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MATH REVIEW KIT
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R1MATH05
MATH REVIEW KIT
Table of Contents
Page
SECTION 1: THE NUMBER SYSTEM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1
ARITHMETIC OF SIGNED (NEGATIVE OR POSITIVE) NUMBERS . . . . . . . . .
1.2
INFINITY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3
PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1-1
1-1
1-1
1-2
SECTION 2: SIMPLE ALGEBRAIC OPERATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1
PROPERTIES OF ADDITION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2
PROPERTIES OF MULTIPLICATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3
DISTRIBUTIVE LAW OF ADDITION AND MULTIPLICATION . . . . . . . . . . . .
2.4
EQUALITY AXIOMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5
PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2-1
2-1
2-2
2-3
2-4
2-4
SECTION 3: COMPOSITE ALGEBRAIC OPERATIONS . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1
ALGEBRAIC EXPRESSIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2
GROUPING SYMBOLS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3
CORRECT USE OF PARENTHESES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4
PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3-1
3-1
3-2
3-5
3-5
SECTION 4: ALGEBRA OF FRACTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1
THE LOWEST FORM OF FRACTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2
CONVERTING FRACTIONS TO A COMMON DENOMINATOR . . . . . . . . . . . .
4.3
ALGEBRA OF FRACTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4
PERCENTAGES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5
ROUNDING OFF A DECIMAL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6
COMMON MISTAKES IN THE USE OF FRACTIONS . . . . . . . . . . . . . . . . . . .
4.7
PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4-1
4-1
4-2
4-4
4-6
4-7
4-8
4-9
SECTION 5: ALGEBRA OF EXPONENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1
MEANING OF EXPONENT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2
ALGEBRA OF EXPONENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3
COMPOUND INTEREST . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.4
PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5-1
5-1
5-3
5-5
5-7
SECTION 6: LINEAR FUNCTIONS AND INTERPOLATION . . . . . . . . . . . . . . . . . . . . . . 6-1
6.1
LINEAR INTERPOLATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6-1
6.2
PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6-1
SECTION 7: SYSTEMS OF LINEAR EQUATIONS AND INEQUALITIES . . . . . . . . . . . . . 7-1
7.1
LINEAR EQUATIONS IN ONE VARIABLE . . . . . . . . . . . . . . . . . . . . . . . . . . 7-1
7.2
PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7-2
SECTION 8: IDENTITIES AND FACTORIZATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.1
THE IDENTITIES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2
FACTORIZATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.3
PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8-1
8-1
8-2
8-3
SOLUTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions-1
SECTION 1
THE NUMBER SYSTEM
1.1
ARITHMETIC OF SIGNED (NEGATIVE OR POSITIVE) NUMBERS
i)
Negative of a negative is its absolute value, i.e., -(-3) = 3. This is because (-3) + 3 = 0.
ii)
To add two numbers of the same sign, we add the absolute values and assign the sign of the
two numbers.
Example 1:
5 + 8 = 13, ⳮ5 + (ⳮ9) = ⳮ14
To add two numbers of different signs, take the difference of the absolute values and attach the sign of the
number with the greater absolute value.
Example 2:
13 + (ⳮ8) = 5, ⳮ15 + 7 = ⳮ8, 12 ⳮ 19 = ⳮ7, and ⳮ11 + 29 = 18.
iii)
To find the product or quotient of two numbers, multiply or divide the absolute values and
attach a positive sign if the two numbers have the same sign; otherwise, attach a negative sign.
Example 3:
3 ⳯ 25 = 75, (ⳮ5) ⳯ (ⳮ3) = 15, 5 ⳯ (ⳮ20) = ⳮ100, (ⳮ8) ⳯ 6 = ⳮ48,
&25
25
&25
= ⳮ5,
= ⳮ5,
= 5.
5
&5
&5
1.2
INFINITY
Definition 1 (Infinity). Infinity denoted as ∞ is a number larger than any given number.
In fact there is no number that would satisfy definition 10. The idea of infinity is to provide some kind of
conceptual bounds to the number line, precisely ⳮ∞ and +∞, the negative and positive infinities. Mathematical
manipulations involving infinite numbers are different from that with finite numbers. The following propositions
depict some characteristics:
Proposition 1:
There are as many even numbers as there are natural numbers.
Consider the natural and even numbers listed as follows:
1
↓
2
↓
3
↓
4
↓
5
↓
...
↓
2
4
6
8
10
...
1-1
Math Review Kit: Section 1
If we associate 1 to 2, 2 to 4, 3 to 6, etc., then for every natural number, we are able to produce an even number
and vice versa.
Proposition 2:
There are infinitely many rational numbers between any two rational numbers.
To illustrate this, take two rationals 0 and 1. Referring to Figure 3, we can generate numbers, 1/2, 1/4, 1/8,
1/16, ... breaking up the lengths to half and moving to the left. There is an infinite number of numbers between
0 and 1.
Figure 3
1.3
PROBLEMS
1.
Indicate whether the following numbers are rational or irrational.
ⳮ265, 2 2 , 1.505050...,
2.
23 &8 B
22
,
,
, .99999..., e 2 ,
7
9
4
7
Solve the following:
i)
13 + (ⳮ6)
ii)
6 + (ⳮ19)
iii)
ⳮ5 + (ⳮ8)
iv)
ⳮ7 ⳮ (ⳮ2)
v)
7 ⳮ (+6)
vi)
(8)(ⳮ13)
vii)
ⳮ12 + (10)
viii)
(ⳮ5)(ⳮ7)
ix)
Note that: [(A)(B)] = [A ⳯ B] = [AB]
1-2
(&6)2 & 3(&7)
(&3)(&1)
SECTION 2
SIMPLE ALGEBRAIC OPERATIONS
Manipulation of numbers comes within the realm of arithmetic. When we state a general rule of arithmetic, we
have progressed to algebra. An arbitrary number may be denoted by a, b, c, d or x, y, z, etc. In this section
we study the properties of basic algebraic operations of addition and multiplication and discuss the algebraic
structure of numbers.
2.1
PROPERTIES OF ADDITION
The operation of addition of two arbitrary numbers “a” and “b” is the number denoted by “a + b”.
Example 1:
i)
If a = 9, b = 15 then a + b = 9 + 15 = 24.
ii)
Adding 2 to the number “a” yields 2 + a.
Definition 1 (Identity of addition). The number 0 is called the identity of addition with the properties:
a+0=a
(1)
0+a=a
(2)
Definition 2 (Additive inverse). The additive inverse of a number “a” is a number “b” such that
and
a+b=0
(3)
b+a=0
(4)
The additive inverse of “a” is “-a”.
Example 2:
Find the additive inverses of 12 and -5.
Since 12 + (ⳮ12) = 0 and (ⳮ12) + 12 = 0, the additive inverse of 12 is ⳮ12. Similarly, the additive inverse
of ⳮ5 is 5.
Thus, the process of subtraction can be interpreted in terms of addition. 6 ⳮ 2 = 4 could be described as the
addition: 6 + (ⳮ2) = 4.
Commutative Law of Addition. If “a” and “b” are any two numbers then
a+b=b+a
Thus:
(5)
2 + 5 = 5 + 2 = 7 and
ⳮ2 + 5 = 5 + (ⳮ2) = 3.
2-1
Math Review Kit: Section 2
Associative Law of Addition. If “a”, “b” and “c” are any three numbers, then:
(a + b) + c = a + (b + c)
(6)
The parentheses in the above equation indicate the order in which the operation (addition) is performed.
Thus:
2 + (5 + 18) = 2 + 23 = 25
and
(2 + 5) + 18 = 7 + 18 = 25.
Example 3:
Consider the following scheme of addition of large numbers:
35
83
29
147
This conventional way of addition involves a combination of commutative and associative laws of addition which
can be paraphrased as:
35 + 83 + 29 = (30 + 5) + (80 + 3) + (20 + 9)
= (30 + 80 + 20) + (5 + 3 + 9)
= 130 + 17 = 147.
2.2
PROPERTIES OF MULTIPLICATION
The operation of multiplication of two arbitrary numbers “a” and “b” is a number denoted by “a ⳯ b” (read
as “a times b”); “a ⳯ b” can be written as “ab”.
Example 4:
i)
If a = 13, b = 5
Then ab = 13 ⳯ 5 = 65.
ii)
If a = 2, b = ⳮ6
then ab = 2 ⳯ (ⳮ6) = ⳮ12.
iii)
3 times a is 3a.
