MATH REVIEW KIT - UBC Sauder School of Business · Math Review Kit: Section 1 1-2 Figure 3 If we...

MATH REVIEW KIT

Reproduced with permission of the Certified General Accountant Association of Canada.

�Copyright 2005 by the Certified General Accountant Association of Canada and the UBC Real Estate Division. Allrights reserved. No part of this book may be reproduced in any form without written permission from the CertifiedGeneral Accountant Association of Canada and the UBC Real Estate Division. Published by the UBC Real EstateDivision. Printed in Vancouver, Canada.

R1MATH05

MATH REVIEW KIT

Table of Contents

Page

SECTION 1: THE NUMBER SYSTEM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-11.1 ARITHMETIC OF SIGNED (NEGATIVE OR POSITIVE) NUMBERS . . . . . . . . . 1-11.2 INFINITY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-11.3 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-2

SECTION 2: SIMPLE ALGEBRAIC OPERATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-12.1 PROPERTIES OF ADDITION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-12.2 PROPERTIES OF MULTIPLICATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-22.3 DISTRIBUTIVE LAW OF ADDITION AND MULTIPLICATION . . . . . . . . . . . . 2-32.4 EQUALITY AXIOMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-42.5 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-4

SECTION 3: COMPOSITE ALGEBRAIC OPERATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . 3-13.1 ALGEBRAIC EXPRESSIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-13.2 GROUPING SYMBOLS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-23.3 CORRECT USE OF PARENTHESES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-53.4 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-5

SECTION 4: ALGEBRA OF FRACTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-14.1 THE LOWEST FORM OF FRACTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-14.2 CONVERTING FRACTIONS TO A COMMON DENOMINATOR . . . . . . . . . . . . 4-24.3 ALGEBRA OF FRACTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-44.4 PERCENTAGES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-64.5 ROUNDING OFF A DECIMAL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-74.6 COMMON MISTAKES IN THE USE OF FRACTIONS . . . . . . . . . . . . . . . . . . . 4-84.7 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-9

SECTION 5: ALGEBRA OF EXPONENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-15.1 MEANING OF EXPONENT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-15.2 ALGEBRA OF EXPONENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-35.3 COMPOUND INTEREST . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-55.4 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-7

SECTION 6: LINEAR FUNCTIONS AND INTERPOLATION . . . . . . . . . . . . . . . . . . . . . . 6-16.1 LINEAR INTERPOLATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6-16.2 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6-1

SECTION 7: SYSTEMS OF LINEAR EQUATIONS AND INEQUALITIES . . . . . . . . . . . . . 7-17.1 LINEAR EQUATIONS IN ONE VARIABLE . . . . . . . . . . . . . . . . . . . . . . . . . . 7-17.2 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7-2

SECTION 8: IDENTITIES AND FACTORIZATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8-18.1 THE IDENTITIES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8-18.2 FACTORIZATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8-28.3 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8-3

SOLUTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions-1

&255

25&5

&25&5

1-1

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 thetwo numbers.

Example 1: 5 + 8 = 13, �5 + (�9) = �14

To add two numbers of different signs, take the difference of the absolute values and attach the sign of thenumber with the greater absolute value.

Example 2: 13 + (�8) = 5, �15 + 7 = �8, 12 � 19 = �7, and �11 + 29 = 18.

iii) To find the product or quotient of two numbers, multiply or divide the absolute values andattach a positive sign if the two numbers have the same sign; otherwise, attach a negative sign.

Example 3: 3 � 25 = 75, (�5) � (�3) = 15, 5 � (�20) = �100, (�8) � 6 = �48,

= �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 ofconceptual bounds to the number line, precisely �∞ and +∞, the negative and positive infinities. Mathematical

manipulations involving infinite numbers are different from that with finite numbers. The following propositionsdepict 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 ...

2 2 237

&89

B4

e 2 227

(&6)2 & 3(&7)(&3)(&1)

Math Review Kit: Section 1

1-2

Figure 3

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 numberand 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 between0 and 1.

1.3 PROBLEMS

1. Indicate whether the following numbers are rational or irrational.

�265, , 1.505050..., , , , .99999..., ,

2. Solve the following:

i) 13 + (�6) ii) 6 + (�19) iii) �5 + (�8)

iv) �7 � (�2) v) 7 � (+6) vi) (8)(�13)

vii) �12 + (10) viii) (�5)(�7) ix)

Note that: [(A)(B)] = [A � B] = [AB]

2-1

SECTION 2

SIMPLE ALGEBRAIC OPERATIONS

Manipulation of numbers comes within the realm of arithmetic. When we state a general rule of arithmetic, wehave progressed to algebra. An arbitrary number may be denoted by a, b, c, d or x, y, z, etc. In this sectionwe study the properties of basic algebraic operations of addition and multiplication and discuss the algebraicstructure 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

a + b = 0 (3)

and b + a = 0 (4)

The additive inverse of “a” is “-a”.

Example 2: Find the additive inverses of 12 and -5.

Since 12 + (�12) = 0 and (�12) + 12 = 0, the additive inverse of 12 is �12. Similarly, the additive inverseof �5 is 5.

Thus, the process of subtraction can be interpreted in terms of addition. 6 � 2 = 4 could be described as theaddition: 6 + (�2) = 4.

Commutative Law of Addition. If “a” and “b” are any two numbers then

a + b = b + a (5)

Thus: 2 + 5 = 5 + 2 = 7 and�2 + 5 = 5 + (�2) = 3.


2-2

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:

358329

147

This conventional way of addition involves a combination of commutative and associative laws of addition whichcan 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” (readas “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 = �6 then ab = 2 � (�6) = �12.

iii) 3 times a is 3a.

Definition 3 (Identity of Multiplication). The number 1 is called the identity of multiplication with theproperties:

a � 1 = a (7)

1 � a = a (8)

Simple Algebraic Operations

2-3

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 �35.

