d
d
45 – d
45
r
14
16
---
t
t
t
---
Why is t the same for both cyclists? Because I am measuring from the time they both started to
the time they meet somewhere in the middle. And how did I get "d" and "45 – d" for the
distances? Because once I'd assigned the slow guy as having covered d miles, that left 45 –
dmiles for the fast guy to cover: the two guys together covered the whole 45 miles.
Using "d = rt", I get d = 14t from the first row, and 45 – d = 16t from the second row. Since
these distances add up to 45, I will add the distance expressions and set equal to the given total:
45 = 14t + 16t
Solve for t.
___________________________________________________________________________________
7) A boat travels for three hours with a current of 3 mph and then returns the same distance
against the current in four hours. What is the boat's speed in calm water? How far did the boat
travel one way?
downstream
upstream
total
d
d
d
2d
r
b+3
b–3
---
t
3
4
7
(It may turn out that I won't need the "total" row.)
I have used "b" to indicate the boat's speed. Why are the rates "b + 3" and "b – 3"? Because I
actually have two speeds combined into one on each trip. The boat has a certain speed (the
"speed in calm water" that I'm looking for; this is the speed that registers on the speedometer),
and the water has a certain speed (this is the "current"). When the boat is going with the current,
the water's speed is added to the boat's speed. This makes sense, if you think about it: even if you
cut the engine, the boat would still be moving, because the water would be carrying it
downstream. When the boat is going against the current, the water's speed is subtracted from the
boat's speed. This makes sense, too: if the water is going fast enough, the boat will still be going
downstream (a "negative" speed, because the boat would be going backwards at this point),
because the water is more powerful than the boat. (Think of a boat in a cartoon heading toward a
waterfall. The guy paddles like crazy, but he still goes over the edge.)
d
d = 3(b + 3)
d = 4(b – 3)
2d
downstream
upstream
total
r
b+3
b–3
---
t
3
4
7
Using "d = rt", the first row (of the revised table above) gives me:
d = 3(b + 3)
The second row gives me:
d = 4(b – 3)
Since these distances are the same, I will set them equal:
3(b + 3) = 4(b – 3)
Solve for b. Then back-solve for d.
In this case, I didn't need the "total" row.
____________________________________________________________________________________
8) With the wind, an airplane travels 1120 miles in seven hours. Against the wind, it takes eight
hours. Find the rate of the plane in still air and the velocity of the wind.
tailwind
headwind
total
d
1120
1120
2240
r
p+w
p–w
---
t
7
8
15
(I probably won't need the "total" row.) Just as with the last problem, I am really dealing with
two rates together: the plane's speedometer reading, and the wind speed. When the plane turns
around, the wind is no longer pushing the plane to go faster, but is instead pushing against the
plane to slow it down. Copyright © Elizabeth Stapel 1999-2009 All Rights Reserved
The first row gives me:
1120 = 7(p + w)
The second row gives me:
1120 = 8(p – w)
The temptation is to just set these equal, like I did with the last problem, but that just gives me:
7(p + w) = 8(p – w)
...which doesn't help much. I need to get rid of one of the variables.
I'll take that first equation:
1120 = 7(p + w)
...and divide through by 7:
160 = p + w
Then, subtracting w from either side, I get that p = 160 – w. I'll substitute "160 – w" for "p" in the
second equation:
1120 = 8([160 – w] – w)
1120 = 8(160 – 2w)
...and solve for w. Then I'll back-solve to find p.
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9) A spike is hammered into a train rail. You are standing at the other end of the rail. You hear the
sound of the hammer strike both through the air and through the rail itself. These sounds arrive at
your point six seconds apart. You know that sound travels through air at 1100 feet per second and
through steel at 16,500 feet per second. How far away is that spike?
air
steel
total
d
d = 1100t
d = 16,500(t – 6)
---
r
1100
16,500
---
t
t
t–6
6
However long the sound took to travel through the air, it took six seconds less to propagate
through the steel. (Since the speed through the steel is faster, then that travel-time must be
shorter.) I multiply the rate by the time to get the values for the distance column. (Once again, I
didn't need the "total" row.)
Since the distances are the same, I set the distance expressions equal to get:
1100t = 16,500(t – 6)
Solve for the time t, and then back-solve for the distance d by plugging t into either expression
for the distance d.
_____________________________________________________________________________________
"Investment" Word Problems
Investment problems usually involve simple annual interest (as opposed to compounded interest), using
the interest formula I = Prt, where I stands for the interest on the original investment, P stands for the
amount of the original investment (called the "principal"), r is the interest rate (expressed in decimal
form), and t is the time.