Definition 3 (Identity of Multiplication). The number 1 is called the identity of multiplication with the
properties:
a⳯1=a
(7)
1⳯a=a
(8)
2-2
Simple Algebraic Operations
Definition 4 (Multiplicative Inverse). The multiplicative inverse of a number “a” is a number “b” such that:
ab = 1
(9)
ba = 1
(10)
The multiplicative inverse of a is 1/a, provided a =
/ 0 (a is not zero).
Example 5:
Find the multiplicative inverses of 8 and ⳮ35.
As 8 ⳯ 1/8 = 1 and 1/8 ⳯ 8 = 1, 1/8 is the multiplicative inverse of 8. Similarly, ⳮ1/35 is the multiplicative
inverse of ⳮ35.
Thus division of “a” by “b” (b =
/ 0) can be identified with the multiplication of a with the multiplicative inverse
of “b”, i.e., “a ⳰ b” the same as “a ⳯ 1/b”.
Proposition 1:
Every real number has a unique additive inverse but only a non-zero number has a (unique)
multiplicative inverse. The multiplicative inverse of 0, which is 1/0 = ∞, is not defined.
Commutative Law of Multiplication. If “a” and “b” are any two numbers then:
ab = ba
Thus:
(11)
2⳯3=3⳯2=6
5 ⳯ (ⳮ9) = (ⳮ9) ⳯ 5 = ⳮ45.
Associative Law of Multiplication. If “a”, “b” and “c” are any three numbers, then:
a(bc) = (ab)c
2.3
(12)
Thus:
4 ⳯ (8 ⳯ 3) = 4 ⳯ 24 = 96
and
(4 ⳯ 8) ⳯ 3 = 32 ⳯ 3 = 96.
DISTRIBUTIVE LAW OF ADDITION AND MULTIPLICATION
Distributive Law. If “a”, “b” and “c” are any three numbers, then:
a(b + c) = ab + ac
(13)
2-3
Math Review Kit: Section 2
Using the rule (11), (13) can also be written as:
(b + c)a = ba + ca
Thus:
Example 6:
(14)
5(3 + 2) = (5 ⳯ 3) + (5 ⳯ 2) = 15 + 10 = 25.
Demonstrate the role of distributive law in the following scheme of multiplication:
39
⳯ 45
195
+ 1560
1755
39 ⳯ 45
2.4
=
=
=
=
=
(30 + 9) ⳯ (40 + 5)
{(30 + 9) ⳯ 5} + {(30 + 9) ⳯ 40}
{(30 ⳯ 5) + (9 ⳯ 5)} + {(30 ⳯ 40) + (9 ⳯ 40)}
195 + 1560
1755
EQUALITY AXIOMS
When “a” is equal to “b”, we write “a = b”. Following are the self-evident facts (axioms) that the equality
relation satisfies:
Reflexive:
a=a
(15)
Symmetric:
if a = b, then b = a
(16)
Transitive:
if a = b and b = c, then a = c
If a = b, then a +
_ c=b+
_ c
If a = b, then ac = bc
If a = b and c = d, then a +
_ c=b+
_ d (20)
If a = b and c = d, then ac = bd
If a + c = b + c, then a = b
If ac = bc and c =
/ 0, then a = b
(17)
(18)
(19)
2.5
PROBLEMS
1.
Find the additive inverse of the following numbers:
8,
91,
ⳮ39,
0,
ⳮ56,
2-4
1.
(21)
(22)
(23)
Simple Algebraic Operations
2.
Find the multiplicative inverse of the numbers:
3,
3.
1,
19,
ⳮ9,
ⳮ25.
Check whether the following statements are true or false.
i)
a(b + c) = ab + ac
ii)
(a +b)5 = 5a + 5b
iii)
a(b + 3) = ab + b + 3
iv)
x(y ⳮ z) = xy ⳮ xz
v)
(y + z ⳮ w)x = xy + xz ⳮ xw
vi)
(x ⳮy)(z ⳮw) = xz ⳮ xw ⳮ yz + yw
vii)
(a + b)(5 ⳮ c) = 5a + 5b ⳮ ca + cb
2-5
SECTION 3
COMPOSITE ALGEBRAIC OPERATIONS
In this section, we discuss more complicated algebraic expressions and the use of grouping symbols.
3.1
ALGEBRAIC EXPRESSIONS
Definition 1 (Constant). A constant is a number or an algebraic symbol that stands for a particular value in a
given context.
For example, 7, ⳮ2, π, e, are all constants. Sometimes a letter “c” may denote a constant.
Definition 2 (Variable). A variable is an algebraic symbol, say “x”, that may take on different values from the
real number system.
Definition 3 (Algebraic Expression). An algebraic expression is created when one or more algebraic operations
are performed upon variables and constants. It may consist of one or more terms, separated from each other
by the signs + and ⳮ.
For example, “3x + 2y + xy ⳮ z” is an algebraic expression consisting of four terms.
When the terms do not differ or differ only in their numerical coefficients, they are called “like” terms. Thus,
3x and 8x; 5ab and ⳮ3ab; xyz and 4xyz are pairs of like terms.
Abbreviation of Some Algebraic Terms
i)
The sum a + a is written as 2a. Similarly, a + a + a = 3a, and so on. The term 2a also means 2 ⳯
a.
ii)
Addition of like terms is obtained by taking the algebraic sum of the coefficients and multiplying by the
term.
Thus, x + 5x = 6x, 8ab ⳮ 2ab + ab = (8 ⳮ 2 + l)ab = 7ab.
iii)
The product “a ⳯ a” means a2 and “a ⳯ a ⳯ a” means a3 and so on.
iv)
Square root of x, denoted by
symbolized as
3
x , is a number y such that y2 = x. Similarly a cube root of x,
x , is a number z such that z3 = x.
For example 2 and -2 are square roots of 4 because 22 = 4 and (-2)2 = 4. The two square roots are
sometimes written as ± 4 .
Example 1:
i)
Find the value of (x2 ⳮ 2x ⳮ 6) if x = 3.
(3)2 ⳮ 2(3) ⳮ 6 = 9 ⳮ 6 ⳮ 6 = 3 ⳮ 6 = ⳮ3
3-1
Math Review Kit: Section 3
ii)
Find the value of (x + y)2 if x = 3 and y = ⳮ5.
(3 ⳮ 5)2 = (ⳮ2)2 = 4
iii)
Find the value of (x 3 &
x % 1) for x = ⳮ1.
(ⳮ1)3 ⳮ &1 % 1 = ⳮ1 ⳮ 0 = ⳮ1
iv)
When multiplying two expressions, the multiplication sign (x) may be dispensed with. The distributive
law can be used to simplify the product.
Example 2:
i)
ii)
Simplify the following:
3(x + 2 ⳮ 1)
= 3x + 6 ⳮ 3
= 3x + 3
[Using the distributive law]
x + (5 ⳮ 3x)
= x + 5 ⳮ 3x [Treating “+ (5 ⳮ 3x)” as “1 ⳯ (5 ⳮ 3x)”, and using the distributive law]
= ⳮ2x + 5
Thus, parentheses preceded by a “+” sign may be removed without changing the expression.
iii)
3y ⳮ 2 ⳮ (x + y ⳮ 1)
= 3y ⳮ 2 ⳮ x ⳮ y + 1
[Treating “-(x + y ⳮ 1)” as “-1(x + y ⳮ 1)”, and using the distributive
law]
= 2y ⳮ x ⳮ 1
Thus, parentheses preceded by a “-” sign may be removed by changing the sign of each term of the expression
within the parentheses.
3.2
GROUPING SYMBOLS
An algebraic expression may involve grouping symbols: parenthesis ( ), brackets [ ], and braces { }. These are
used to indicate the order in which the basic operations of “+, ⳮ, ⳯, and ⳰" are performed.
We now consider the rules of simplifying expressions involving composite algebraic operations.
3-2
Composite Algebraic Operations
Expressions with no Grouping Symbol
Rule 1: In the order of operations, multiplication and division belong to the same hierarchy followed by the
hierarchy of addition and subtraction. When two operations in the same hierarchy appear one after the
other, merely proceed from left to right.
Example 3:
i)
Simplify:
12 ⳮ 2 ⳮ 30 ⳰ 5 ⳯ 2
= 12 ⳮ 2 ⳮ 6 ⳯ 2
= 12 ⳮ 2 ⳮ 12
= 10 ⳮ 12
= ⳮ2
ii)
18 ⳰ 3 ⳯ 2 + 1 ⳮ 6 ⳯ 4 ⳰ 2
= 6 ⳯ 2 + 1 ⳮ 24 ⳰ 2
= 12 + 1 ⳮ 12
= 13 ⳮ 12
= 1
Expressions with more than one Grouping Symbol
Rule 2: First solve the innermost parentheses, then brackets and then outermost braces.
