Using Algebra to Solve Problems

1
Using Algebra to Solve Problems
3
Mathematics in Workplaces
3.1 Introduction to Algebra
3.2 Formulas and Substitution
3.3 Solving Equations in One Unknown
3.4 More about Solving Equations in One Unknown
3.5 Applications of Equations in One Unknown
Chapter Summary
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Mathematics in Workplaces
Programmer Do you have a computer at home? Do
you know how a computer works?
A computer operates with its own language called
a program. A program is a set of instructions,
which are coded involving many mathematical
calculations, expressions and symbols. It
instructs the computer to perform a sequence of
operations.
A programmer is responsible for designing and
writing the lines of code with a simple and clear
computer language, to ensure that the computer
operates efficiently.
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3.1 Introduction to Algebra
A. Representing Unknowns by Symbols
In primary school, we learnt that x is called the
unknown.
Besides x, we can use other letters or symbols to
represent unknowns.
We can present the relationship between numbers
by connecting unknowns in an expression. This
helps people understand the relations among
different unknowns.
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3.1 Introduction to Algebra
B. Algebraic Expressions
We can use symbols and the 4 basic arithmetic
operations ?, ?, ?, ? to form algebraic
expressions, to represent relation among
different quantities.
Suppose that the letters h and k represent any 2
numbers.
1. Addition Add h to kor the sum of h and k
can be represented as h ? k.
2. Subtraction Subtract k from h or h minus
k can be represented as h ? k.
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3.1 Introduction to Algebra
B. Algebraic Expressions
3. Multiplication Multiply h by k, the
product of h and k or h times k can be
represented as h ? k or hk.
(1) The order of multiplication can be
interchanged. For example, hk ? kh.
(2) Numbers should be written first, such as 6h
and 4k.
Notes In algebraic expressions, upper case
letters and lower case letters represent
different numbers. For example, 3 ? A ? 3 ? a.
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Example 3.1T
3.1 Introduction to Algebra
B. Algebraic Expressions

• Write an algebraic expression for each of the
  following word sentences.
• (a) Subtract 3 times a from b.
• Divide x by the sum of 1 and y.

Solution
(a) Since 3 times a ? 3 ? a ? 3a
(b) Since the sum of 1 and y ? 1 ? y,
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Example 3.2T
3.1 Introduction to Algebra
B. Algebraic Expressions
The cost of a bottle of milk and a carton of
orange juice are p and q respectively. (a) Find
the total cost of 3 bottles of milk and 3 cartons
of orange juice. (b) If the total cost in (a) is
equal to 10 cartons of strawberry juice,
find the cost of each carton of strawberry
juice. Express your answer in terms of p and q.
Solution
(a) Cost of 3 bottles of milk ? (p ? 3) ?
3p
Cost of 3 cartons of orange juice ? (q ? 3)
3q
(b) Cost of each canton of strawberry juice
? (3p ? 3q) ? 10
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3.1 Introduction to Algebra
B. Algebraic Expressions
Notes 1. The 4 basic operations (?, ?, ?, ?)
in algebra are similar to those in arithmetic.
2. When using a and b to represent unknowns,
their values vary in different situations. We
say that they are variables.
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3.1 Introduction to Algebra
B. Algebraic Expressions
In an algebraic expression, each part separated
by the symbol ? or ? is called a term.
If 2 terms in an expression contain the same
unknown(s), they are called like terms, such as
4b and ?2b.
To simplify an algebraic expression, we
should (1) remove the ? and ? signs
(2) add or subtract the like terms.
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Example 3.3T
3.1 Introduction to Algebra
B. Algebraic Expressions
Simplify the expression 6p ? 9q ? r.
Solution
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3.2 Formulas and Substitution
Consider a rectangle of length x cm and width y
cm.
If we use variables P cm and A cm2 to represent
the perimeter and the area respectively, we have
the following formulas
P 2(x y) and A xy
Once the values of x and y are known, we can find
the corresponding unknown values of P and A from
the formulas by the method of substitution.
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Example 3.4T
3.2 Formulas and Substitution
Solution
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Example 3.5T
3.2 Formulas and Substitution
The figure on the right shows a
triangle. (a) Write down the formula for
calculating the perimeter P (in cm) of the
triangle. (b) Find the perimeter if a ? 8, b ? 6
and c ? 10.
Solution
(a) Perimeter of the triangle ? sum of the
lengths of each side
? P ? a ? 2b ? c
(b) Substitute a ? 8, b ? 6 and c ? 10 into the
formula, we have P ? 8 ? 2(6) ? 10
? 8 ? 12 ? 10 ? 30
? The perimeter is 30 cm.