For annual interest, the time t must be in years. If they give you a time of, say, nine months, you must
first convert this to 9/12 = 3/4 = 0.75 years. Otherwise, you'll get the wrong answer. The time units must
match the interest-rate units. If you got a loan from your friendly neighborhood loan shark, where the
interest rate is monthly, rather than yearly, then your time must be measured in terms of months.
Investment word problems are not generally terribly realistic; in "real life", interest is pretty much always
compounded somehow, and investments are not generally all for whole numbers of years. But you'll get
to more "practical" stuff later; this is just warm-up, to prepare you for later.
In all cases of these problems, you will want to substitute all known information into the "I = Prt"
equation, and then solve for whatever is left.
_____________________________________________________________________________________
1) You put $1000 into an investment yielding 6% annual interest; you left the money in for two
years. How much interest do you get at the end of those two years?
In this case, P = $1000, r = 0.06 (because I have to convert the percent to decimal form), and the
time is t = 2. Substituting, I get:
I = (1000)(0.06)(2) = 120
I will get $120 in interest.
Another example would be: Copyright © Elizabeth Stape6-2008 All Rights Reserved
_____________________________________________________________________________________
2) You invested $500 and received $650 after three years. What had been the interest rate?
For this exercise, I first need to find the amount of the interest. Since interest is added to the
principal, and since P = $500, then I = $650 – 500 = $150. The time is t = 3. Substituting all of
these values into the simple-interest formula, I get:
150 = (500)(r)(3)
150 = 1500r
150
/1500 = r = 0.10
Of course, I need to remember to convert this decimal to a percentage.
I was getting 10% interest.
The hard part comes when the exercises involve multiple investments. But there is a trick to these that
makes them fairly easy to handle.
_____________________________________________________________________________________
3) You have $50,000 to invest, and two funds that you'd like to invest in. The You-Risk-It Fund
(Fund Y) yields 14% interest. The Extra-Dull Fund (Fund X) yields 6% interest. Because of college
financial-aid implications, you don't think you can afford to earn more than $4,500 in interest
income this year. How much should you put in each fund?"
The problem here comes from the fact that I'm splitting that $50,000 in principal into two
smaller amounts. Here's how to handle this:
Fund X
Fund Y
total
I
?
?
4,500
P
?
?
50,000
r
0.06
0.14
---
t
1
1
---
How do I fill in for those question marks? I'll start with the principal P. Let's say that I put "x"
dollars into Fund X, and "y" dollars into Fund Y. Then x + y = 50,000. This doesn't help much,
since I only know how to solve equations in one variable. But then I notice that I can
solve x + y= 50,000 to get y = $50,000 – x.
THIS TECHNIQUE IS IMPORTANT! The amount in Fund Y is (the total) less (what we've
already accounted for in Fund X), or 50,000 – x. You will need this technique, this "how much is
left" construction, in the future, so make sure you understand it now.
Fund X
Fund Y
total
I
?
?
4,500
P
x
50,000 – x
50,000
r
0.06
0.14
---
t
1
1
---
Now I will show you why I set up the table like this. By organizing the columns according to the
interest formula, I can now multiply across (right to left) and fill in the "interest" column.
Fund X
Fund Y
total
I
0.06x
0.14(50,000 – x)
4,500
P
x
50,000 – x
50,000
r
0.06
0.14
---
t
1
1
---
Since the interest from Fund X and the interest from Fund Y will add up to $4,500, I can add
down the "interest" column, and set this sum equal to the given total interest:
0.06x + 0.14(50,000 – x) = 4,500
0.06x + 7,000 – 0.14x = 4,500
7,000 – 0.08x = 4,500
–0.08x = –2,500
x = 31,250
Then y = 50,000 – 31,250 = 18,750.
I should put $31,250 into Fund X, and $18,750 into Fund Y.
Note that the answer did not involve "neat" values like "$10,000" or "$35,000". You should understand
that this means that you cannot always expect to be able to use "guess-n-check" to find your answers. You
really do need to know how to do these exercises.
f you set up your investment word problems so everything is labeled and well-organized, they should all
work out fairly easily. Just take your time and do things in an orderly fashion. I've done the set-up (but
not the complete solutions) for a few more examples:
_____________________________________________________________________________________
4) An investment of $3,000 is made at an annual simple interest rate of 5%. How much additional
money must be invested at an annual simple interest rate of 9% so that the total annual interest
earned is 7.5% of the total investment?
first
additional
total
I
(3,000)(0.05) = 150
0.09 x
(3,000 + x)(0.075)
P
3,000
x
3,000 + x
r
0.05
0.09
0.075
t
1
1
1
First I fill in the P, r, and t columns with the given values.