Example 4:
i)
Simplify:
[3 ⳮ (10 ⳮ 9)] ⳯ 6 ⳰ 4
= [3 ⳮ (1)] ⳯ 6 ⳰ 4
= [2] ⳯ 6 ⳰ 4
= 12 ⳰ 4
= 3
3-3
Math Review Kit: Section 3
ii)
5x ⳮ [4x + x(3x ⳮ 2)]
= 5x ⳮ [4x + 3x2 ⳮ 2x]
= 5x ⳮ [2x + 3x2]
= 5x ⳮ 2x ⳮ 3x2
= 3x ⳮ 3x2
iii)
2{8 ⳯ 5 + 2[13 ⳮ 9(5 + 3 ⳮ 6)] ⳰ 5}
= 2{40 + 2[13 ⳮ 9(2)] ⳰ 5}
= 2{40 + 2[13 ⳮ 18] ⳰ 5}
= 2{40 + 2[-5] ⳰ 5}
= 2{40 ⳮ 10 ⳰ 5}
= 2{40 ⳮ 2}
= 2{38}
= 76
iv)
2a ⳮ 4{a ⳮ 3[2b + 5(a ⳮ 1) ⳮ 2(3b ⳮ 5)]}
= 2a ⳮ 4{a ⳮ 3[2b + 5a ⳮ 5 ⳮ 6b + 10]}
= 2a ⳮ 4{a ⳮ 3[5a ⳮ 4b + 5]}
= 2a ⳮ 4{a ⳮ 15a + 12b ⳮ 15}
= 2a ⳮ 4{-14a + 12b ⳮ 15}
= 2a + 56a ⳮ 48b + 60
= 58a ⳮ 48b + 60
3-4
Composite Algebraic Operations
3.3
CORRECT USE OF PARENTHESES
The following examples depict some common mistakes in using or not using parentheses.
Example 5:
i)
From x subtract 1 less than a number y.
Answer: x ⳮ y ⳮ 1 (incorrect)
The correct answer is x ⳮ (y ⳮ 1).
ii)
Find the area of a rectangle with sides a and a ⳮ 2 units.
Answer: (a) ⳯ (a) ⳮ 2 (incorrect)
The correct answer is a(a ⳮ 2).
iii)
x ⳮ (y + 1) = x ⳮ y + 1 (incorrect)
The correct solution is x ⳮ (y + 1) = x ⳮ y ⳮ 1.
iv)
(
1
1
1
x)( y) = xy (incorrect)
3
3
3
The correct solution is: (
v)
1
1
1
x)( y) = xy
3
3
9
(x + l)(y + 2) = x + y + 2 (incorrect)
The correct solution is: (x + 1)(y + 2) = (x + 1)y + (x + 1)2
= xy + y + 2x + 2
3.4
PROBLEMS
1.
Find the values of the following expressions at the given variable values.
i)
2x2 + 7x ⳮ 2
at x = ⳮ2
ii)
(x + y)2 ⳮ xy + y2
at x = 1, y = 3
3-5
Math Review Kit: Section 3
iii)
iv)
v)
2.
3.
2
2
(x % y)2
& x3 % y2 & 2
3a 2bc
ab
at x = 3, y = ⳮ1
at x = 1, y =3
at a = 2, b = 8, c = ⳮ1
Simplify the following:
i)
3ⳮ9+3⳯4⳰2
ii)
9 ⳰ 3 ⳯ 4 ⳮ 2 ⳯ (ⳮ6)
iii)
2 + 8[6 ⳮ 9(8 ⳮ 10)]
iv)
10 ⳮ {7 + 3[5 ⳮ (9 + 2)]}
v)
2{(2 ⳮ 3 + 9) ⳮ 2[2 ⳮ (3 + 6 ⳮ 12) + 6(3 ⳮ 7)]}
Simplify the following:
i)
x + 3x ⳮ (4x ⳮ 6x)
ii)
2a ⳮ (3a + 2b) + (a ⳮ 3b)
iii)
x + y ⳮ z ⳮ (x ⳮ y + z) ⳮ (2x + y) + (x ⳮ 3z)
iv)
2x{x ⳮ y[z + (x ⳮ 3y)]}
v)
{x + 3y ⳮ [2x ⳮ (x + y)]} ⳮ x ⳮ y
vi)
xy + [3x ⳮ 2y + (x + y)(x ⳮ 3y)]
3-6
SECTION 4
ALGEBRA OF FRACTIONS
In this section, the algebraic operations discussed in section 3 will be applied to fractions. Percentages and
rounding procedure will be discussed as well.
4.1
THE LOWEST FORM OF FRACTIONS
/ 0, is called a fraction.
Definition 1 (Fraction). An expression of the form a/b, b =
Division by zero is not allowed. Since a/b = c means a = bc, a fraction like a/0 is not defined as we cannot
find a unique number c such that a = 0 ⳯ c. In the notation of infinity, we may say that if a > 0 then a/0 =
∞ and -a/0 = -∞. The fractions 0/0, ∞/∞, -∞/-∞, -∞/∞, and ∞/-∞, are undefined. They are known as
indeterminate forms. A few other indeterminate forms are 0 ⳯ ∞ and ∞ ⳯ ∞ . It will be convenient to place
fractions over a common denominator for addition or subtraction purposes.
Definition 2 (Equality of Fractions). We define a/b = c/d if ad = bc.
For example
2
4
=
because 2 ⳯ 6 = 4 ⳯ 3.
3
6
Definition 2 gives rise to the following rule of fractions.
Rule 1: Given a fraction a/b, the numerator (a) and denominator (b) can be multiplied or divided by a non-zero
number without changing its value, i.e., a/b = ka/kb, k =
/ 0.
Example 1:
i)
2
2 × 2
4
=
=
3
3 × 2
6
ii)
20
20 ÷ 10
2
=
=
50
50 ÷ 10
5
Definition 3 (Lowest Form of a Fraction). A fraction, a/b, is said to be of lowest form if a and b have no
common factor. For example, 2/3, 1/7, 2/9, 11/21 are fractions in lowest form, whereas, 6/4, 10/50, ab/bd,
ab2/a2c are not. (A “Factor” is a number or variable that can be divided into both the numerator and the
denominator such as to reduce the value of each.)
Example 2:
i)
12
,
16
Reduce the fractions to their lowest form:
ii)
50
,
125
iii)
6a 2b
9ab 2
, iv)
7a 2c
28ab 2c 2
4-1
Math Review Kit: Section 4
i)
ii)
iii)
iv)
4.2
12
16
=
50
125
6a 2b
9ab
2
[Writing the numerator and denominator as a product of prime
factors]
=
3
4
=
2 × 5 × 5
2
=
5 × 5 × 5
5
=
2 × 3 × a × a × b
2a
=
3 × 3 × a × b × b
3b
7a 2bc
2 2
28ab c
2 × 2 × 3
2 × 2 × 2 × 2
=
[Dividing numerator and denominator by common factors 2 ⳯ 2]
7 × a × a × b × c
a
=
2 × 2 × 7 × a × b × b × c × c
4bc
CONVERTING FRACTIONS TO A COMMON DENOMINATOR
Rule 2: The fractions a/b and c/d can be converted to an equivalent pair of fractions: ad/bd and bc/bd, with a
common denominator bd. Sometimes it is convenient to convert into fractions with lowest common
denominator.
Example 3:
i)
i)
Place the following fractions over a common denominator.
1 1
,
2 3
ii)
2 5
,
3 6
iii)
3
5
,
11 22
2 ⳯ 3 = 6 is a common denominator. The fractions are then:
3 2
1
3
3 1
2
2
,
(i.e.,
×
' ,
×
' )
6 6
2
3
6 3
2
6
ii)
3 ⳯ 6 = 18
The fractions are
12 15 2
6
12 5
3
15
,
( ×
,
)
'
×
'
18 18 3
6
18 6
3
18
4-2
Algebra of Fractions
iii)
11 ⳯ 22 = 242
The fractions are
Example 4:
66
55
3
22
66
5
11
55
×
'
×
'
,
,(
), (
)
242 242
11
22
242
22
11
242
Place the fractions in example 3 over the lowest common denominator.
i)
3 2
,
6 6
ii)
2 5
,
have 6 as a common denominator which is lowest.
3 6
The fractions are
iii)
4 5
, .