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Example 3.6T
3.2 Formulas and Substitution
Jane got some red packets at this Lunar New Year,
and each of them contained either a 20 or 50
banknote. Suppose x of them contained a 20
banknote and y of them contained a 50
banknote. (a) Write down the formula for the
total value C of these red packets. (b) Find the
total value of the red packets when x ? 12 and y
? 7.
Solution
(a) C ? 20x ? 50y
(b) C ? 20(12) ? 50(7)
? 240 ? 350 ? 590
? The total value is 590.
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3.3 Solving Equations in One Unknown
A. Introduction to Equations
If we spend 60 to buy n apples at 3 each, then
this situation can be described by the following
statement
3 ? n ? 60 i.e., 3n ? 60
The above statement is called an algebraic
equation.
Since it involves only one unknown n, it is
called an algebraic equation in one unknown.
The solution for the equation is the value for
the unknown that makes both sides of the equation
equal, such as n ? 20.
The process of finding the solutions for
equations is called solving the equations.
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3.3 Solving Equations in One Unknown
B. Basic Methods for Solving Equations
Rules for solving equations
For any numbers m, n and k, if m ? n, then
1. m ? k ? n ? k
2. m ? k ? n ? k
3. m ? k ? n ? k
The above steps in solving equations can be
simplified.
The method of moving terms and reversing their
signs is called transposing terms, as illustrated
in Example 3.7T and Example 3.8T.
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Example 3.7T
3.3 Solving Equations in One Unknown
B. Basic Methods for Solving Equations
Solution
x ? 6 ? ?3 ? 5 ? ?15
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Example 3.7T
3.3 Solving Equations in One Unknown
B. Basic Methods for Solving Equations
Solution
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Example 3.8T
3.3 Solving Equations in One Unknown
B. Basic Methods for Solving Equations
Solve the equation ?3(x ? 9) ? 7 ? 19.
Solution
?3(x ? 9) ? 7 ? 19 ?3(x ? 9) ? 19 ? 7 ?
12
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3.4 More about Solving Equations in One
Unknown
A. Grouping Like Terms
In Section 3.1, we studied how to group like
terms in an algebraic expression.
For equations in the form 6x ? 2x ? 8, we
should first simplify by

• transposing all the like terms to one side
• of the equation

• grouping them together

6x ? 2x ? 8 6x ? 2x ? 8
4x ? 8
x ? 2
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Example 3.9T
3.4 More about Solving Equations in One
Unknown
A. Grouping Like Terms
Solve the equation ?2x ? 15 ? ?5x ? 21.
Solution
?2x ? 15 ? ?5x ? 21 ?2x ? 5x ? 21 ? 15
3x ? 6
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3.4 More about Solving Equations in One
Unknown
B. Removing Brackets
Examples for removing brackets
3(x ? 1)
3x
3x ? 3
3(x ? 1)
3(x ? 1)
3(x ? 1)
4(2x ? 3)
8x ? 12
?2(x ? 3)
?2x ? 6
?3(2x ? 1)
?6x ? 3
If the equation involves brackets, we should
remove the brackets by expanding the terms inside
them first.
Then we can simplify the equation by grouping all
the like terms.
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Example 3.10T
3.4 More about Solving Equations in One
Unknown
B. Removing Brackets
Solve the equation ?2(x ? 5) ? 8(x ? 2) ? 12.
Solution
?2(x ? 5) ? 8(x ? 2) ? 12 ?2x ? 10 ? 8x ? 16 ?
12
?2x ? 8x ? 12 ? 10 ? 16 6x ? 18
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Example 3.11T
3.4 More about Solving Equations in One
Unknown
B. Removing Brackets
Solve the equation 56p ? (2p ? 5) ? 11(2p ? 1).
Solution
56p ? (2p ? 5) ? 11(2p ? 1) 5(6p ? 2p ? 5) ?
22p ? 11
5(4p ? 5) ? 22p ? 11
20p ? 25 ? 22p ? 11 25 11 ? 22p 20p
14 ? 2p
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3.4 More about Solving Equations in One
Unknown
C. Simplifying Fractions
To solve this kind of equation, we should convert
all fractions to a common denominator to simplify
the equation.