Then I multiply across the rows (from the right to the left) in order to fill in the I column.
Then add down the I column to get the equation 150 + 0.09 x = (3,000 + x)(0.075).
To find the solution, I would solve for the value of x.
_____________________________________________________________________________________
5) A total of $6,000 is invested into two simple interest accounts. The annual simple interest rate on
one account is 9%; on the second account, the annual simple interest rate is 6%. How much should
be invested in each account so that both accounts earn the same amount of annual interest?
9% account
6% account
total
I
0.09x
(6,000 – x)(0.06)
---
P
x
6,000 – x
6,000
r
0.09
0.06
---
t
1
1
---
In this problem, I don't actually need the "total" row at all.
First I'll fill in the P, r, and t columns, and multiply to the left to fill in the I column.
From the interest column, I then get the equation 0.09x = ($6,000 – x)(0.06), because the yields
are required to be equal.
Then I'd solve for the value of x, and back-solve to find the value invested in the 6% account.
(This exercise's set-up used that "how much is left" construction, mentioned earlier.)
_____________________________________________________________________________________
6) An investor deposited an amount of money into a high-yield mutual fund that returns
a9% annual simple interest rate. A second deposit, $2,500 more than the first, was placed in a
certificate of deposit that returns a 5% annual simple interest rate. The total interest earned on
both investments for one year was $475. How much money was deposited in the mutual fund?
The amount invested in the CD is defined in terms of the amount invested in the mutual fund,so I
will let "x" be the amount invested in the mutual fund.
mutual fund
cert. of deposit
total
I
0.09x
(x + 2,500)(0.05)
475
P
x
x + $2,500
2x + $2,500
r
0.09
0.05
---
In this problem, I don't actually need the "total" for the "rate" or "time" columns.
First I'll fill in the P, r, and t columns, multiplying to the left to fill in the I column.
Then I'll add down the I column to get the equation 0.09x + (x + 2,500)(0.05) = 475.
t
1
1
---
Then I'd solve for the value of x. Copyright © Elizabeth Stape999-2010 All Rights Reserved
____________________________________________________________________________________
7) The manager of a mutual fund placed 30% of the fund's available cash in a 6% simple interest
account, 25% in 8% corporate bonds, and the remainder in a money market fund that
earns 7.5% annual simple interest. The total annual interest from the investments was $35,875.
What was the total amount invested?
For this problem, I'll let "x" stand for the total amount invested.
6% account
8% bonds
7.5% fund
total
I
(0.30x)(0.06) = 0.018x
(0.25x)(0.08) = 0.02x
(0.45x)(0.075) = 0.03375x
$35,875
P
0.30x
0.25x
0.45x
x
r
0.06
0.08
0.075
---
t
1
1
1
---
Once 30% and 25% was accounted for in the 6% and 8% accounts, then there is 100% – 30% –
25% = 45% left for the third account. I can use this information to fill in the "principal" column.
Then I'll fill out the "rate" and "time" columns, and multiply to the left to fill in the "interest"
column.
From the interest column, I get the equation 0.018x + 0.02x + 0.03375x = 35,875.
Then I'd solve for the value of x.
_____________________________________________________________________________________
"Number" Word Problems
_____________________________________________________________________________________
1) The sum of two consecutive integers is 15. Find the numbers.
"The numbers are 7 and 8."
_____________________________________________________________________________________
2) The product of two consecutive negative even integers is 24. Find the numbers.
They have told me quite a bit about these two numbers: the numbers are even and they are
negative. (The fact that they are negative may help if I come up with two solutions — a positive
and a negative — so I'll know which one to pick.) Since even numbers are two apart (for
example,–4 and –2 or 10 and 12), then I also know that the second number is two greater than
the first. I also know that, when I multiply the two numbers, I will get 24. In other words, letting
the first number be "n" and the second number be "n + 2", I have:
(n)(n + 2) = 24
n2 + 2n = 24
n2 + 2n – 24 = 0
(n + 6)(n – 4) = 0
Then the solutions are n = –6 and n = 4. Since the numbers I am looking for are negative, I can
ignore the "4" and take n = –6. Then the next number is n + 2 = –4, and the answer is
The numbers are –6 and –4.
In the exercise above, one of the answers was one of the solutions to the equation; the other answer was
the negative of the other solution to the equation. Warning: Do not assume that you can use both solutions
if you just change the signs to be whatever you feel like. While this often "works", it does
notalways work, and it's sure to annoy your teacher. Throw out invalid results, and solve properly for
valid ones.
_____________________________________________________________________________________
3) Twice the larger of two numbers is three more than five times the smaller, and the sum of four
times the larger and three times the smaller is 71. What are the numbers?