6 6
3
5
,
have 22 as the lowest common denominator.
11 22
The fractions are
6
5
,
.
22 22
Definition 4 (Proper and Improper Fractions). A fraction in which the denominator is greater than the numerator
is called a proper fraction. If the denominator is less than the numerator, it is called an improper fraction.
An improper fraction can be changed into a mixed number which is the sum of an integer and a proper fraction.
Example 5:
i)
Convert the following into mixed numbers.
21
5
4
5 21
20
1
So that
21
1
1
=4+
or 4
5
5
5
4-3
Math Review Kit: Section 4
ii)
40
3
13
3 40
3
10
9
1
So that
4.3
40
1
= 13
3
3
ALGEBRA OF FRACTIONS
Rule 3: i)
Sum (Difference) of Fractions.
a
c
ad ± bc
+
_
=
b
d
bd
ii)
To find the sum (difference) of two fractions, place them over the lowest common denominator
and add (subtract) the numerator.
Example 5:
i)
2
5
+
5
4
=
ii)
iii)
Solve the following:
5
12
8
%
20
&
'
8 % 15
'
20
23
20
' 1
3
20
17
18
'
15
&
36
'
15 & 34
36
2a
%
3b
15
20
34
36
'
[Note that 36 is the lowest common denominator]
&19
36
ac
4bd
4-4
Algebra of Fractions
Though Rule 3 (i) can be used, we may also see that 12bd is the lowest common denominator.
2a
×
3b
4d
%
4d
2a × 4d
%
12bd
'
ac
4bd
×
3
3
3 × ac
12bd
8ad % 3ac
'
12bd
a(8d % 3c)
12bd
Rule 4: Multiplication of Fractions
a
×
b
Example 7:
i)
5
19
'
ii)
ac
bd
Solve the following:
×
6
8
5 × 6
'
19 × 8
2a
×
3b
'
c
'
d
30
'
152
15
76
b2
a2
2a × b 2
3b × a
2
'
2a × b × b
'
3b × a × a
2b
3a
(Note here the use of Section 4.1 [definition 3] in reducing to a lowest-form fraction.)
Rule 5: Division of Fractions
a
÷
b
c
'
d
a
×
b
d
'
c
ad
bc
4-5
Math Review Kit: Section 4
Example 8:
i)
Solve the following.
2
÷
3
2
×
3
'
ii)
2a
÷
3b
'
4.4
5
8
8
'
5
16
15
' 1
1
15
2c
a 2b
2a
×
3b
a 2b
'
2c
2a 3b
'
6bc
a3
3c
PERCENTAGES
Definition (Percentage). Percent, denoted by %, is a fraction with 100 as the denominator.
Example 9:
i)
8%
8%
Also,
ii)
Convert the following percentages into fractions and decimals.
=
8
= .08
100
1/2%
1
% '
2
Also,
Example 10:
i)
8
2
=
100
25
1
÷
2
100
'
1
1
×
2
1
'
100
1
200
1
÷ 100 ' .05 ÷ 100 ' .005
2
Convert the following fractions to percentages:
3
5
4-6
Algebra of Fractions
3
'
5
ii)
3 × 20
5 × 20
'
60
' 60%
100
5
6
In this case, it is difficult to convert the denominator to 100. Therefore convert the fraction to a decimal then
multiply and divide by 100.
5
'
6
.8333̇ × 100
'
1 × 100
83.33̇
1
' 83 %
100
3
_
(Note that a “@” above the last figure of decimal indicates that figure repeats to infinity; thus “.8333”
is really
“.8333333333...”).
4.5
ROUNDING OFF A DECIMAL
Rule of Rounding. To round off the decimal at any position, add one if the next decimal digit to the right is 5
or more, or add nothing if the next decimal digit to the right is less than 5.
For example:
Rounding off 5.769 to two decimal places gives 5.77.
Rounding off 13.8734 to three decimal places gives 13.873.
Rounding off 8.3849 to two decimal places gives 8.38.
Rounding off to nearest cents
Example 11:
i)
Round off to nearest cents.
5
of $100.00
6
5
' .83̇ Multiplication by 100 will shift the decimal two places to the right. Therefore, round
6
off at the fourth decimal place.
Now
Hence
5
⳯ 100 = .8333 ⳯ 100 = $83.33
6
4-7
Math Review Kit: Section 4
ii)
1
of $1000.00
7
Now,
1
= .142857
7
Multiplying by 1000 will shift the decimal three places to the right. Therefore, round off at the fifth
place.
1
× 1000 ' .14286 × 1000 ' $142.86
7
(Note: In Mortgage Finance calculations, payments are always rounded up, even if the decimal digit is less than
5.)
4.6
i)
COMMON MISTAKES IN THE USE OF FRACTIONS
x % y
'
5
x
% y
5
(wrong)
The correction solution is:
a
ii)
b
c
'
d
a × c
bd
a
In fact,
b
c
d
iii)
'
x % y
'
5
x
y
%
5
5
(wrong)
a
c
÷
'
b
d
a
d
×
'
b
c
7
'
28
7
'
2 × 2 × 7
0
' 0
4
7
'
28
7
'
2 × 2 × 7
1
7
Since
'
4
7
ad
bc
(wrong)
1
1
4-8
Algebra of Fractions
4.7
PROBLEMS
1.
Reduce the following to lowest form:
i)
2.
3.
36
64
iii)
i)
1
2
,
7
13
ii)
2
7
4
,
,
5
15
25
iv)
a
c
,
b
bd
v)
x
y
,
yx
xz
iv)
abc 2
bcd
16
12
,
39
65
iii)
Insert one of <, = or > in the following:
Hint:
Note:
5.
21
49
ii)
Place the fractions over their lowest common denominator:
i)
4.
4
12
1
2
5
8
ii)
3
10
2
7
iii)
&
2
5
&
3
7
5
19
10
38
iv)
&
iv)
5
3
1
%
&
42
56
21
&
Convert to common denominator and compare numerators.
“>” means “greater than”; “<” means “less than”.
Solve the following and reduce to lowest form:
i)
1
3
%
5
7
ii)
2
3
%
7
14
v)
a
c
%
b
bd
vi)
x % y
x & y
&
z
z
viii)
3
1
4
×
×
10
2
11
xi)
5/6
8
xii)
5
3/7
iii)
ix)
a
x
b
×
×
bc
y
d
xiii)
a 2x
a
÷
b
bx
5
6
&
8
11
vii)
x)
xiv)
Convert the following percentages to fractions:
i)
5%
ii)
1
%
4
iii)
3
%
4
4-9
iv)
2
7
×
5
8
125%
2
1
÷
9
3
xy
x2
÷
zu
u2
Math Review Kit: Section 4
6.
Convert the following fractions to percentages:
i)
7.
8.
1
25
ii)
2
5
iii)
2
3
iv)
8
15
Round off the following to the nearest cent:
i)
1
of $1000
8
ii)
2
over $1250
3
iii)
1
of a million dollars
9
iv)
1
of $10,000
6
Simplify the following:
i)
iii)
2 8
1
1
[ & ( & )]
3 9
2
3
2{
ii)
x
1
1
% [x % y & (x & y)]}
2
2
3
1 1
1
1
{ % [(1 & ) & (2 & )]}
2 4
3
6
iv)
4-10
(
a
b 1
a
& )( & )
2
3 4
6
SECTION 5
ALGEBRA OF EXPONENTS
In section 3, we introduced terms like x2 and x , referred to as exponential expressions. In this section we study
the algebra of exponential expressions and numbers. The compound interest formula is discussed as well.
5.1
MEANING OF EXPONENT
An expression, symbolized by an, is called an exponential number or expression. The non-zero number “a” is
called the base, and the number “n” is called the power or exponent. The interpretation of “n” depends on
whether “n” is a positive integer, zero, negative integer, fractional or irrational number.
Definition 1:
If “n” is a positive integer, then an means “a” multiplied by itself “n” times. That is,
an = a ⳯ a ⳯ a ⳯ ... ⳯ a [n times]
For example, 34 = 3 ⳯ 3 ⳯ 3 ⳯ 3 = 81. Similarly (ⳮ1)5 = ⳮ1; (
1 4
1
1
) = 1/81; (ⳮ )7 = ⳮ
; and
3
2
228
16 = 1.
Definition 2:
If the exponent is a negative integer, then we define aⳮn =
For example, 2ⳮ3 =
Similarly 2ⳮ1 =
1
;
2
1
3
2
=
an
1
8
1
( ) -2 =
2
1
1
22
Definition 3:
1
=
1
1
4
=4
a0 = 1.