As 8 � 1/8 = 1 and 1/8 � 8 = 1, 1/8 is the multiplicative inverse of 8. Similarly, �1/35 is the multiplicativeinverse of �35.

Thus division of “a” by “b” (b =/ 0) can be identified with the multiplication of a with the multiplicative inverseof “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 (11)

Thus: 2 � 3 = 3 � 2 = 6

5 � (�9) = (�9) � 5 = �45.

Associative Law of Multiplication. If “a”, “b” and “c” are any three numbers, then:

a(bc) = (ab)c (12)

Thus: 4 � (8 � 3) = 4 � 24 = 96

and (4 � 8) � 3 = 32 � 3 = 96.

2.3 DISTRIBUTIVE LAW OF ADDITION AND MULTIPLICATION

Distributive Law. If “a”, “b” and “c” are any three numbers, then:

a(b + c) = ab + ac (13)


2-4

Using the rule (11), (13) can also be written as:

(b + c)a = ba + ca (14)

Thus: 5(3 + 2) = (5 � 3) + (5 � 2) = 15 + 10 = 25.

Example 6: Demonstrate the role of distributive law in the following scheme of multiplication:

39� 45

195+ 1560

1755

39 � 45 = (30 + 9) � (40 + 5)= {(30 + 9) � 5} + {(30 + 9) � 40}= {(30 � 5) + (9 � 5)} + {(30 � 40) + (9 � 40)}= 195 + 1560= 1755

2.4 EQUALITY AXIOMS

When “a” is equal to “b”, we write “a = b”. Following are the self-evident facts (axioms) that the equalityrelation satisfies:

Reflexive: a = a (15)

Symmetric: if a = b, then b = a (16)

Transitive: if a = b and b = c, then a = c (17)If a = b, then a +_ c = b +_ c (18)If a = b, then ac = bc (19)If a = b and c = d, then a +_ c = b +_ d (20)If a = b and c = d, then ac = bd (21)If a + c = b + c, then a = b (22)If ac = bc and c =/ 0, then a = b (23)

2.5 PROBLEMS

1. Find the additive inverse of the following numbers:

8, 91, �39, 0, �56, 1.

Simple Algebraic Operations

2-5

2. Find the multiplicative inverse of the numbers:

3, 1, 19, �9, �25.

3. 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 � z) = xy � xz

v) (y + z � w)x = xy + xz � xw

vi) (x �y)(z �w) = xz � xw � yz + yw

vii) (a + b)(5 � c) = 5a + 5b � ca + cb

x3

x

± 4

3-1

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 agiven context.

For example, 7, �2, 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 thereal number system.

Definition 3 (Algebraic Expression). An algebraic expression is created when one or more algebraic operationsare performed upon variables and constants. It may consist of one or more terms, separated from each otherby the signs + and �.

For example, “3x + 2y + xy � 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 �3ab; 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 theterm.

Thus, x + 5x = 6x, 8ab � 2ab + ab = (8 � 2 + l)ab = 7ab.

iii) The product “a � a” means a and “a � a � a” means a and so on.2 3

iv) Square root of x, denoted by , is a number y such that y = x. Similarly a cube root of x,2

symbolized as , is a number z such that z = x.3

For example 2 and -2 are square roots of 4 because 2 = 4 and (-2) = 4. The two square roots are2 2

sometimes written as .

Example 1:

i) Find the value of (x � 2x � 6) if x = 3.2

(3) � 2(3) � 6 = 9 � 6 � 6 = 3 � 6 = �32

(x 3 & x % 1)

&1 % 1 0


3-2

ii) Find the value of (x + y) if x = 3 and y = �5.2

(3 � 5) = (�2) = 42 2

iii) Find the value of for x = �1.

(�1) � = �1 � = �13

iv) When multiplying two expressions, the multiplication sign (x) may be dispensed with. The distributivelaw can be used to simplify the product.

Example 2: Simplify the following:

i) 3(x + 2 � 1)= 3x + 6 � 3 [Using the distributive law]= 3x + 3

ii) x + (5 � 3x)= x + 5 � 3x [Treating “+ (5 � 3x)” as “1 � (5 � 3x)”, and using the distributive law]= �2x + 5

Thus, parentheses preceded by a “+” sign may be removed without changing the expression.

iii) 3y � 2 � (x + y � 1)= 3y � 2 � x � y + 1 [Treating “-(x + y � 1)” as “-1(x + y � 1)”, and using the distributive

law]= 2y � x � 1

Thus, parentheses preceded by a “-” sign may be removed by changing the sign of each term of the expressionwithin the parentheses.

3.2 GROUPING SYMBOLS

An algebraic expression may involve grouping symbols: parenthesis ( ), brackets [ ], and braces { }. These areused to indicate the order in which the basic operations of “+, �, �, and �" are performed.

We now consider the rules of simplifying expressions involving composite algebraic operations.

Composite Algebraic Operations

3-3

Expressions with no Grouping Symbol

Rule 1: In the order of operations, multiplication and division belong to the same hierarchy followed by thehierarchy of addition and subtraction. When two operations in the same hierarchy appear one after theother, merely proceed from left to right.

Example 3: Simplify:

i) 12 � 2 � 30 � 5 � 2

= 12 � 2 � 6 � 2

= 12 � 2 � 12

= 10 � 12

= �2

ii) 18 � 3 � 2 + 1 � 6 � 4 � 2

= 6 � 2 + 1 � 24 � 2

= 12 + 1 � 12

= 13 � 12

= 1

Expressions with more than one Grouping Symbol

Rule 2: First solve the innermost parentheses, then brackets and then outermost braces.

