Download Document

1A_Ch2(1)
2.1 Introduction to Algebra
1A_Ch2(2)
A Using Letters to
Represent Numbers
B The Advantages of Using
Algebraic Language
C More about Algebraic
Expressions
Index
2.2 Algebraic Equations in One
Unknown
1A_Ch2(3)
A Formulate Algebraic
Equations in One Unknown
B Properties of Equations
C Solving Algebraic
Equations in One Unknown
Index
2.1 Introduction to Algebra
1A_Ch2(4)
 Example
A)
Using Letters to Represent Numbers
1. In algebra, we can use letters to represent numbers.
The number represented by a letter is called an
unknown.
2. Expressions formed by numbers, unknowns and the
signs +, –, ×, ÷ etc. are called algebraic
expressions.
 Index 2.1
Index
2.1 Introduction to Algebra
1A_Ch2(5)
Jimmy buys 4 boxes of Pokamon model and 10
pieces of trading cards. If each box of Pokamon
model already has m pieces of trading cards
inside, use algebraic language to write the total
number of trading cards that Jimmy now has.
The total number of trading cards Jimmy now has
= 4 × m + 10
Index
2.1 Introduction to Algebra
1A_Ch2(6)
Peggy is t years old now and her brother is 4
years older than her. Use algebraic language to
write the sum of the ages of Peggy and her
brother.
The age of her brother
=t+4
The sum of the ages of Peggy and her brother
=t+t+4
 Key Concept 2.1.1
Index
2.1 Introduction to Algebra
B)
1A_Ch2(7)
The Advantages of Using Algebraic Language
1. It is simple and direct, so it is easy to read
and understand.
2. It can be an international language.
Index
2.1 Introduction to Algebra
1A_Ch2(8)
 Example
B)
Word phase
Algebraic expression
1. Add 9 to an unknown number.
2. Subtract 6 from an unknown
number and multiply the
difference by 2.
3. Divide an unknown even
number by 2 and then add 1 to
the quotient.
x+9
(y – 6) × 2
z÷2+1
 Index 2.1
Index
2.1 Introduction to Algebra
1A_Ch2(9)
Represent each of the following word phrases by an algebraic
expression. Use the letter “x” to represent the unknown.
(a) Multiply a number by 6 and add 3 to the product.
(b) Divide a number by 4 and then subtract 5 from the quotient.
(a) 6x + 3
x
(b)  5
4
 Key Concept 2.1.3
Index
2.1 Introduction to Algebra
1A_Ch2(10)
C) More about Algebraic Expressions
1. For the algebraic expression x + 2y – 3x + 5,
i. x, 2y, –3x, 5 are the terms of the algebraic
expression,
ii. 5 is the constant term,
iii. x and –3x are like terms,
iv. x and 2y is a pair of unlike terms.
Index
2.1 Introduction to Algebra
C)
1A_Ch2(11)
More about Algebraic Expressions
2. In an algebraic expression,
i. numbers and letters that are separated by the signs
‘×’ or ‘÷’ are not considered as two different terms,
E.g. +2 × a (can be written as +2a) is considered
as one term.
ii. the ‘+’ sign before a single term or the first
term is usually skipped.
E.g. +2a is written as 2a, +7b – 6 is written as
7b – 6.
Index
2.1 Introduction to Algebra
1A_Ch2(12)
 Example
C)
More about Algebraic Expressions
3. In an algebraic expression,
i.
like terms can be combined by addition,
subtraction, multiplication or division,
ii. unlike terms, on the other hand, can be
combined by multiplication or division.
 Index 2.1
Index
2.1 Introduction to Algebra
1A_Ch2(13)
How many terms are there in the algebraic expression
2 × a + 5 × a ÷ 6 – 4a × b + 5 × 7 ?
Identify the terms as constant term, like terms and unlike terms.
There are 4 terms in the algebraic expression
2 × a + 5 × a ÷ 6 – 4a × b + 5 × 7.
They are +2 × a (or simply 2a),
5a
),
6
(or simply 35).
+5 × a ÷ 6 (or simply
–4a × b (or simply –4ab), +5 × 7
Index
2.1 Introduction to Algebra
1A_Ch2(14)
 Back to Question
Constant term : 35 ,
5a
like terms : 2a and
6
,
unlike terms : 2a and –4ab ;
2a and 35 ;
5a and –4ab ;
6
5a and 35 ;
6
–4ab and 35.
Index
2.1 Introduction to Algebra
1A_Ch2(15)
Dick intends to save $100 every week. Use an algebraic
expression to represent each of the following answers.
(a) After x weeks, how much has he saved?
 Soln
(b) If Dick takes out $25 each week and spends the
money on food, how much has he saved after x
weeks?