Definition 3 is a consequence of definitions 1 and 2, as we shall see later in this section.
Definition 4:
If “n” is a positive integer, then we define a1/n as a number “b” such that bn = a. In other
words, a1/n is nth root of “a”, and is denoted by
Thus,
2
a , which is also written as
n
a.
a , is called a square root of a,
called a fourth root of a and so on. Note that we called
n
3
a is called a cube root of a and
a is
a as “a” nth root and not “the” nth root because there
may be more than one nth root. For example 4 has two roots, 2 = 4 and ⳮ2 = ⳮ 4 .
5-1
4
Math Review Kit: Section 5
Example 1:
Simplify the following:
i)
81/3
811/4
i)
81/3 = 2 because 23 = 8
ii)
811/4 = 3 because 34 = 81
iii)
125ⳮ1/3 =
iv)
32ⳮ1/5 =
ii)
1
1/3
125
1
32
1/5
=
iii)
=
125ⳮ1/3 iv)
32ⳮ1/5
1
because 53 = 125
5
1
because 25 = 32.
2
As in parts iii) and iv), the same rule applies to negative fractional exponents as to negative integral exponents.
(Refer to definition 2.)
Care has to be taken when, in a1/n, “a” is negative. Then the roots are not defined when “n” is even. However,
when “n” is odd, a root may exist. For example, (-1)1/2 or &1 is not defined in the context of real numbers
3
as there is no number “b” such that b2 = ⳮ1. But &1 = -1 as (-1) 3 = -1. For the sake of simplifying
matters, we shall assume that a > 0 whenever finding a root of “a” is involved. Also, a1/n would mean the
positive nth root of “a”. Talking of nth roots, the following theorem, which generates many irrational numbers,
will not be out of place.
If “a” is a prime number, then for every “n” where n = 1, 2, 3, 4, ... ,
number.
Theorem 1:
Thus, 2, 3,
Example 2:
iii)
iv)
7,
4
a is an irrational
11, 13 are all irrational numbers.
Use your calculator, if possible, to check the following:
3 ' 1.73205080 ... • 1.7321
i)
ii)
3
n
3
4
3
7 ' 1.912931183 ... • 1.9129
11 ' 1.821160287 • 1.8212
&29 ' &3.072316826 ... • &3.0723
Definition 5:
Let p/q be a fraction, then ap/q is defined as the qth root of a raised to the power p.
q
Symbolically, ap/q = ( a )p or (a1/q)p
5-2
Algebra of Exponents
Example 3:
i)
43/2 = ( 4 )3 = 23 = 8
ii)
274/3 = ( 27 )4 = 34 = 81
iii)
25-3/2 =
iv)
(0.09)3/2 = ( 0.09 )3 = (0.3)3 = .027
3
1
25
3/2
=
1
3
( 25)
=
1
3
5
=
1
125
The exponential expression, an, with an irrational exponent “n” is difficult to compute because of infinite decimal
expansion of irrational numbers. However, an approximation of the irrational exponent by a finite decimal, the
number of digits depending upon the degree of approximation desired, can give a near value of an.
Example 4:
2
= 51.41421... ≅ 51.414 = 9.735
i)
5
ii)
2B = 23.14159... ≅ 23.142 = 8.827
5.2
ALGEBRA OF EXPONENTS
[Using a calculator and rounding off at third decimal]
[Using a calculator and rounding off at third decimal]
Now that an is defined for any real number “n” and a =
/ 0, we are set to go into the algebra of exponents.
Recall that whenever “n” is a fraction, we assume that a > 0. Let “m” and “n” be arbitrary numbers and “a”
and “b” be non-zero numbers.
Rule 1: aman
= am+n
Rule 2: (am)n
= amn
Rule 3: (ab)n
= anbn
Rule 4: (
Rule 5:
a n
)
b
am
a
n
=
an
bn
= am ⳮ n
5-3
Math Review Kit: Section 5
Example 5:
i)
Simplify the following:
23 × 16 × 25
8 × 24 × 2
23 × 24 × 25
'
23 × 24 × 2
'
= 212-8
212
28
[using rule 1]
[using rule 5]
= 24 = 16
ii)
32 × 9 × 33
27 × 34
32 × 32 × 33
'
33 × 34
37
'
[using rule 1]
37
2187
=1
2187
'
This example gives us an opportunity to check that a0 = 1 for a = 3.
37
7
3
' 37&7 ' 30
[using rule 5]
Hence = 30 = 1
iii)
(a 2b 2) × (a 3) × (b)
(2ab)4
'
'
'
a 2a 3bb 2
24a 4b 4
a 5b 3
16a 4b 4
1
ab &1
16
[using rule 3]
[using rule 1]
[using rule 5]
5-4
Algebra of Exponents
a
16b
'
iv)
(2x2yz3)2(x3yz4)ⳮ1
(2x 2yz 3)2
'
x 3yz 4
22(x 2)2y 2(z 3)2
'
x 3yz 4
4x 4y 2z 6
'
x 3yz 4
= 4xyz2
ab 2c 3
v)
[using rule 2]
[using rule 5]
2
x 3y 2
'
'
5.3
[using rule 3]
(ab 2c 3)2
(x 3y 2)2
a 2b 4c 6
x 6y 4
[using rule 4]
[using rules 3 and 2]
COMPOUND INTEREST
The compound interest formula is an application of exponential expressions.
Example 6:
A person borrows $1000 at 8% interest, compounded annually. What is the amount due after
3 years?
Figure 1
5-5
Math Review Kit: Section 5
The interest rate is $0.08 per dollar per year.
Interest during first year =
1000(.08)
Amount due after one year
=
=
=
Principal + interest
1000 + 1000(.08)
1000(1 + .08)
Now, this amount becomes the principal at the beginning of the second year, then, interest during second year
= 1000(1 + .08)(.08).
Amount due after two years
=
=
=
1000(1 + .08) + 1000(1 +.08)(.08)
1000(1 + .08)[1 + .08]
1000(1 + .08)2
This amount is the principal for the third year. Then, interest during the third year = 1000(1 + .08)2(.08).
Amount due after 3 years
=
=
=
=
=
=
1000(1 + .08)2 + 1000(1 + .08)2(.08)
1000(1 + .08)2[1 + .08]
1000(1 + .08)3
1000(1.08)3
1000 ⳯ 1.25971 (using a calculator)
$1259.71
In general, if an amount P, called principal, is invested at the interest rate of i per dollar per year compounded
yearly, then the amount A due after n years is given by:
A = P(1 +i)n
This is the compound interest formula.
Example 7:
Here
Therefore,
Solve example 6 using the compound interest formula.
P = 1000, i = .08, n = 3
A = 1000(1 + .08)3
= 1000 ⳯ 1.25971
= $1259.71
5.4
PROBLEMS
1.
Compute the following (do not give decimal answers):
i)
iv)
(
243
ii) 272/3
25 &3/2
)
49
v)
iii)
(0.0625)1/4
5-6
32ⳮ4/5
Algebra of Exponents
2.
Evaluate the following:
32 × 27
9 × 81
i)
(&8)2/3 (&27)4/3
iv)
3.
v)
(
2 2 9 2
) ( )
3
4
iii)
53/2 × 5&1/2
55/2 × 51/2
2 × 3&1 × 2
2&5/2 × 2&3/2
Approximate the irrational exponent by a three-digit decimal and compute using a calculator:
i)
4.
43/2 (&125)2/3
ii)
3
2
ii)
5B
iii)
2e
ii)
(
x 2yz 3
)
axb
iii)
v)
(x-2/5 y )5
vi)
Simplify:
i)
iv)
(x 2) × (x 5)
(x) × (x 5)
x[x1/2 + 2(x2 ⳮ x)]
(a 2)3(b 3)2
(ab)3
(x2y)2/3 (xyz)ⳮ1/3
5.
If $1,000 is invested for 5 years at 8% compounded yearly, what will this investment be worth at the
end of this time?
6.
On January 1, 1975, a man incurs a debt of $5,000 at 12% compounded annually. If the lender
demands payment on January 1, 1979, how much must the borrower pay?
7.
As a part of his retirement savings program, a person sets aside $8,000 in a deferred savings account
paying 9% compounded annually. What is the maturity value of this account after 10 years when he
becomes 65?
5-7
SECTION 6
LINEAR FUNCTIONS AND INTERPOLATION
6.1
LINEAR INTERPOLATION
Suppose two points (x1,y1) and (x2,y2) lie on a straight line and we want to find the value of y3 for a given x3
between x1 and x2, such that the point (x3 ,y3 ) lies on the straight line. This process is called linear interpolation.