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Example 3.12T
3.4 More about Solving Equations in One
Unknown
C. Simplifying Fractions
Solution
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Example 3.12T
3.4 More about Solving Equations in One
Unknown
C. Simplifying Fractions
Solution
Alternative Solution
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Example 3.13T
3.4 More about Solving Equations in One
Unknown
C. Simplifying Fractions
Solution
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3.5 Applications of Equations in One Unknown
The working steps for solving a problem
Step 1 Understand the problem and identify
the unknown in the problem.
Step 2 Assign a symbol/letter to represent
the unknown quantity.
Step 3 Use the information given to set up an
equation.
Step 4 Solve the equation and check the solution.
Step 5 Write down the answer clearly.
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Example 3.14T
3.5 Applications of Equations in One
Unknown
Becky has some stickers. Karen has 6 more
stickers than Becky. If they have 30 stickers
altogether, how many stickers does Becky have?
Solution
Let n be the number of stickers that Becky
has. Then the number of stickers that Karen has ?
n ? 6.
Since their total number of stickers is 30, we
have n ? (n ? 6) ? 30
? The number of stickers that Becky has is 12.
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Example 3.15T
3.5 Applications of Equations in One
Unknown
Jenny packs sweets for a Christmas party. If she
puts 12 sweets more in each package, then there
will be 720 sweets in 15 packages. How many
sweets are there in each package originally?
Solution
Let n be the number of sweets in each package
originally. Then the number of sweets in each new
package ? n ? 12.
Since there will be 720 sweets in 15 new
packages, we have 15(n ? 12) ? 720
? The number of sweets in each original package
is 36.
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Example 3.16T
3.5 Applications of Equations in One
Unknown
Lily has some 5 and 10 coins. The total number
of coins is 20 and the total value of the coins
is 160. How many 5 coins does Lily have?
Solution
Let n be the number of 5 coins that Lily
has. Then the number of 10 coins that Lily has ?
20 ? n.
Since the total value is 160, we have
5n ? 10(20 ? n) ? 160
? Lily has 8 5 coins.
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Chapter Summary
3.1 Introduction to Algebra
1. Basic operations of algebra
Addition x ? y
Subtraction x ? y
Multiplication x ? y ? xy
2. Simplification of algebraic expressions
Algebraic expressions can be simplified by
(a) omitting ? and ? in the expressions.
(b) grouping like terms.
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Chapter Summary
3.2 Formulas and Substitution
Formulas are expressions which relate different
variables. We can find the unknown in a formula
with known values of other variables by using the
method of substitution.
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Chapter Summary
3.3 Solving Equations in One Unknown
We can use the following rules to solve for
equations in one unknown.
For any numbers m, n and k, if m ? n, then 1. m ?
k ? n ? k
2. m ? k ? n ? k
3. m ? k ? n ? k
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Chapter Summary
3.4 More about Solving Equations in One Unknown
We can use the following skills to solve
complicated equations
1. Grouping like terms
2. Removing brackets
3. Simplifying fractions
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Chapter Summary
3.5 Applications of Equations in One Unknown
We can follow the steps below to solve a word
problem
1. Study the problem carefully and identify the
unknown in the problem.
2. Assign a symbol/letter to the unknown.
3. Set up an equation.
4. Solve the equation and check the solution.
5. Write down the answer clearly.
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Follow-up
3.1 Introduction to Algebra
B. Algebraic Expressions
2 unknowns m and n are given. Express the word
sentence m is subtracted from n, and the
difference is then multiplied by 3 as an
algebraic expression.
Solution
Since the subtraction of m from n ? n ? m,
the required result ? (n ? m) ? 3
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Follow-up
3.1 Introduction to Algebra
B. Algebraic Expressions
A van can carry x kg of apples, while a lorry can
carry y kg of apples. (a) How heavy the apples
can be carried by 5 vans and 3 lorries? (b) If
the transportation cost in (a) is 600, what is
the transportation cost for each kg of apple?
Solution
(a) Weight of apples that can be carried by 5
vans ? 5x kg
Weight of apples that can be carried by 3
lorries ? 3y kg
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Follow-up
3.1 Introduction to Algebra
B. Algebraic Expressions
Simplify the expression 6p ? 4q ? 12r.
Solution
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Follow-up
3.2 Formulas and Substitution
Consider k ? 4a ? 1. Find the value of k for the
following values of a by using the method of
substitution. (a) a ? 3 (b) a ? ?2
Solution
(a) k ? 4a ? 1 ? 4(3) ? 1
(b) k ? 4a ? 1 ? 4(?2) ? 1
? 12 ? 1
? ?8 ? 1
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Follow-up
3.2 Formulas and Substitution
The figure on the right shows a
trapezium. (a) Write down the formula for
calculating the area A (in cm2) of the
trapezium. (b) Find the area if b ? 7, d ? 3 and
h ? 5.
