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Algebra
Important Terms
Algebra
Algebraic Equation
Algebraic Expression
Constant Term
Like Terms
代數
代數方程
代數式
常數項
同類項
Solution
Solve Equation
Term
Transpose Terms
Unknown
解
解方程
項
移項
未知數
Principle of Balance
天秤法
Unlike Terms
異類項
2.1
Algebraic Expressions
Unknowns are mostly represented by letters such as a, b, c, x, y, z, etc. Operations of unknowns are
studied under a branch of Mathematics called Algebra.
Note
An expression where number(s) and letter(s) are connected by
operation sign(s) ‘+’, ‘–’, ‘’, ‘’ is called an algebraic expression.
Below are some examples of using algebraic expressions to express sentences.
Sentence
Algebraic expression
16 – x
1.
x is subtracted from 16.
2.
Three times y is added to 8.
8+3×y
3.
Half of a is multiplied by 4, then the result is added
to b.
b + 4(
a
)
2
4.
Subtract the result of dividing B by 6 from the
product of A and 2.
2A –
B
6
In this algebraic expression, 3 × y
can also be written as 3  y or 3y.
Example
Mary has just bought 3 boxes of cakes from a cake shop and each box contains
K cakes. She gives 5 cakes to John. Use an algebraic expression to represent the
number of cakes that Mary still has.
Solution
The number of cakes that Mary still has = 3K – 5
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Class Practice
Express the following sentences by using algebraic expressions.
Algebraic expression
1.
Add 35 to a.
____________________
2.
Multiply A by 4 and B by 2 respectively, and then add the two
products.
____________________
3.
Four times S is subtracted from half of T.
____________________
4.
K is divided by 12, and then the result is subtracted from five
times P.
____________________
The difference of two numbers is divided by the product of
these two numbers.
____________________
5.
An algebraic expression can be separated into several parts by either ‘+’ sign or ‘–‘ sign. Each part
together with the sign in front of it is called a term.
e.g. For algebraic expression 3x + 5 – 4y + 2z, each of 3x, +5, –4y and +2z is a term.
When writing the terms +5 and +2z, we may omit ‘+’.
Note
(a)
(b)
(c)
Example
Solution
Constant terms are terms which contain numbers only.
Like terms are terms which contain same letter(s) to the same power(s).
Unlike terms are terms which are not like terms.
How many terms are there in the algebraic expression 2x + 3y – 5 – x? Identify the
constant term, a pair of like terms and a pair of unlike terms.
There are 4 terms in the algebraic expression 2x + 3y – 5 – x, namely, 2x, 3y, –5
and –x.
Constant term: –5
A pair of like terms: 2x and –x
There are 1 pair of like terms
A pair of unlike terms: 2x and 3y
and 5 pairs of unlike terms.
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Class Practice
For each of the algebraic expressions in the table below, list the constant term, a pair of like terms
and a pair of unlike terms.
Algebraic expression
1.
a + 2b – 6 – 3b
2.
5y + 2 + 4x – y
3.
c – 3d + 8 –
Constant term
A pair of like terms
A pair of unlike terms
d
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Exercise 2.1
How many terms are there in each of the following algebraic expressions? (1 – 3)
1.
7x – 8 + 2y + 9
Number of terms = ________
2.
6h – 4k – k + 5hk
Number of terms = ________
3.
a
+ 5 – 6a
3
Number of terms = ________
Write down a pair of like terms in each of the following algebraic expressions. (4 – 5)
4.
5p – 2 + 9q – p
5.
3hk + 2k +
A pair of like terms: ____________________
k
–2
2
A pair of like terms: ____________________
Write down a pair of unlike terms in each of the following algebraic expressions. (6 – 7)
6.
10a – a + 2b
A pair of unlike terms: ___________________
7.
4m – 2 – 3mn
A pair of unlike terms: ___________________
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2.2
Algebraic Equations in One Unknown
2.2.1
Forming Algebraic Equations
An algebraic equation consists of two expressions (algebraic or arithmetic) connected by an equal
sign ‘=’ where at least one of the two expressions contains one or more unknowns.
x
+ y = 10, 4m – 5 = n, 2a + 4 = 8 are examples of algebraic equations.
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Is 8x – 9 + 2 an algebraic
equation?
Note
An algebraic equation that contains one unknown only is called an
algebraic equation in one unknown.
e.g. 2a + 4 = 8 is an algebraic expression in one unknown while 4m – 5 = n is not.
Example
Express the following sentences by using algebraic equations.
(a) Half of d is subtracted from 30, the result is 6.
(b) John has x $2 coins. After buying an ice-cream which costs $5, he has $45 left.
Solution
(a) 30 –
d
=6
2
(b) 2x – 5 = 45
Class Practice
Express the following sentences by using algebraic equations.
Algebraic equation
1.
Five times k is equal to the sum of k and 8.
____________________
2.
Mr Lam’s hourly wage is $150. He worked for m hours last week
and earned $2250.
____________________
3.
Nancy has 105 stamps and Bob has y stamps. They have 230 stamps
altogether.
____________________
4.
In an English language test, Anson got x marks. The mark of Sam
was twice that of Anson. George got 85 marks and his mark was 5
more than Sam’s.
____________________
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2.2.2
Solving Algebraic Equations
A value of the unknown that satisfies an equation is called a solution of that equation. The process
of finding the solution of an equation is called solving equation. We can solve equations by
different methods.