If x3 is outside x1 and x2, then it is called intrapolation. In both the cases, the “two point formula” as indicated
in the following example is used.
Example 1:
An investment of $1000 for 5 years at 6% interest compounded annually yields $1338.23 and
at 8% compounded annually yields $1469.33. Use linear interpolation to find the amount at
7% compounded annually.
Consider the points (6, 1338.23) and (8, 1469.33). Substituting these points for (x1,y1) and (x2,y2) into the two
point formula and letting x = 7, we get:
y2 & y1
(x & x1)
y ⳮ y1
=
y ⳮ 1338.23
=
1469.33 & 1338.23
(7 ⳮ 6)
8 & 6
=
131.10
(1)
2
x2 & x1
[The two point formula]
= 65.55
Hence y
= 65.55 + 1338.23
= 1403.78
Therefore, at 7%, the interpolated amount is $1403.78.
6.2
PROBLEMS
1.
On a piece of land, the application of 1 gallon of fertilizer yields 2 bushels of a certain crop and the
application of 3 gallons of fertilizer yields 5 bushels of the crop. Estimate the yield if 2 gallons of
fertilizer were used, using linear (two-point) interpolation.
6-1
SECTION 7
SYSTEMS OF LINEAR EQUATIONS AND INEQUALITIES
In section 6, we discussed linear equations in two variables. In this section, we will discuss the algebraic aspects
of linear equations.
7.1
LINEAR EQUATIONS IN ONE VARIABLE
A linear equation in one variable, x, is of the form:
ax + b = 0, a =
/ 0.
(A)
Definition 1 (Solution). A value of x is said to be the solution of equation (A) if it satisfies the equation.
Example 1:
i)
Solve 5x ⳮ 3 = 0 and check to see that the solution satisfies the equation.
Using the laws of equality, we have:
5x ⳮ 3 + 3 = 3 (adding 3 on both sides)
5x = 3
This is also known as transposing ⳮ 3 on the right hand side of the equation by changing its sign.
Simplifying further, we get:
x=
3
(dividing both sides by 5)
5
Substituting x =
5(
3
in the original equation, we get:
5
3
)ⳮ3=0
5
3ⳮ3=0
0=0
Hence, x =
3
is the solution of the given equation.
5
7-1
Math Review Kit: Section 7
ii)
Solve 3x + 5 = x + 15 and check the solution.
Adding (-x ⳮ 5) on both sides we get:
3x + 5 ⳮ x ⳮ 5 = x + 15 ⳮ x ⳮ 5
2x = 10
x = 5 (dividing the equation by 2)
Substituting x = 5 in the original equation, we get:
3(5) + 5 = 5 + 15
15 + 5 = 20
20 = 20
Hence, x = 5 is the solution of the given equation.
7.2
PROBLEMS
1.
Solve for x:
i)
5(x +3) ⳮ 2 = 23
ii)
x +2 ⳮ 3x + 5 = x + 10
iii)
3(x + 5) +5(2 ⳮ x) = 25
7-2
SECTION 8
IDENTITIES AND FACTORIZATION
In this section, we introduce identities and use them in the factorization of algebraic expressions. Factorization
of quadratic expressions is discussed in detail.
8.1
THE IDENTITIES
Definition 1 (Identity). An identity is an equation involving two algebraic expressions which is satisfied by all
possible values of the variables in the equation.
For example, 2x = x + x;
x2 + x = x(x + 1).
We list below a few common identities which can be confirmed by using the distributive law upon the left hand
side (L.H.S.):
(x + y)2 = x2 + 2xy + y2
(1)
(x ⳮ y)2 = x2 ⳮ 2xy + y2
(2)
(x + y)(x ⳮ y) = x2 ⳮ y2
(3)
(x + y)3 = x3 + 3x2y + 3xy2 + y3
(4)
(x ⳮ y)3 = x3 ⳮ 3x2y + 3xy2 ⳮ y3
(5)
(x + y)(x2 ⳮ xy + y2) = x3 + y3
(6)
(x - y)(x2 + xy + y2) = x3 ⳮ y3
(7)
As an illustration, we check identity (1)
L.H.S. = (x + y)2 = (x + y)(x + y)
= x(x + y) + y(x + y)
[using the distributive law]
= x2 + xy + yx + y2
= x2 + 2xy + y2 = R.H.S.
Example 1:
i)
Simplify the following:
(2x + 3)2,
ii)
(3a ⳮ 5)2,
iii)
8-1
(m + 1)(m ⳮ 1),
iv)
(x + 3)3
Math Review Kit: Section 8
i)
Using identity (1), we have:
(2x + 3)2
= (2x)2 + 2[(2x) ⳯ (3)] + 32
= 4x2 + 12x + 9
ii)
Using identity (2), we get:
(3a ⳮ 5)2
= (3a)2 ⳮ 2[(3a) ⳯ (5)] + 52
= 9a2 ⳮ 30a + 25
iii)
In this case, identity (3) applies. Thus:
(m + 1)(m ⳮ 1)
= m2 ⳮ 12
= m2 ⳮ 1
iv)
Using identity (4), we have:
(x + 3)3
= x3 + 3(3x2) + 3(9x) + 33
= x3 + 9x2 + 27x + 27
8.2
FACTORIZATION
Definition 2 (Factorization). By factorization of an algebraic expression, we mean writing the expression as a
product of two or more terms, called its factors.
Obviously, the factors 1 and the algebraic expression itself are not used in factorization. For example:
2x + 10 = 2(x + 5) and
x2 ⳮ y2 = (x + y)(x ⳮ y)
Note that these two examples are also identities. The identities and the distributive law are extensively used in
factorization.
Example 2:
Factorize the following expressions:
i)
xy + xz + 2x
ii)
a2b + 3ab + ab2
iii)
3x3y + 6x2y2 + 27x2y3 + 3x2y
8-2
Identities and Factorization
iv)
27m3 ⳮ 8n3
i)
We note that x is a common factor in each term. Invoking the distributive law, we factor out x:
xy + xz + 2x = x(y + z + 2)
ii)
Factoring out ab and using the laws of exponents:
a2b + 3ab + ab2 = ab(a + 3 + b)
iii)
Factoring out 3x2y and using the laws of exponents:
3x3y + 6x2y2 + 27x2y3 + 3x2y = 3x2y(x + 2y + 9y2 + 1)
iv)
27m3 ⳮ 8n3
= (3m)3 ⳮ (2n)3
= (3m ⳮ 2n)[(3m)2 + (3m ⳯ 2n) + (2n)2]
= (3m ⳮ 2n)(9m2 + 6mn + 4n2)
8.3
PROBLEMS
1.
Factorize the following:
i)
81m2 ⳮ 25n2
ii)
a6 + 27
iii)
8a3 ⳮ 125b3
iv)
25abc + 5a2b + 10abc2
v)
1 3 2
1
x y % x 2y 3
2
4
8-3
[using the identity (7)]
Solutions
SOLUTIONS
Solutions to the Problems in Section 1
1.
The rational numbers are: ⳮ265, 1.505050...,
2.
i)
13 + (ⳮ6) = 7
ii)
6 + (ⳮ19) = ⳮ13
iii)
ⳮ5 + (ⳮ8) = ⳮ13
iv)
ⳮ7 ⳮ (ⳮ2) = ⳮ5
v)
7 ⳮ (+6) = 1
vi)
(8)(ⳮ13) = ⳮ104
vii)
ⳮ12 + (10) = ⳮ2
viii)
(ⳮ5)(ⳮ7) = 35
ix)
(&6)2 & 3(&7)
'
(&3)(&1)
&12 & (&21)
'
3
23
8
22
, ⳮ , .99999...,
7
9
7
9
' 3
3
Solutions to the Problems in Section 2
1.
The additive inverse (a.i.) of 8 is ⳮ8 because 8 + (ⳮ8) = 0 and ⳮ8 + 8 = 0.
Similarly the a.i. of 91 is ⳮ91.
The a.i. of 0 is 0.
The a.i. of ⳮ56 is 56.
The a.i. of 1 is ⳮ1.
The a.i. of ⳮ39 is 39.
2.
1
1
1
because 3 ⳯
= 1 and
⳯ 3 = 1. Similarly the m.i.
3
3
3
1
1
1
of 1 is 1, the m.i. of 19 is
, the m.i. of ⳮ9 is ⳮ and the m.i. of ⳮ25 is ⳮ
.