The point of exercises like this is to give you practice in unwrapping and unwinding these words, and
turning the words into algebraic equations. The point is in the solving, not in the relative "reality" of the
problem. That said, how do you solve this? The best first step is to start labelling:
the larger number: x
the smaller number: y
twice the larger: 2x
three more than five times the smaller: 5y + 3
relationship between ("is"): 2x = 5y + 3
four times the larger: 4x
three times the smaller: 3y
relationship between ("sum of"): 4x + 3y = 71
Now I have two equations in two variables:
2x = 5y + 3
4x + 3y = 71
I will solve, say, the first equation for x:
x = (5/2)y + (3/2)
Then I'll plug the right-hand side of this into the second equation in place of the "x":
4[ (5/2)y + (3/2) ] + 3y = 71
10y + 6 + 3y = 71
13y + 6 = 71
13y = 65
y = 65/13 = 5
Now that I have the value for y, I can solve for x:
x = (5/2)y + (3/2)
x = (5/2)(5) + (3/2)
x = (25/2) + (3/2)
x = 28/2 = 14
As always, I need to remember to answer the question that was actually asked. The solution here
is not "x = 14", but is the following sentence:
The larger number is 14, and the smaller number is 5.
The trick to doing this type of problem is to label everything very explicitly. Until you become used to
doing these, do not attempt to keep track of things in your head. Do as I did in this last example: clearly
label every single step. When you do this, these problems generally work out rather easily.
_____________________________________________________________________________________
"Percent of" Word Problems
_____________________________________________________________________________________
1) A golf shop pays its wholesaler $40 for a certain club, and then sells it to a golfer for$75. What is
the markup rate?
First, I'll calculate the markup in absolute terms:
75 – 40 = 35
Then I'll find the relative markup over the original price, or the markup rate: ($35) is (some
percent) of ($40), or: Copyright © Elizabeth Stapel 1999-2009 All Rights Reserved
35 = (x)(40)
...so the relative markup over the original price is:
35 ÷ 40 = x = 0.875
Since x stands for a percentage, I need to remember to convert this decimal value to the
corresponding percentage.
The markup rate is 87.5%.
_____________________________________________________________________________________
2) A shoe store uses a 40% markup on cost. Find the cost of a pair of shoes that sells for$63.
This problem is somewhat backwards. They gave me the selling price, which is cost plus
markup, and they gave me the markup rate, but they didn't tell me the actual cost or markup. So I
have to be clever to solve this.
I will let "x" be the cost. Then the markup, being 40% of the cost, is 0.40x. And the selling price
of $63 is the sum of the cost and markup, so:
63 = x + 0.40x
63 = 1x + 0.40x
63 = 1.40x
63 ÷ 1.40 = x= 45
The shoes cost the store $45.
_____________________________________________________________________________________
3) An item originally priced at $55 is marked 25% off. What is the sale price?
First, I'll find the markdown. The markdown is 25% of the original price of $55, so:
x = (0.25)(55) = 13.75
By subtracting this markdown from the original price, I can find the sale price:
55 – 13.75 = 41.25
The sale price is $41.25.
_____________________________________________________________________________________
4) An item that regularly sells for $425 is marked down to $318.75. What is the discount rate?
First, I'll find the amount of the markdown:
425 – 318.75 = 106.25
Then I'll calculate "the markdown over the original price", or the markdown rate: ($106.25) is
(some percent) of ($425), so:
106.25 = (x)(425)
...and the relative markdown over the original price is:
x = 106.25 ÷ 425 = 0.25
Since the "x" stands for a percentage, I need to remember to convert this decimal to percentage
form.
The markdown rate is 25%.
_____________________________________________________________________________________
5) An item is marked down 15%; the sale price is $127.46. What was the original price?
This problem is backwards. They gave me the sale price ($127.46) and the markdown rate(15%),
but neither the markdown amount nor the original price. I will let "x" stand for the original price.
Then the markdown, being 15% of this price, was 0.15x. And the sale price is the original price,
less the markdown, so I get:
x – 0.15x = 127.46
1x – 0.15x = 127.46
0.85x = 127.46
x = 127.46 ÷ 0.85 = 149.952941176...
This problem didn't state how to round the final answer, but dollars-and-cents is always written
with two decimal places, so:
The original price was $149.95.
Note in this last problem that I ended up, in the third line of calculations, with an equation that said
"eighty-five percent of the original price is $127.46". You can save yourself some time if you think of
discounts in this way: if the price is 15% off, then you're only actually paying 85%. Similarly, if the price
is 25% off, then you're paying 75%; if the price is 30% off, then you're paying 70%; and so on.
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Mixture Word Problems
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Work Word Problems
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