Example 4: Simplify:

i) [3 � (10 � 9)] � 6 � 4= [3 � (1)] � 6 � 4= [2] � 6 � 4= 12 � 4= 3


3-4

ii) 5x � [4x + x(3x � 2)]

= 5x � [4x + 3x � 2x]2

= 5x � [2x + 3x ]2

= 5x � 2x � 3x2

= 3x � 3x2

iii) 2{8 � 5 + 2[13 � 9(5 + 3 � 6)] � 5}

= 2{40 + 2[13 � 9(2)] � 5}

= 2{40 + 2[13 � 18] � 5}

= 2{40 + 2[-5] � 5}

= 2{40 � 10 � 5}

= 2{40 � 2}

= 2{38}

= 76

iv) 2a � 4{a � 3[2b + 5(a � 1) � 2(3b � 5)]}

= 2a � 4{a � 3[2b + 5a � 5 � 6b + 10]}

= 2a � 4{a � 3[5a � 4b + 5]}

= 2a � 4{a � 15a + 12b � 15}

= 2a � 4{-14a + 12b � 15}

= 2a + 56a � 48b + 60

= 58a � 48b + 60

13

13

13

13

13

19

Composite Algebraic Operations

3-5

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 � y � 1 (incorrect)

The correct answer is x � (y � 1).

ii) Find the area of a rectangle with sides a and a � 2 units.

Answer: (a) � (a) � 2 (incorrect)

The correct answer is a(a � 2).

iii) x � (y + 1) = x � y + 1 (incorrect)

The correct solution is x � (y + 1) = x � y � 1.

iv) ( x)( y) = xy (incorrect)

The correct solution is: ( x)( y) = xy

v) (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) 2x + 7x � 2 at x = �22

ii) (x + y) � xy + y at x = 1, y = 32 2

2

(x 2 % y)2

& x 3 % y 2 & 2

3a 2bc

ab


3-6

iii) at x = 3, y = �1

iv) at x = 1, y =3

v) at a = 2, b = 8, c = �1

2. Simplify the following:

i) 3 � 9 + 3 � 4 � 2

ii) 9 � 3 � 4 � 2 � (�6)

iii) 2 + 8[6 � 9(8 � 10)]

iv) 10 � {7 + 3[5 � (9 + 2)]}

v) 2{(2 � 3 + 9) � 2[2 � (3 + 6 � 12) + 6(3 � 7)]}


i) x + 3x � (4x � 6x)

ii) 2a � (3a + 2b) + (a � 3b)

iii) x + y � z � (x � y + z) � (2x + y) + (x � 3z)

iv) 2x{x � y[z + (x � 3y)]}

v) {x + 3y � [2x � (x + y)]} � x � y

vi) xy + [3x � 2y + (x + y)(x � 3y)]

23

46

23

2 × 23 × 2

46

2050

20 ÷ 1050 ÷ 10

25

1216

50125

6a 2b

9ab 2

7a 2c

28ab 2c 2

4-1

SECTION 4

ALGEBRA OF FRACTIONS

In this section, the algebraic operations discussed in section 3 will be applied to fractions. Percentages androunding procedure will be discussed as well.

4.1 THE LOWEST FORM OF FRACTIONS

Definition 1 (Fraction). An expression of the form a/b, b =/ 0, is called a fraction.

Division by zero is not allowed. Since a/b = c means a = bc, a fraction like a/0 is not defined as we cannotfind 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 = because 2 � 6 = 4 � 3.

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-zeronumber without changing its value, i.e., a/b = ka/kb, k =/ 0.

Example 1:

i) = =

ii) = =

Definition 3 (Lowest Form of a Fraction). A fraction, a/b, is said to be of lowest form if a and b have nocommon factor. For example, 2/3, 1/7, 2/9, 11/21 are fractions in lowest form, whereas, 6/4, 10/50, ab/bd,ab /a c are not. (A “Factor” is a number or variable that can be divided into both the numerator and the2 2

denominator such as to reduce the value of each.)

Example 2: Reduce the fractions to their lowest form:

i) , ii) , iii) , iv)

1216

2 × 2 × 32 × 2 × 2 × 2

34

50125

2 × 5 × 55 × 5 × 5

25

6a 2b

9ab 2

2 × 3 × a × a × b3 × 3 × a × b × b

2a3b

7a 2bc

28ab 2c 2

7 × a × a × b × c2 × 2 × 7 × a × b × b × c × c

a4bc

12

13

23

56

311

522

36

26

12

× 33

'36

13

× 22

'26

1218

1518

23

× 66

'1218

56

× 33

'1518


4-2

i) = [Writing the numerator and denominator as a product of prime

factors]

= [Dividing numerator and denominator by common factors 2 � 2]

ii) = =

iii) = =

iv) = =

4.2 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 acommon denominator bd. Sometimes it is convenient to convert into fractions with lowest commondenominator.

Example 3: Place the following fractions over a common denominator.

i) , ii) , iii) ,

i) 2 � 3 = 6 is a common denominator. The fractions are then:

, (i.e., , )

ii) 3 � 6 = 18

The fractions are , ( , )

66242

55242

311

× 2222

'66242

522

× 1111

'55242

36

26

23

56

46

56

311

522

622

522

215

45 21201

215

15

15

Algebra of Fractions

4-3

iii) 11 � 22 = 242

The fractions are , , ( ), ( )

Example 4: Place the fractions in example 3 over the lowest common denominator.

i) ,

ii) , have 6 as a common denominator which is lowest.

The fractions are , .

iii) , have 22 as the lowest common denominator.

The fractions are , .

Definition 4 (Proper and Improper Fractions). A fraction in which the denominator is greater than the numeratoris 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: Convert the following into mixed numbers.

i)

So that = 4 + or 4

403

133 40

31091

403

13

ab

cd

ad ± bcbd

25

54

820

%1520

'8 % 15

20'

2320

' 1 320

512

&1718

'1536

&3436

'15 & 34

36'

&1936

2a3b

%ac

4bd


4-4

ii)

So that = 13

4.3 ALGEBRA OF FRACTIONS

Rule 3: i) Sum (Difference) of Fractions.