19
9
25
The multiplicative inverse (m.i.) of 3 is
Solutions-1
Math Review Kit: Solutions
3.
i)
ii)
iii)
iv)
v)
vi)
vii)
True
True
False because a(b + 3) = ab + 3a
True
True
True
False because (a + b)(5 ⳮ c) = a(5 ⳮ c) + b(5 ⳮ c) = 5a ⳮ ac + 5b ⳮ bc
Solutions to the Problems in Section 3
1.
i)
2(ⳮ2)2 + 7(ⳮ2) ⳮ 2
= 2(ⳮ2)(ⳮ2) + (ⳮ14) ⳮ 2
= 2 ⳯ 4 ⳮ 14 ⳮ 2
= 8 ⳮ 16
= ⳮ8
ii)
(1 + 3)2 ⳮ 1(3) + 32
= 42 ⳮ 3 + 9
= 16 ⳮ 3 + 9 = 25 ⳮ 3 = 22
iii)
iv)
v)
2.
2
2
2
[3 % (&1)]
'
2
2
(9&1)
'
2
8
2
'
2
1
'
64
32
& 13 % 32 & 2 ' & 1 % 9 & 2 ' & 8
3 × 22 × 8 × (&1)
2 × 8
'
3 × 4 × 8(&1)
i)
3ⳮ9+3⳯4⳰2
= 3 ⳮ 9 + 12 ⳰ 2
=3ⳮ9+6
= ⳮ6 + 6 = 0
ii)
9 ⳰ 3 ⳯ 4 ⳮ 2 ⳯ (ⳮ6)
= 3 ⳯ 4 ⳮ (ⳮ12)
= 12 + 12
= 24
iii)
2 + 8[6 ⳮ 9(8 ⳮ 10)]
= 2 + 8[6 ⳮ 9(ⳮ2)]
= 2 + 8[6 ⳮ (ⳮ18)]
= 2 + 8[6 + 18] = 2 + 8[24]
= 2 + 192 = 194
16
Solutions-2
'
3 × 4 × &8
16
'
&96
' &24
4
Math Review Kit: Solutions
3.
iv)
10 ⳮ {7 + 3[5 ⳮ (9 + 2)]}
= 10 ⳮ {7 + 3[ⳮ6]}
= 10 ⳮ {7 ⳮ 18}
= 10 ⳮ {ⳮ11}
= 10 + 11 = 21
v)
2{(2 ⳮ 3 + 9 (ⳮ2[2 ⳮ 3(3 + 6 ⳮ 12) + 6(3 ⳮ 7)]}
= 2 {(ⳮ1 + 9) ⳮ 2[2 ⳮ (9 ⳮ 12) + 6(ⳮ4)]}
= 2{8 ⳮ 2[2 ⳮ (ⳮ3) ⳮ 24]}
= 2{8 ⳮ 2[ⳮ19]}
= 2{8 + 38}
= 2{46} = 92
i)
x + 3x ⳮ (4x ⳮ6x)
= 4x ⳮ 4x + 6x
= 6x
ii)
2a ⳮ (3a + 2b) + (a ⳮ 3b)
= 2a ⳮ 3a ⳮ 2b + a ⳮ 3b
= ⳮ5b
iii)
x + y ⳮ z ⳮ (x ⳮ y + z) ⳮ (2x + y) + (x ⳮ 3z)
= x + y ⳮ z ⳮ x + y ⳮ z ⳮ 2x ⳮ y + x ⳮ 3z
= x ⳮ x ⳮ 2x + x + y + y ⳮ y ⳮ z ⳮ z ⳮ 3z
= ⳮx + y ⳮ 5z
iv)
2x{x ⳮ y[z + (x ⳮ 3y)]}
= 2x{x ⳮ y[z + x ⳮ 3y]}
= 2x{x ⳮ yz ⳮ yx + 3y2}
= 2x2 ⳮ 2xyz ⳮ 2x2y + 6xy2
v)
{x + 3y ⳮ [2x ⳮ (x + y]} ⳮ x ⳮ y
= {x + 3y ⳮ [2x ⳮ x ⳮ y]} ⳮ x ⳮ y
= {x + 3y ⳮ [x ⳮ y]} ⳮ x ⳮ y
= {x + 3y ⳮ x + y} ⳮ x ⳮ y
= 4y ⳮ x ⳮ y
= 3y ⳮ x
vi)
xy + [3x ⳮ 2y + (x + y)(x ⳮ 3y)]
= xy + [3x ⳮ 2y + x(xⳮ 3y) + y(x ⳮ 3y)]
= xy + [3x ⳮ 2y + x2 ⳮ 3xy + xy ⳮ 3y2]
= xy + 3x ⳮ 2 y + x2 ⳮ 3xy + xy ⳮ 3y2
= 3x ⳮ 2y + xy ⳮ 3xy + xy + x2 ⳮ 3y2
= 3x ⳮ 2y ⳮ xy + x2 ⳮ 3y2
Solutions-3
Math Review Kit: Solutions
Solutions to the Problems in Section 4
1.
2.
i)
4
2 × 2
1
'
'
12
2 × 2 × 3
3
ii)
21
3 × 7
3
'
'
49
7 × 7
7
iii)
36
2 × 2 × 3 × 3
9
'
'
64
2 × 2 × 2 × 2 × 2 × 2
16
iv)
abc 2
a × b × c × c
ac
'
'
b × c × d
d
bcd
i)
Lowest common demoninator (l.c.d.) is 7 ⳯ 13 = 91, hence the fractions are:
1 × 13
2 × 7
13
14
and
or
and
7 × 13
13 × 7
91
91
ii)
The l.c.d. is 75. The fractions are:
2 × 15 7 × 5
4 × 3
30 35 12
,
,
or
,
,
5 × 15 15 × 5 25 × 3
75 75 75
iii)
The l.c.d. is 195. The fractions are:
16 × 5 12 × 3
80
36
,
or
,
39 × 5 65 × 3
195 195
iv)
The l.c.d. is bd. The fractions are:
a × b c
ad c
,
or
,
b × d bd
bd bd
v)
The l.c.d. is xyz. The fractions are:
x × x
y × y
x2
y2
,
or
,
yz × x xz × y
xyz xyz
3.
i)
1
5
1
4
<
as
'
2
8
2
8
ii)
3
2
3
21
2
20
>
as
'
and
'
10
7
10
70
7
70
Solutions-4
Math Review Kit: Solutions
4.
iii)
&
2
3
2
14
3
15
> & as & ' &
and & ' &
5
7
5
35
7
35
iv)
&
5
10
5
5 × 2
10
' &
as &
' &
' &
19
38
19
19 × 2
38
i)
1
3
7
15
7 % 15
22
%
'
%
'
'
5
7
35
35
35
35
ii)
2
3
4
3
7
1
%
'
%
'
'
7
14
14
14
14
2
iii)
5
6
55
48
55 & 48
7
&
'
&
'
'
8
11
88
88
88
88
iv)
5
3
1
20
9
8
20 % 9 & 8
21
1
%
&
'
%
&
'
'
'
8
42
56
21
168
168
168
168
168
v)
a
c
ad
c
ad % c
%
'
%
'
b
bd
bd
bd
bd
vi)
x % y
x & y
x % y & (x & y)
x % y & x % y
2y
&
'
'
'
z
z
z
z
z
vii)
2
7
14
7
×
'
'
5
8
40
20
viii)
3
1
4
12
3
×
×
'
'
10
2
11
220
55
ix)
a
x
b
axb
ax
×
×
'
'
bc
y
d
bcyd
cyd
x)
2
1
2
3
6
2
%
'
×
'
'
9
3
9
1
9
3
xi)
5/6
5
5
1
5
'
÷ 8 '
×
'
8
6
6
8
48
xii)
5
3
7
35
2
' 5 ÷
' 5 ×
'
' 11
3/7
7
3
3
3
xiii)
a 2x
a
a 2x
bx
a 2x 2b
ax 2
÷
'
×
'
'
' ax 2
b
bx
b
a
ab
1
Solutions-5
Math Review Kit: Solutions
5.
6.
7.