A. Principle of balance
Whatever we do on one side of an equation, we do the same on the other side in order to keep the
values of both sides equal.
Example
Solve the following equations by the principle of balance.
(a) a + 12 = 30
(b) k – 25 = 7
(c) 3x = 24
(d)
n
=9
5
(e) 4p – 2 = 22
Solution
(a)
a + 12 = 30
Both sides minus 12
a + 12 – 12 = 30 – 12
a = 18
(b)
k – 25 = 7
Both sides plus 25
k – 25 + 25 = 7 + 25
k = 32
(c)
3x = 24
3x 24
=
3
3
Both sides divided by 3
x=8
(d)
n
=9
5
n
5=95
5
Both sides times 5
n = 45
(e)
4p – 2 = 22
Both sides plus 2
4p – 2 + 2 = 22 + 2
4p = 24
4 p 24
=
4
4
Both sides are divided by 4
p=6
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Class Practice
Solve the following equations by the principle of balance.
1.
y + 14 = 26
x
=8
4
3.
5.
2m + 5 = 17
2.
b – 7 = 11
4.
9a = 63
6.
n
– 12 = 0
3
B. Transposing terms
We can move a term in an equation from one side to the other side by changing its sign (‘+’, ‘–’, ‘’,
‘’). We call this process transposing terms.
Example
Solve the following equations by transposing terms.
(a) y + 5 = 12
(b) x – 6 = 24
(c) 4b = 36
(d)
Solution
a
= 11
8
(a) y + 5 = 12
y = 12 – 5
+5–5
y=7
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(b) x – 6 = 24
x = 24 + 6
x = 30
–6+6
(c) 4b = 36
b=
36
4
×44
b=9
(d)
a
= 11
8
a = 11  8
a = 88
8×8
Class Practice
Solve the following equations by transposing terms.
1.
x – 18 = 19
2.
6a = 30
3.
p
=6
8
4.
k + 2 = 21
Exercise 2.2
Express the following sentences by using algebraic equations. (1 – 2)
Algebraic equation
1.
The result of subtracting 5d from 30 is equal to 16.
____________________
2.
The height of Joey is y cm and she is 15 cm taller than Sue. The
height of Sue is 110 cm.
____________________
Solve the following equations.
3.
6x = 18
x = ______
4.
a – 5 = 20
a = ______
5.
b + 9 = 21
b = ______
6.
n
=9
3
n = ______
7.
8 + y = 23
y = ______
8.
k ÷ 2 = 20
k = ______
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2.3
More Complicated Algebraic Equations in One
Unknown
2.3.1
Addition and Subtraction of Like Terms
For a pair of like terms like 3x and 2x, 3x can be regarded as three x’s while 2x can be regarded as
two x’s.
Thus 3x + 2x can be regarded as putting three x’s and two x’s together. From which, we have five
x’s.
i.e. 3x + 2x = (3 + 2)x = 5x
Similarly, 3x – 2x can be regarded as taking two x’s away from three x’s. From which, we have one
x.
i.e. 3x – 2x = (3 – 2)x = x
Example
Simplify the following algebraic expressions.
(a) 2.5x + x
(b) 4y – 3y + 2
Solution
(a) 2.5x + x = (2.5 + 1)x
= 3.5x
(b) 4y – 3y + 2 = (4 – 3)y + 2
=y+2
Class Practice
Simplify the following algebraic expressions.
1.
5x + 2x
____________________
2.
4.3x + x
____________________
3.
8y – 3y
____________________
4.
12.5y – 0.5y
____________________
5.
7a + 3 + 3a
____________________
6.
9d – 4d – 1
____________________
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2.3.2
Solving More Complicated Algebraic Equations
For a more complicated algebraic equation like 2x + 3x = 10, we can solve it by first simplifying the
expression 2x + 3x as 5x, and then solve the resulting equation 5x = 10.
Example
Solve the following equations by the principle of balance.
(a) 2a – a = 7
(b) 3.5k + 1.5k = 15
(c) 3x = 16 + x
(d) 4p + 4 = 9 – p
Solution
(a) 2a – a = 7
2a – a = (2 – 1)a
a=7
(b) 3.5k + 1.5k = 15
3.5k + 1.5k = (3.5 + 1.5)k
5k = 15
5k 15
=
5
5
k=3
(c)
3x = 16 + x
3x – x = 16
+x–x
2x = 16
2x 16
=
2
2
x=8
(d) 4p + 4 = 9 – p
4p + p = 9 – 4
5p = 5
5p 5
=
5
5
p=1
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Class Practice
Solve the following equations by the principle of balance.
1.
4.2m – 3.2m = 9
2.
8p – 2p = 18
3.
5n = 5 + 4n
4.
6w – 10 = 5 + w
Exercise 2.3
Simplify the following algebraic expressions. (1 – 4)
1.
3.2x + 2.3x
____________________
2.
10y – 7y
____________________
3.
6k + 8k – 5
____________________
4.
9 + 12h – 4h
____________________
Solve the following equations by the principle of balance. (5 – 8)
5.
5.5a – 1.5a = 8
a = ______
6.
b = 20 – 4b
b = ______
7.
9t – 12 = 5t
t = ______
8.
15 – 7k = 4k – 7
k = ______
Set up an equation and answer the following question.
9.
The price of a skirt is $C and the price of a coat is six times that of the skirt. Jane needs to pay
$490 to buy one skirt and one coat. How much is the skirt?
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