+_ =

ii) To find the sum (difference) of two fractions, place them over the lowest common denominatorand add (subtract) the numerator.

Example 5: Solve the following:

i) +

=

ii)

[Note that 36 is the lowest common denominator]

iii)

2a3b

× 4d4d

%ac

4bd× 3

3

2a × 4d12bd

%3 × ac12bd

'8ad % 3ac

12bd'

a(8d % 3c)12bd

ab

× cd

'acbd

519

× 68

'5 × 619 × 8

'30152

'1576

2a3b

× b 2

a 2

'2a × b 2

3b × a 2'

2a × b × b3b × a × a

'2b3a

ab

÷ cd

'ab

× dc

'adbc


4-5

Though Rule 3 (i) can be used, we may also see that 12bd is the lowest common denominator.

Rule 4: Multiplication of Fractions

Example 7: Solve the following:

i)

ii)

(Note here the use of Section 4.1 [definition 3] in reducing to a lowest-form fraction.)

Rule 5: Division of Fractions

23

÷ 58

'23

× 85

'1615

' 1 115

2a3b

÷ 2c

a 2b

'2a3b

× a 2b2c

'2a 3b6bc

'a 3

3c

8100

225

8100

12

% '12

÷ 1001

'12

× 1100

'1

200

12

÷ 100 ' .05 ÷ 100 ' .005

35


4-6

Example 8: Solve the following.

i)

ii)

4.4 PERCENTAGES

Definition (Percentage). Percent, denoted by %, is a fraction with 100 as the denominator.

Example 9: Convert the following percentages into fractions and decimals.

i) 8%

8% = =

Also, = .08

ii) 1/2%

Also,

Example 10: Convert the following fractions to percentages:

i)

35

'3 × 205 × 20

'60100

' 60%

56

56

'.8333̇ × 100

1 × 100'

83.33̇100

' 83 13

%

56

56

' .83̇

56


4-7

ii)

In this case, it is difficult to convert the denominator to 100. Therefore convert the fraction to a decimal thenmultiply and divide by 100.

(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 5or 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: Round off to nearest cents.

i) of $100.00

Now Multiplication by 100 will shift the decimal two places to the right. Therefore, round

off at the fourth decimal place.

Hence � 100 = .8333 � 100 = $83.33

17

17

.142857

17

× 1000 ' .14286 × 1000 ' $142.86

x % y5

'x5

% y

x % y5

'x5

%y5

a

b

c

d

'a × c

bd

a

b

c

d

'ab

÷ cd

'ab

× dc

'adbc

728

'7

2 × 2 × 7'

04

' 0

728

'7

2 × 2 × 7'

14

77

'11


4-8

ii) of $1000.00

Now, =

Multiplying by 1000 will shift the decimal three places to the right. Therefore, round off at the fifthplace.

(Note: In Mortgage Finance calculations, payments are always rounded up, even if the decimal digit is less than5.)

4.6 COMMON MISTAKES IN THE USE OF FRACTIONS

i) (wrong)

The correction solution is:

ii) (wrong)

In fact,

iii) (wrong)

Since

412

2149

3664

abc 2

bcd

17

, 213

25

, 715

, 425

1639

, 1265

ab

, cbd

xyx

, yxz

12

58

310

27

&25

&37

&519

&1038

15

%37

27

%314

58

&611

542

%356

&121

ab

%cbd

x % yz

&x & y

z25

× 78

310

× 12

× 411

abc

× xy

× bd

29

÷ 13

5/68

53/7

a 2xb

÷ abx

xyzu

÷ x 2

u 2

14

34


4-9

4.7 PROBLEMS

1. Reduce the following to lowest form:

i) ii) iii) iv)

2. Place the fractions over their lowest common denominator:

i) ii) iii)

iv) v)

3. Insert one of <, = or > in the following:

i) ii) iii) iv)

Hint: Convert to common denominator and compare numerators.Note: “>” means “greater than”; “<” means “less than”.

4. Solve the following and reduce to lowest form:

i) ii) iii) iv)

v) vi) vii)

viii) ix) x)

xi) xii) xiii) xiv)

5. Convert the following percentages to fractions:

i) 5% ii) % iii) % iv) 125%

125

25

23

815

18

23

19

16

23

[ 89

& ( 12

&13

)] 12

{ 14

% [(1 &13

) & (2 &16

)]}

2{ x2

%12

[x % y &13

(x & y)]} ( a2

&b3

)( 14

&a6

)


4-10

6. Convert the following fractions to percentages:

i) ii) iii) iv)

7. Round off the following to the nearest cent:

i) of $1000 ii) over $1250

iii) of a million dollars iv) of $10,000


i) ii)

iii) iv)

x

13

12

1228

1

a n

1

23

18

12

( 12

) 11

22

114

na

2a a

3a

4a

na

4 4

5-1

SECTION 5

ALGEBRA OF EXPONENTS

In section 3, we introduced terms like x and , referred to as exponential expressions. In this section we study2

the algebra of exponential expressions and numbers. The compound interest formula is discussed as well.

5.1 MEANING OF EXPONENT

An expression, symbolized by a , is called an exponential number or expression. The non-zero number “a” isn

called the base, and the number “n” is called the power or exponent. The interpretation of “n” depends onwhether “n” is a positive integer, zero, negative integer, fractional or irrational number.