 Soln
(c) If Dick takes out $100 from his total savings after
x weeks and spends the money on toys, find the
actual amount that Dick has saved after x weeks.
 Soln
Index
2.1 Introduction to Algebra
1A_Ch2(16)
 Back to Question
(a) After x weeks, the amount that Dick has saved
= $100x
(b) Since Dick takes out $25 each week on food,
the actual amount that he saves each week
= $(100 – 25) = $75
∴ The amount that he has saved after x weeks
= $75x
Index
2.1 Introduction to Algebra
1A_Ch2(17)
 Back to Question
(c) From part (a), the amount that Dick
has saved after x weeks
= $100x
But he takes out $100 to buy toys,
∴ the actual amount that he has saved after x weeks
= $(100x – 100)
Fulfill Exercise Objective
 Express the meaning of sentences or the
 Key Concept 2.1.4
answers of questions in algebraic language.
Index
2.2 Algebraic Equations in One Unknown
1A_Ch2(18)
 Example
A) Formulate Algebraic Equations in One Unknown
1. An equation contains an equal sign ‘=’ and also one or
more unknowns.
2. The equation that contains one unknown only is called an
algebraic equation in one unknown.
3. To solve an equation is to find the value of the unknown
in that equation. That value is called the solution of the
equation.
4. To find the solution of an equation is ‘to solve the
equation’.
 Index 2.2
Index
2.2 Algebraic Equations in One Unknown
1A_Ch2(19)
In each of the following, formulate an algebraic equation to find
the value of the unknown.
(a) The cost of a doll is $x and that of a ball is $25. It is known
that the total cost of the doll and the ball is $43.
(b) There are 150 candies in a jar. Kenny has y jars of candies
and it is known that he has 600 candies.
(a) x + 25 = 43, ∴ x = 18
(b) 150y = 600, ∴ y = 4
 Key Concept 2.2.1
Index
2.2 Algebraic Equations in One Unknown
1A_Ch2(20)
 Example
B) Properties of Equations
For three numbers a, b and c, if a = b, then
1. a + c = b + c
2. a – c = b – c
3. ac = bc
a
b
4.
(where c  0)
=
c
c
 Index 2.2
Index
2.2 Algebraic Equations in One Unknown
1A_Ch2(21)
There are a red flowers and b yellow flowers in the vase. Mrs
Law picks c red flowers and c yellow flowers from the vase. How
many red flowers and yellow flowers in the vase now? If the
original numbers of red flowers and yellow flowers are the same,
write down the relationship between the numbers
of the flowers by using the algebraic expression.
Numbers of flowers in the vase:
Red flowers = a – c ; Yellow flowers = b – c
If the original numbers of red flowers and yellow flowers
are the same, then a – c = b – c .
Index
2.2 Algebraic Equations in One Unknown
1A_Ch2(22)
Helen has a pears and b oranges. She divides the pears and oranges
among her brothers. It is known that Helen has c brothers. How many
pears and oranges that each of Helen’s brothers can get? If the
numbers of pears and oranges are the same, write
down the relationship between the numbers of
the fruits by using the algebraic expression.
Numbers of fruits that each of Helen’s brothers can get:
a
Pears =
c
; Oranges =
b
c
If the numbers of pears and oranges are the same,
then a  b .
c
c
 Key Concept 2.2.2
Index
2.2 Algebraic Equations in One Unknown
1A_Ch2(23)
 Example
C) Solving Algebraic Equations in One Unknown
1. We may apply guessing and checking or apply the same
operation to the two sides of the equation to solve
algebraic equations.
Index
2.2 Algebraic Equations in One Unknown
1A_Ch2(24)
C) Solving Algebraic Equations in One Unknown
2. Skills to solve equations
i.
Transposing Terms
 Example
The terms in an equation can be transposed from one
side of the equal sign to the other. At the same time,
the signs of the terms must be changed.
For example:
‧If x – a = b, after transposing terms, we have x = b + a.
‧If x + a = b, after transposing terms, we have x = b – a.
‧If a = b – x, after transposing terms, we have x + a = b.
Index
2.2 Algebraic Equations in One Unknown
1A_Ch2(25)
C) Solving Algebraic Equations in One Unknown
2. Skills to solve equations
ii. Removing Brackets  Example
When brackets are involved in an equation, they
should be removed first in order to simplify the
equation.
In general, we can remove brackets in the following way:
a(b + c) = ab + ac
or (b + c)a = ba + ca
Index
2.2 Algebraic Equations in One Unknown
1A_Ch2(26)
C) Solving Algebraic Equations in One Unknown
2. Skills to solve equations
iii. Cancelling Denominators  Example
When there are denominators in an equation, we
can multiply the two sides of the equation at the
same time by the L.C.M. of the denominators, so
that the equation is simplified.