Solution
43
Follow-up
3.2 Formulas and Substitution
The figure on the right shows a
trapezium. (a) Write down the formula for
calculating the area A (in cm2) of the
trapezium. (b) Find the area if b ? 7, d ? 3 and
h ? 5.
Solution
(b) Substitute b ? 7, d ? 3 and h ? 5 into the
formula, we have
? The area is 25 cm2.
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Follow-up
3.2 Formulas and Substitution
A supermarket sells 2 different packs of soft
drink. Each small pack contains 10 cans of soft
drink, while each large pack contains 24 cans.
Suppose Ricky bought x small packs and y large
packs. (a) Write down the formula for the total
number of cans N of soft drink that Ricky
bought. (b) Find the total number of cans of soft
drink if he bought 5 small packs and 3 large
packs.
Solution
(a) N ? 10x ? 24y
(b) N ? 10(5) ? 24(3)
50 72 ? 122
? The total number of cans of soft drink is 122.
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Follow-up
3.3 Solving Equations in One Unknown
B. Basic Methods for Solving Equations
Solution
x ? 8 ? 4 ? 4 ? 16
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Follow-up
3.3 Solving Equations in One Unknown
B. Basic Methods for Solving Equations
Solution
47
Follow-up
3.3 Solving Equations in One Unknown
B. Basic Methods for Solving Equations
Solve the equation 4(x ? 7) ? 5 ? 21.
Solution
4(x ? 7) ? 5 ? 21 4(x ? 7) ? 21 ?
5 ? 16
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Follow-up
3.4 More about Solving Equations in One
Unknown
A. Grouping Like Terms
Solve the equation 9x ? 10 ? 3x ? 8.
Solution
9x ? 10 ? 3x ? 8 9x ? 3x ? ?8 ? 10
6x ? ?18
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Follow-up
3.4 More about Solving Equations in One
Unknown
B. Removing Brackets
Solve the equation 3(x ? 4) ? 5(x ? 3) ? 11.
Solution
3(x ? 4) ? 5(x ? 3) ? 11 3x ? 12 ? 5x ? 15 ? 11
3x ? 5x ? 11 ? 12 ? 15 ?2x ? ?16
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Follow-up
3.4 More about Solving Equations in One
Unknown
B. Removing Brackets
Solve the equation 4(7n ? 6) ? 38n ? (3n ? 5).
Solution
4(7n ? 6) ? 38n ? (3n ? 5) 28n ? 24 ? 3(8n ?
3n ? 5)
28n ? 24 ? 3(5n ? 5)
28n ? 24 ? 15n ? 15
28n ? 15n ? 15 ? 24
13n ? 39
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Follow-up
3.4 More about Solving Equations in One
Unknown
C. Simplifying Fractions
Solution
52
Follow-up
3.4 More about Solving Equations in One
Unknown
C. Simplifying Fractions
Alternative Solution
53
Follow-up
3.4 More about Solving Equations in One
Unknown
C. Simplifying Fractions
Solution
54
Follow-up
3.5 Applications of Equations in One Unknown
There are 36 students in class 1A. The number of
students who wear glasses is 8 more than the
number who do not wear glasses. Find the number
of students who wear glasses.
Solution
Let n be the number of students who wear
glasses. Then the number of students who do not
wear glasses ? 36 ? n.
Since the number of students who wear glasses is
8 more than the number who do not wear glasses,
we have n ? (36 ? n) ? 8
? The number of students who wear glasses is 22.
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Follow-up
3.5 Applications of Equations in One Unknown
Mr. Lee drove 360 km from town A to town B for a
summer vacation. On his return trip, he drove 8
km/h faster and reached town A in 4 hours. Find
his driving speed in the trip from town A to town
B.
Solution
Let x km/h be the driving speed in the trip from
town A to town B. Then the driving speed in the
return trip ? (x ? 8) km/h.
? The driving speed in the trip from town A to
town B is 82 km/h.
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Follow-up
3.5 Applications of Equations in One Unknown
Ricky has 4 stamp albums with the same number of
stamps in each. If he wants to use one album
less, then 12 more stamps should be put in each
of the other 3 albums and 10 stamps will be left
over. How many stamps does he have in total?
Solution
Let x be the number of stamps in each album
originally. Then the number of stamps in each
album after using one album less? x ? 12.
4x ? 3(x ? 12) ? 10
? Total number of stamps ? 4 ? 46