Definition 1: If “n” is a positive integer, then a means “a” multiplied by itself “n” times. That is,n

a = a � a � a � ... � a [n times]n

For example, 3 = 3 � 3 � 3 � 3 = 81. Similarly (�1) = �1; ( ) = 1/81; (� ) = � ; and4 5 4 7

1 = 1.6

Definition 2: If the exponent is a negative integer, then we define a = �n

For example, 2 = = �3

Similarly 2 = ; = = = 4�1 -2

Definition 3: a = 1.0

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 a as a number “b” such that b = a. In other1/n n

words, a is n root of “a”, and is denoted by .1/n th

Thus, , which is also written as , is called a square root of a, is called a cube root of a and is

called a fourth root of a and so on. Note that we called as “a” n root and not “the” n root because thereth th

may be more than one n root. For example 4 has two roots, 2 = and �2 = � .th

1

1251/3

15

1

321/5

12

&13&1

na

2, 3,3

7,4

11, 13

3 ' 1.73205080 ... • 1.7321

37 ' 1.912931183 ... • 1.9129

411 ' 1.821160287 • 1.8212

3&29 ' &3.072316826 ... • &3.0723

qa


5-2


i) 8 ii) 81 iii) 125 iv) 321/3 1/4 1/3 1/5� �

i) 8 = 2 because 2 = 81/3 3

ii) 81 = 3 because 3 = 811/4 4

iii) 125 = = because 5 = 125�1/3 3

iv) 32 = = because 2 = 32.�1/5 5

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 a , “a” is negative. Then the roots are not defined when “n” is even. However,1/n

when “n” is odd, a root may exist. For example, (-1) or is not defined in the context of real numbers1/2

as there is no number “b” such that b = �1. But = -1 as (-1) = -1. For the sake of simplifying2 3

matters, we shall assume that a > 0 whenever finding a root of “a” is involved. Also, a would mean the1/n

positive n root of “a”. Talking of n roots, the following theorem, which generates many irrational numbers,th th

will not be out of place.

Theorem 1: If “a” is a prime number, then for every “n” where n = 1, 2, 3, 4, ... , is an irrationalnumber.

Thus, are all irrational numbers.

Example 2: Use your calculator, if possible, to check the following:

i)

ii)

iii)

iv)

Definition 5: Let p/q be a fraction, then a is defined as the q root of a raised to the power p.p/q th

Symbolically, a = ( ) or (a )p/q p 1/q p

4

327

1

253/2

1

( 25)3

1

53

1125

0.09

5 2

2B

ab

a n

b n

a m

a n

Algebra of Exponents

5-3

Example 3:

i) 4 = ( ) = 2 = 83/2 3 3

ii) 27 = ( ) = 3 = 814/3 4 4

iii) 25 = = = = -3/2

iv) (0.09) = ( ) = (0.3) = .0273/2 3 3

The exponential expression, a , with an irrational exponent “n” is difficult to compute because of infinite decimaln

expansion of irrational numbers. However, an approximation of the irrational exponent by a finite decimal, thenumber of digits depending upon the degree of approximation desired, can give a near value of a .n

Example 4:

i) = 5 ≅ 5 = 9.735 [Using a calculator and rounding off at third decimal]1.41421... 1.414

ii) = 2 ≅ 2 = 8.827 [Using a calculator and rounding off at third decimal]3.14159... 3.142

5.2 ALGEBRA OF EXPONENTS

Now that a is defined for any real number “n” and a =/ 0, we are set to go into the algebra of exponents.n

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: a a = am n m+n

Rule 2: (a ) = am n mn

Rule 3: (ab) = a bn n n

Rule 4: ( ) = n

Rule 5: = am n�

23 × 16 × 25

8 × 24 × 2

'23 × 24 × 25

23 × 24 × 2'

212

28

32 × 9 × 33

27 × 34

'32 × 32 × 33

33 × 34

'37

37

'21872187

37

37' 37&7 ' 30

(a 2b 2) × (a 3) × (b)

(2ab)4

'a 2a 3bb 2

24a 4b 4

'a 5b 3

16a 4b 4

'116

ab &1


5-4


i)

[using rule 1]

= 2 [using rule 5]12-8

= 2 = 164

ii)

[using rule 1]

= 1

This example gives us an opportunity to check that a = 1 for a = 3.0

[using rule 5]

Hence = 3 = 10

iii)

[using rule 3]

[using rule 1]

[using rule 5]

'a

16b

'(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

ab 2c 3

x 3y 2

2

'(ab 2c 3)2

(x 3y 2)2

'a 2b 4c 6

x 6y 4


5-5

Figure 1

iv) (2x yz ) (x yz )2 3 2 3 4 1�

[using rule 3]

[using rule 2]

= 4xyz [using rule 5]2

v)

[using rule 4]

[using rules 3 and 2]

5.3 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 after3 years?

243

( 2549

)&3/2


5-6

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) (.08).2

Amount due after 3 years = 1000(1 + .08) + 1000(1 + .08) (.08)2 2

= 1000(1 + .08) [1 + .08]2

= 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 compoundedyearly, then the amount A due after n years is given by:

A = P(1 +i)n

This is the compound interest formula.

Example 7: Solve example 6 using the compound interest formula.

Here P = 1000, i = .08, n = 3

Therefore, A = 1000(1 + .08) = 1000 � 1.259713

= $1259.71

5.4 PROBLEMS

1. Compute the following (do not give decimal answers):

i) ii) 27 iii) 322/3 4/5�

iv) v) (0.0625)1/4

32 × 279 × 81

23

94

53/2 × 5&1/2

55/2 × 51/2

(&8)2/3 (&27)4/3

43/2 (&125)2/3

2 × 3&1 × 2

2&5/2 × 2&3/2

3 2 5B 2e

(x 2) × (x 5)

(x) × (x 5)

x 2yzaxb

(a 2)3(b 3)2

(ab)3

y


5-7

2. Evaluate the following:

i) ii) ( ) ( ) iii)2 2

iv) v)

3. Approximate the irrational exponent by a three-digit decimal and compute using a calculator:

i) ii) iii)

4. Simplify:

i) ii) ( ) iii)3

iv) x[x + 2(x � x)] v) (x ) vi) (x y) (xyz)1/2 2 -2/5 5 2 2/3 1/3�

5. If $1,000 is invested for 5 years at 8% compounded yearly, what will this investment be worth at theend of this time?

6. On January 1, 1975, a man incurs a debt of $5,000 at 12% compounded annually. If the lenderdemands 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 accountpaying 9% compounded annually. What is the maturity value of this account after 10 years when hebecomes 65?

y2 & y1

x2 & x1

(x & x1)

1469.33 & 1338.238 & 6

131.102

6-1

SECTION 6

LINEAR FUNCTIONS AND INTERPOLATION

6.1 LINEAR INTERPOLATION

Suppose two points (x ,y ) and (x ,y ) lie on a straight line and we want to find the value of y for a given x1 1 2 2 3 3

between x and x such that the point (x ,y ) lies on the straight line. This process is called linear interpolation.1 2, 3 3

If x is outside x and x , then it is called intrapolation. In both the cases, the “two point formula” as indicated3 1 2

in the following example is used.