 Index 2.2
Index
2.2 Algebraic Equations in One Unknown
1A_Ch2(27)
Use the method of guessing and checking to solve the equation
x – 5 = 15.
i.
Try
x = 10,
then x – 5 = 10 – 5
= 5  15
ii. Try
x = 15,
then x – 5 = 15 – 5
= 10  15
iii. Try
x = 20,
then x – 5 = 20 – 5
= 15
∴ 20 is the required solution of the equation.
 Key Concept 2.2.3
Index
2.2 Algebraic Equations in One Unknown
1A_Ch2(28)
Solve the following equations.
(a) x – 8 = 14
(c) 10 = 5 – x
(b) x + 8 = 14
(a) After transposing terms, x = 14 + 8
= 22
(b) After transposing terms, x = 14 – 8
= 6
(c) After transposing terms, x + 10 = 5
Again, after transposing terms, x = 5 – 10
= –5
Index
2.2 Algebraic Equations in One Unknown
1A_Ch2(29)
Solve the equation 1 = 30 + 4x – 5.
1 = 30 + 4x – 5
1 – 30 + 5 = 4x
–24 = 4x
 24 4x
=
4
4
–6 = x
i.e.
Fulfill Exercise Objective
 Solve equations by using the
technique of transposing terms.
x = –6
Index
2.2 Algebraic Equations in One Unknown
1A_Ch2(30)
Solve the equation 8x + 18 – 5x = 24.
8x + 18 – 5x = 24
8x – 5x + 18 = 24
3x + 18 = 24
3x = 24 – 18
3x = 6
3x
6
=
3
3
x= 2
Fulfill Exercise Objective
 Solve equations by using the
technique of transposing terms.
Index
2.2 Algebraic Equations in One Unknown
1A_Ch2(31)
Solve the equation 8m + 11 – 2m = 16 – 4m.
8m + 11 – 2m = 16 – 4m
6m + 11 = 16 – 4m
4m + 6m + 11 = 16
10m + 11 = 16
10m = 16 – 11
10m = 5
10m 5
=
10
10
1
m=
2
Fulfill Exercise Objective
 Solve equations by using the
technique of transposing terms.
 Key Concept 2.2.4
Index
2.2 Algebraic Equations in One Unknown
1A_Ch2(32)
Solve the equation 3(x – 2) = 9.
3(x – 2) = 9
3x – 6 = 9
3x = 9 + 6
3x = 15
3x 15
=
3
3
x= 5
Index
2.2 Algebraic Equations in One Unknown
1A_Ch2(33)
Solve the equation 3(x – 3) – 2(x + 5) = 6.
3(x – 3) – 2(x + 5) = 6
3x – 9 – 2x – 10 = 6
x – 19 = 6
x = 6 + 19
= 25
Fulfill Exercise Objective
 Remove brackets and use the technique of
transposing terms to solve equations.
Index
2.2 Algebraic Equations in One Unknown
1A_Ch2(34)
Solve the equation 6[2(y + 6) – 3] = 1.5y – 30.
6[2(y + 6) – 3] = 1.5y – 30
6[2y + 12 – 3] = 1.5y – 30
6[2y + 9] = 1.5y – 30
12y + 54 = 1.5y – 30
12y – 1.5y = –30 – 54
10.5y = –84
 84
y
10.5
= –8
Fulfill Exercise Objective
 Remove brackets and use
the technique of transposing
terms to solve equations.
 Key Concept 2.2.5
Index
2.2 Algebraic Equations in One Unknown
1A_Ch2(35)
3x x 1
  .
Solve the equation
4 3 2
3x x 1
 
4 3 2
3x x
1
12(  )  12 
4 3
2
9x  4x  6
5x  6
6
x
5
Index
2.2 Algebraic Equations in One Unknown
1A_Ch2(36)
2
Solve the equation x  x  21 .
5
2
x  x  21
5
2
5( x  x)  21 5
5
2 x  5 x  105
7 x  105
105
x
7
 15
Fulfill Exercise Objective
 Solve equations that involve
fractions without finding the
L.C.M. of the denominators.
Index
2.2 Algebraic Equations in One Unknown
1A_Ch2(37)
3t  1 t 1
  .
Solve the equation
4
2 3
3t  1 t 1
 
4
2 3
3t  1
t 1
12(
)  (  )12
4
2 3
3(3t  1)  6t  4
9t  3  6t  4
9t  6t  4  3
3t  1
1
t
3
1
 
3
Fulfill Exercise Objective
 Solve equations that involve
fractions by finding the L.C.M.
of the denominators first.