8.
xiv)
xy
x2
xy
u2
xyu 2
yu
÷
×
'
'
'
2
2
2
zu
zu
zx
u
x
zux
i)
5% '
ii)
1
1/4
1
1
1
% '
'
×
'
4
100
4
100
400
iii)
3
3/4
3
1
3
% '
'
×
'
4
100
4
100
400
iv)
125% '
i)
1
1
=
⳯ 100% = 4%
25
25
ii)
2
2
=
⳯ 100% = 40%
5
5
iii)
2
2
2
=
⳯ 100% = 66.6% = 66 %
3
3
3
iv)
8
8
1
=
⳯ 100% = 53.3% = 53 %
15
15
3
i)
1
of $1000 = .125 ⳯ 1000 = $125.00
8
ii)
2
of $1250 = .66667 ⳯ 1250 = $833.34
3
iii)
1
of $1,000,000 = .11111111 ⳯ 1,000,000 = $111,111.11
9
iv)
1
of $10,000 = .166667 ⳯ 10,000 = $1,666.67
6
i)
5
1
'
100
20
125
5 × 5 × 5
5
1
'
'
' 1
100
2 × 2 × 5 × 5
4
4
2 8
1
1
[ & ( & )]
3 9
2
3
2 8
1
1
1
3
2
1
[ & ] as
&
'
&
'
=
6
3 9
6
2
3
6
6
Solutions-6
Math Review Kit: Solutions
2 16
3
[
&
]
3 18
18
2 13
26
= [ ] '
3 18
54
13
=
27
=
ii)
iii)
1 1
1
1
{ % [(1 & ) & (2 & )]}
2 4
3
6
1 1
2
11
3
12
{ % [ &
]} as 1 '
and 2 '
=
2 4
3
6
3
6
1 1
7
2
4
{ & } as
'
=
2 4
6
3
6
1 3
14
1 11
{
&
} ' (& )
=
2 12
12
2 12
11
= &
24
=
=
=
=
=
iv)
x
1
1
% [x % y & (x & y)]}
2
2
3
x
1
1
1
2 { % [x % y & x % y]}
2
2
3
3
x
1 2
4
2{ % [ x % y]}
2
2 3
3
x
1
2
2{ % x % y}
2
3
3
5
2
2{ x % y}
6
3
5
4
x % y
3
3
= 2{
a
b 1
a
& )( & )
2
3 4
6
a 1
a
b 1
= ( & ) & ( &
2 4
6
3 4
2
a
a
b
ab
&
%
&
=
8
12
18
12
1
1 2
1
a &
b %
= a &
8
12
12
(
a
)
6
1
ab
18
Solutions to the Problems in Section 5
1.
i)
243 ' (243)
1
2
1
5 2
' (3 )
' 3
5
2
2
' 3 × 3
1
2
Solutions-7
' 9 3
Math Review Kit: Solutions
2
ii)
iii)
2
6
27 3 ' (33) 3 ' 3 3 ' 32 ' 9
32
&
4
5
32
1
'
4/5
&3/2
25
49
iv)
1
'
(2 )
1
'
3/2
25
49
1
v)
2.
i)
iv)
4
2
'
493/2
'
25
'
3/2
1
16
(72)3/2
2 3/2
(5 )
'
76/2
6/2
5
'
73
5
3
343
125
'
1
32 × 27
32 × 33
35
1
'
'
' 3
'
2
4
6
9 × 81
3 × 3
3
3&1
5
2
3
2
9
4
53/2 × 5&1/2
5/2
5
× 5
25
'
1/2
3
(&8)2/3(&27)4/3
'
2 × &5
2
5
5
3
2 × 3
2
&5/2
× 2
× 2&3/2
2(
'
(2 )
' 5&2 '
25
3
5
'
×
34
2
4
' 21 × 3&1 '
2
3
1
25
'
&26/3 × &312/3
26/2× &56/3
4 × 81
324
81
31
'
'
' 1
8 × 25
200
50
50
1
× 21/2
25/2 × 23/2 × 2 × 21/2
3
'
1
1
3
×
5/2
3/2
2
2
2 ×
'
5
8
1
%
% 1 % )
211/2
25 × 2
32 2
2
2
2
'
'
'
3
3
3
3
* Alternatively:
2
2 2
'
(22)3/2(&53)2/3
v)
&1
(32)2
(&23)2/3(&33)4/3
'
&22 × &34
3
×
51
'
43/2(&125)2/3
3.
1
(0.0625) 4 ' [(.5)4] 4 ' .51 ' .5
ii)
iii)
'
5 4/5
(23)2/3(&33)4/3
(22)3/2 & (&53)2/3
'
(&2)2(&3)4
(23)(&52)
= 31.41421... = 31.414 = 4.728 (rounded value)
i)
3
ii)
5B ' 53.14159... ' 53.142 ' 157.096
Solutions-8
Math Review Kit: Solutions
iii)
4.
(i)
(ii)
(iii)
2e ' 22.71828... ' 22.718 ' 6.580
(x 2)(x 5)
5
(x)(x )
x 2y
axb
'
(a 2)3(b 3)2
3
(ab)
1
2
x7
'
x
6
' x1 ' x
x 6y 3z 3
a 3x 3b 3
'
a 6b 6
3 3
a b
'
x 3y 3z 3
a 3b 3
xyz
ab
'
3
' a 3b 3
2
% 2(x & x)] ' x[x
1
2
2
% 2x & 2x] ' x
3
2
% 2x 3 & 2x 2
(iv)
x[x
(v)
(x &2/5 y)5 ' [x &2/5y 1/2]5 ' [x &2/5]5 [y 1/2]5 ' x &2y 5/2 '
y 5/2
x2
(vi)
(x 2y)2/3(xyz)&1/3 ' [(x 2)2/3y 2/3] [x &1/3y &1/3z &1/3] ' [x 4/3y 2/3] [x &1/3y &1/3z &1/3] ' xy 1/3z &1/3 '
5.
P
A
= 100, i = .08, n = 5
= 1000(1 + .08)5
= 1000 ⳯ 1.46933
= $1469.33
6.
P
A
= 5000, i = .12, n = 4
= 5000(1 + .12)4
= 5000 ⳯ 1.57352
= $7867.60
7.
P
= 8000, i = .09, n = 10
= 8000(1 + .09)10
= 8000 ⳯ 2.36736
= $18,983.91
xy 1/3
z 1/3
Solutions to the Problems in Section 6
1.
Let x denote the amount of fertilizer and y, the crop yield. Then the two points are (1,2) and (3,5).
The question of the line passing through these points is:
Solutions-9
Math Review Kit: Solutions
y & y1'
y & 2 '
y2 & y1
x2 & x1
(x & x1)
[The Two Point Formula]
5 & 2
3
(x & 1) ' (x & 1)
3 & 1
2
y '
3
3
x &
% 2
2
2
y '
3
1
x %
2
2
[adding 2 to both sides]
for x = 2,
y '
3
1
6
1
7
1
(2) %
'
%
'
' 3
2
2
2
2
2
2
Hence, the estimated yield corresponding to 2 gallons of fertilizer is 3 1/2 bushels.
Solutions to the Problems in Section 7
1.
i)
5(x + 3) ⳮ 2 = 23
5x + 15 ⳮ 2 = 23
5x + 13 = 23
5x = 10 [adding (ⳮ13) to both sides]
x=2
ii)
x + 2 ⳮ 3x + 5 = x + 10
ⳮ2x + 7 = x + 10
Adding ⳮ7 and ⳮx both sides,
ⳮ2x + 7 ⳮ 7 ⳮ x = x + 10 ⳮ 7 ⳮ x
ⳮ3x = 3
x = ⳮ1
iii)
3(x + 5) + 5(2 ⳮ x) = 25
3x + 15 + 10 ⳮ 5x = 25
ⳮ2x + 25 = 25 [subtracting 25 from both sides]
ⳮ2x = 0
x=0
Solutions to the Problems in Section 8
1.
i)
81m2 ⳮ 25n2
= (9m)2 ⳮ (5n)2
= (9m + 5n)(9m ⳮ 5n) [using identity (3)]
Solutions-10
Math Review Kit: Solutions
ii)
a6 + 27
= (a2)3 + 33
Using identify (6) we get:
(a2 + 3)[(a2)2 ⳮ a2 ⳯ 3 + 32]
= (a2 + 3)(a4 ⳮ 3a2 + 9)
iii)
8a3 ⳮ 125b2
= (2a)3 ⳮ (5b)3
= (2a ⳮ 5b)[(2a)2 + 2a ⳯ 5b + (5b)2] [using identity (7)]
= (2a ⳮ 5b)(4a2 + 10ab + 25b2)
iv)
25abc + 5a2b + 10abc2
Factoring out 5ab, we get:
5ab(5c + a + 2c2)
v)
1 3 2
1
x y + x2y3
2
4
Factoring out
1 2 2
x y , we get:
4
1 2 2
x y (2x + y)
4
Solutions-11