Example 1: An investment of $1000 for 5 years at 6% interest compounded annually yields $1338.23 andat 8% compounded annually yields $1469.33. Use linear interpolation to find the amount at7% compounded annually.

Consider the points (6, 1338.23) and (8, 1469.33). Substituting these points for (x ,y ) and (x ,y ) into the two1 1 2 2

point formula and letting x = 7, we get:

y � y = [The two point formula]1

y � 1338.23 = (7 � 6)

= (1)

= 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 theapplication of 3 gallons of fertilizer yields 5 bushels of the crop. Estimate the yield if 2 gallons offertilizer were used, using linear (two-point) interpolation.

35

35

35

35

7-1

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 aspectsof 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 � 3 = 0 and check to see that the solution satisfies the equation.

Using the laws of equality, we have:

5x � 3 + 3 = 3 (adding 3 on both sides)

5x = 3

This is also known as transposing � 3 on the right hand side of the equation by changing its sign.Simplifying further, we get:

x = (dividing both sides by 5)

Substituting x = in the original equation, we get:

5( ) � 3 = 0

3 � 3 = 0

0 = 0

Hence, x = is the solution of the given equation.


7-2

ii) Solve 3x + 5 = x + 15 and check the solution.

Adding (-x � 5) on both sides we get:

3x + 5 � x � 5 = x + 15 � x � 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) � 2 = 23

ii) x +2 � 3x + 5 = x + 10

iii) 3(x + 5) +5(2 � x) = 25

8-1

SECTION 8

IDENTITIES AND FACTORIZATION

In this section, we introduce identities and use them in the factorization of algebraic expressions. Factorizationof 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 allpossible values of the variables in the equation.

For example, 2x = x + x; x + x = x(x + 1).2

We list below a few common identities which can be confirmed by using the distributive law upon the left handside (L.H.S.):

(x + y) = x + 2xy + y (1)2 2 2

(x � y) = x � 2xy + y (2)2 2 2

(x + y)(x � y) = x � y (3)2 2

(x + y) = x + 3x y + 3xy + y (4)3 3 2 2 3

(x � y) = x � 3x y + 3xy � y (5)3 3 2 2 3

(x + y)(x � xy + y ) = x + y (6)2 2 3 3

(x - y)(x + xy + y ) = x � y (7)2 2 3 3

As an illustration, we check identity (1)

L.H.S. = (x + y) = (x + y)(x + y)2

= x(x + y) + y(x + y) [using the distributive law]

= x + xy + yx + y2 2

= x + 2xy + y = R.H.S.2 2


i) (2x + 3) , ii) (3a � 5) , iii) (m + 1)(m � 1), iv) (x + 3)2 2 3


8-2

i) Using identity (1), we have:

(2x + 3) = (2x) + 2[(2x) � (3)] + 32 2 2

= 4x + 12x + 92

ii) Using identity (2), we get:

(3a � 5) = (3a) � 2[(3a) � (5)] + 52 2 2

= 9a � 30a + 252

iii) In this case, identity (3) applies. Thus:

(m + 1)(m � 1) = m � 12 2

= m � 12

iv) Using identity (4), we have:

(x + 3) = x + 3(3x ) + 3(9x) + 33 3 2 3

= x + 9x + 27x + 273 2

8.2 FACTORIZATION

Definition 2 (Factorization). By factorization of an algebraic expression, we mean writing the expression as aproduct 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

x � y = (x + y)(x � y)2 2

Note that these two examples are also identities. The identities and the distributive law are extensively used infactorization.

Example 2: Factorize the following expressions:

i) xy + xz + 2x

ii) a b + 3ab + ab2 2

iii) 3x y + 6x y + 27x y + 3x y3 2 2 2 3 2

12

x 3y 2 %14

x 2y 3

Identities and Factorization

8-3

iv) 27m � 8n3 3

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:

a b + 3ab + ab = ab(a + 3 + b)2 2

iii) Factoring out 3x y and using the laws of exponents:2

3x y + 6x y + 27x y + 3x y = 3x y(x + 2y + 9y + 1)3 2 2 2 3 2 2 2

iv) 27m � 8n = (3m) � (2n)3 3 3 3

= (3m � 2n)[(3m) + (3m � 2n) + (2n) ] [using the identity (7)]2 2

= (3m � 2n)(9m + 6mn + 4n )2 2

8.3 PROBLEMS

1. Factorize the following:

i) 81m � 25n2 2

ii) a + 276

iii) 8a � 125b3 3

iv) 25abc + 5a b + 10abc2 2

v)

Solutions

237

89

227

(&6)2 & 3(&7)(&3)(&1)

'&12 & (&21)

3'

93

' 3

13

13

13

119

19

125

Solutions-1

SOLUTIONS

Solutions to the Problems in Section 1

1. The rational numbers are: �265, 1.505050..., , � , .99999...,

2. i) 13 + (�6) = 7

ii) 6 + (�19) = �13

iii) �5 + (�8) = �13

iv) �7 � (�2) = �5

v) 7 � (+6) = 1

vi) (8)(�13) = �104

vii) �12 + (10) = �2

viii) (�5)(�7) = 35

ix)


1. The additive inverse (a.i.) of 8 is �8 because 8 + (�8) = 0 and �8 + 8 = 0.

Similarly the a.i. of 91 is �91.