 Key Concept 2.2.6
Index
2.3 More on Forming and Solving Algebraic Equations
1A_Ch2(38)
Forming and Solving Algebraic Equations in One Unknown
According to a given situation, we can form an equation to solve
a problem. The steps are:
1. Identify the unknown in the question.
2. Choose a letter to represent the unknown.
3. Form an equation according to the information of the
question.
4. Solve the equation.
5. Write down the answer to the question clearly.
Index
2.3 More on Forming and Solving Algebraic Equations
1A_Ch2(39)
 Example
Note:
i.
Before writing down the answer, we should check
the solution obtained to see if it satisfies the
original question.
ii. We should consider whether an appropriate unit
is required in the answer.
Index
2.3 More on Forming and Solving Algebraic Equations
1A_Ch2(40)
The total weight of Amy and Peter is represented by the
algebraic expression (45 + p) kg, where p kg stands for
the weight of Peter. If the total weight is 108 kg, find the
weight of Peter.
Step 1 : The unknown has already been identified here.
Step 2 : The letter p is chosen as the unknown.
Index
2.3 More on Forming and Solving Algebraic Equations
1A_Ch2(41)
 Back to Question
Step 3 : According to the question, the total
weight of Amy and Peter is 108 kg.
We can set up the equation as
45 + p = 108.
Step 4 : Solve the equation :
45 + p = 108
p = 108 – 45
= 63
Step 5 : ∴ The weight of Peter is 63 kg.
Index
2.3 More on Forming and Solving Algebraic Equations
1A_Ch2(42)
The cost of spare battery and memory cards
is represented by the algebraic expression
$(800 + 300C), where C stands for the
number of memory cards that the French
student Joan wants to buy. If Joan has to pay
$2 000 in total, how many memory cards has
she bought?
Step 1 : The unknown has already been identified here.
Step 2 : The letter C is chosen as the unknown.
Index
2.3 More on Forming and Solving Algebraic Equations
1A_Ch2(43)
 Back to Question
Step 3 : According to the question, Joan has to pay $2 000
for her purchase. We can set up the equation as
800 + 300C = 2 000.
Step 4 : Solve the equation :
800 + 300C = 2 000
300C = 2 000 – 800
300C = 1 200
1 200
C
300
= 4
Fulfill Exercise Objective
 Solve equations without
brackets or fractions in
word problems.
Step 5 : ∴ Joan has bought 4 memory cards.
Index
2.3 More on Forming and Solving Algebraic Equations
1A_Ch2(44)
In a concert, the price of an adult ticket was $150 while
that of a student ticket was $80. If altogether 1 500 tickets
were sold and the income was $204 000,
how many adult tickets have been sold?
Let x be the number of adult tickets sold.
Then, the number of student tickets sold is 1 500 – x.
The income from adult tickets = $150x
The income from student tickets = $80(1 500 – x)
Index
2.3 More on Forming and Solving Algebraic Equations
1A_Ch2(45)
 Back to Question
According to the question, we obtain the equation
150x + 80(1500 – x) = 204 000
150x + 120 000 – 80x = 204 000
150x – 80x = 204 000 – 120 000
70x = 84 000
84 000
x
70
= 1 200
∴ The number of adult tickets sold = 1 200
Fulfill Exercise Objective
 Solve equations involving brackets in word problems.
Index
2.3 More on Forming and Solving Algebraic Equations
1A_Ch2(46)
A Nature Education Track is 4 km long. It is divided into
4
two sections A and B. Martin has walked
of Section A
5
while Bobby has walked 1 of Section B. It is known that
3
Martin has walked 1 1 km more than Bobby. Find the
2
lengths of Section A and Section B.
Let l km be the length of Section A.
Then (4 – l) km is the length of Section B.
Index
2.3 More on Forming and Solving Algebraic Equations
1A_Ch2(47)
 Back to Question
The distance Martin has walked
4
= of Section A
5
4
 l km
5
The distance Bobby has walked
1
of Section B
3
1
 (4  l ) km
3
=
Index
2.3 More on Forming and Solving Algebraic Equations
1A_Ch2(48)
 Back to Question
According to the question, we obtain the equation
4 1
1
l  (4  l )  1
5 3
2
4 1
3
30[ l  (4  l )]   30
5 3
2
24l – 10(4 – l) = 45
24l – 40 + 10l = 45
24l + 10l = 45 + 40
34l = 85
85
34
= 2.5
Fulfill Exercise Objective
 Solve equations involving
fractions in word problems.
l
∴
The length of Section A = 2.5 km
The length of Section B = (4 – 2.5) km
= 1.5 km
 Key Concept 2.3
Index