The a.i. of 0 is 0.

The a.i. of �56 is 56.

The a.i. of 1 is �1.

The a.i. of �39 is 39.

2. The multiplicative inverse (m.i.) of 3 is because 3 � = 1 and � 3 = 1. Similarly the m.i.

of 1 is 1, the m.i. of 19 is , the m.i. of �9 is � and the m.i. of �25 is � .

2

[32 % (&1)]2'

2

(9&1)2'

2

82'

264

'132

& 13 % 32 & 2 ' & 1 % 9 & 2 ' & 8

3 × 22 × 8 × (&1)

2 × 8'

3 × 4 × 8(&1)

16'

3 × 4 × &8

16'

&964

' &24

Math Review Kit: Solutions

Solutions-2

3. i) Trueii) Trueiii) False because a(b + 3) = ab + 3aiv) Truev) Truevi) Truevii) False because (a + b)(5 � c) = a(5 � c) + b(5 � c) = 5a � ac + 5b � bc


1. i) 2(�2) + 7(�2) � 22

= 2(�2)(�2) + (�14) � 2= 2 � 4 � 14 � 2= 8 � 16= �8

ii) (1 + 3) � 1(3) + 32 2

= 4 � 3 + 92

= 16 � 3 + 9 = 25 � 3 = 22

iii)

iv)

v)

2. i) 3 � 9 + 3 � 4 � 2= 3 � 9 + 12 � 2= 3 � 9 + 6= �6 + 6 = 0

ii) 9 � 3 � 4 � 2 � (�6)= 3 � 4 � (�12)= 12 + 12= 24

iii) 2 + 8[6 � 9(8 � 10)]= 2 + 8[6 � 9(�2)]= 2 + 8[6 � (�18)]= 2 + 8[6 + 18] = 2 + 8[24]= 2 + 192 = 194


Solutions-3

iv) 10 � {7 + 3[5 � (9 + 2)]}= 10 � {7 + 3[�6]}= 10 � {7 � 18}= 10 � {�11}= 10 + 11 = 21

v) 2{(2 � 3 + 9 (�2[2 � 3(3 + 6 � 12) + 6(3 � 7)]}= 2 {(�1 + 9) � 2[2 � (9 � 12) + 6(�4)]}= 2{8 � 2[2 � (�3) � 24]}= 2{8 � 2[�19]}= 2{8 + 38}= 2{46} = 92

3. i) x + 3x � (4x �6x)= 4x � 4x + 6x= 6x

ii) 2a � (3a + 2b) + (a � 3b)= 2a � 3a � 2b + a � 3b= �5b

iii) x + y � z � (x � y + z) � (2x + y) + (x � 3z)= x + y � z � x + y � z � 2x � y + x � 3z= x � x � 2x + x + y + y � y � z � z � 3z= �x + y � 5z

iv) 2x{x � y[z + (x � 3y)]}= 2x{x � y[z + x � 3y]}= 2x{x � yz � yx + 3y }2

= 2x � 2xyz � 2x y + 6xy2 2 2

v) {x + 3y � [2x � (x + y]} � x � y= {x + 3y � [2x � x � y]} � x � y= {x + 3y � [x � y]} � x � y= {x + 3y � x + y} � x � y= 4y � x � y= 3y � x

vi) xy + [3x � 2y + (x + y)(x � 3y)]= xy + [3x � 2y + x(x� 3y) + y(x � 3y)]= xy + [3x � 2y + x � 3xy + xy � 3y ]2 2

= xy + 3x � 2 y + x � 3xy + xy � 3y2 2

= 3x � 2y + xy � 3xy + xy + x � 3y2 2

= 3x � 2y � xy + x � 3y2 2

412

'2 × 2

2 × 2 × 3'

13

2149

'3 × 77 × 7

'37

3664

'2 × 2 × 3 × 3

2 × 2 × 2 × 2 × 2 × 2'

916

abc 2

bcd'

a × b × c × cb × c × d

'acd

1 × 137 × 13

and 2 × 713 × 7

or 1391

and 1491

2 × 155 × 15

, 7 × 515 × 5

, 4 × 325 × 3

or 3075

, 3575

, 1275

16 × 539 × 5

, 12 × 365 × 3

or 80195

, 36195

a × bb × d

, cbd

or adbd

, cbd

x × xyz × x

, y × yxz × y

or x 2

xyz, y 2

xyz

12

< 58

as 12

'48

310

> 27

as 310

'2170

and 27

'2070


Solutions-4


1. i)

ii)

iii)

iv)

2. i) Lowest common demoninator (l.c.d.) is 7 � 13 = 91, hence the fractions are:

ii) The l.c.d. is 75. The fractions are:

iii) The l.c.d. is 195. The fractions are:

iv) The l.c.d. is bd. The fractions are:

v) The l.c.d. is xyz. The fractions are:

3. i)

ii)

&25

> &37

as &25

' &1435

and &37

' &1535

&519

' &1038

as &519

' &5 × 219 × 2

' &1038

15

%37

'735

%1535

'7 % 15

35'

2235

27

%314

'414

%314

'714

'12

58

&611

'5588

&4888

'55 & 48

88'

788

542

%356

&121

'20168

%9

168&

8168

'20 % 9 & 8

168'

21168

'18

ab

%cbd

'adbd

%cbd

'ad % c

bd

x % yz

&x & y

z'

x % y & (x & y)z

'x % y & x % y

z'

2yz

25

× 78

'1440

'720

310

× 12

× 411

'12220

'355

abc

× xy

× bd

'axbbcyd

'axcyd

29

%13

'29

× 31

'69

'23

5/68

'56

÷ 8 '56

× 18

'548

53/7

' 5 ÷ 37

' 5 × 73

'353

' 11 23

a 2xb

÷ abx

'a 2xb

× bxa

'a 2x 2b

ab'

ax 2

1' ax 2


Solutions-5

iii)

iv)

4. i)

ii)

iii)

iv)

v)

vi)

vii)

viii)

ix)

x)

xi)

xii)

xiii)

xyzu

÷ x 2

u 2'

xyzu

× u 2

x 2'

xyu 2

zux 2'

yuzx

5% '5

100'

120

14

% '1/4100

'14

× 1100

'1

400

34

% '3/4100

'34

× 1100

'3

400

125% '125100

'5 × 5 × 5

2 × 2 × 5 × 5'

54

' 1 14

125

125

25

25

23

23

66.6% 66 23

%

815

815

53.3% 53 13

%

18

23

19

16

23

[ 89

& ( 12

&13

)]

23

[ 89

&16

] as 12

&13

'36

&26

'16


Solutions-6

xiv)

5. i)

ii)

iii)

iv)

6. i) = � 100% = 4%

ii) = � 100% = 40%

iii) = � 100% = =

iv) = � 100% = =

7. i) of $1000 = .125 � 1000 = $125.00

ii) of $1250 = .66667 � 1250 = $833.34

iii) of $1,000,000 = .11111111 � 1,000,000 = $111,111.11

iv) of $10,000 = .166667 � 10,000 = $1,666.67

8. i)

=

23

[ 1618

&318

]

23

[ 1318

] '2654

1327

12

{ 14

% [(1 &13

) & (2 &16

)]}

12

{ 14

% [ 23

&116

]} as 1 '33

and 2 '126

12

{ 14

&76

} as 23

'46

12

{ 312

&1412

} '12

(& 1112

)

&1124

2{ x2

%12

[x % y &13

(x & y)]}

2{ x2

%12

[x % y &13

x %13

y]}

2{ x2

%12

[ 23

x %43

y]}

2{ x2

%13

x %23

y}

2{ 56

x %23

y}

53

x %43

y

( a2

&b3

)( 14

&a6

)

a2

( 14

&a6

) &b3

( 14

&a6

)

a8

&a 2

12&

b12

%ab18

18

a &112

a 2 &112

b %118

ab

243 ' (243)12 ' (35)

12 ' 3

52 ' 32 × 3

12 ' 9 3


Solutions-7

=

=

=

ii)

=

=

=

=

iii) =

=

=

=

=

=

iv)

=

=

=


1. i)

2723 ' (33)

23 ' 3

63 ' 32 ' 9

32&

45 '

1

324/5'

1

(25)4/5'

1

24'

116

2549

&3/2

'1

2549

3/2'

493/2

253/2'

(72)3/2

(52)3/2'

76/2

56/2'

73

53'

343125

(0.0625)14 ' [(.5)4]

14 ' .51 ' .5

32 × 279 × 81

'32 × 33

32 × 34'

35

36'

1

3&1' 3

23

5 94

2

'25

35× (32)2

(22)2'

25

35× 34

24' 21 × 3&1 '

23

53/2 × 5&1/2

55/2 × 51/2'

51

53' 5&2 '

125

(&8)2/3(&27)4/3

43/2(&125)2/3'

(&23)2/3(&33)4/3

(22)3/2(&53)2/3'

&26/3 × &312/3

26/2× &56/3

'&22 × &34

23 × &52'

4 × 818 × 25

'324200

'8150

' 1 3150

2 × 3&1 × 2

2&5/2 × 2&3/2'

2 × 13

× 21/2

1

25/2× 1

23/2

'25/2 × 23/2 × 2 × 21/2

3

'

2( 52

%82

% 1 %12

)

3'

211/2

3'

25 × 23

'32 2

3

(23)2/3(&33)4/3

(22)3/2 & (&53)2/3'

(&2)2(&3)4

(23)(&52)

3 2

5B ' 53.14159... ' 53.142 ' 157.096


Solutions-8

ii)

iii)

iv)

v)

2. i)

ii)

iii)

iv)

v)

* Alternatively:

3. i) = 3 = 3 = 4.728 (rounded value)1.41421... 1.414

ii)

2e ' 22.71828... ' 22.718 ' 6.580

(x 2)(x 5)

(x)(x 5)'

x 7

x 6' x 1 ' x

x 2yaxb

'x 6y 3z 3

a 3x 3b 3'

x 3y 3z 3

a 3b 3'

xyzab

3

(a 2)3(b 3)2

(ab)3'

a 6b 6

a 3b 3' a 3b 3

x[x12 % 2(x 2 & x)] ' x[x

12 % 2x 2 & 2x] ' x

32 % 2x 3 & 2x 2

(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

x 2

(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 'xy 1/3

z 1/3


Solutions-9

iii)

4. (i)

(ii)

(iii)

(iv)

(v)

(vi)

5. P = 100, i = .08, n = 5A = 1000(1 + .08)5

= 1000 � 1.46933= $1469.33

6. P = 5000, i = .12, n = 4A = 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


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:

y & y1'y2 & y1

x2 & x1

(x & x1)

y & 2 '5 & 23 & 1

(x & 1) '32

(x & 1)

y '32

x &32

% 2

y '32

x %12

y '32

(2) %12

'62

%12

'72

' 3 12


Solutions-10

[The Two Point Formula]

[adding 2 to both sides]

for x = 2,

Hence, the estimated yield corresponding to 2 gallons of fertilizer is 3 1/2 bushels.


1. i) 5(x + 3) � 2 = 235x + 15 � 2 = 235x + 13 = 235x = 10 [adding (�13) to both sides]x = 2

ii) x + 2 � 3x + 5 = x + 10�2x + 7 = x + 10Adding �7 and �x both sides,�2x + 7 � 7 � x = x + 10 � 7 � x�3x = 3x = �1

iii) 3(x + 5) + 5(2 � x) = 253x + 15 + 10 � 5x = 25�2x + 25 = 25 [subtracting 25 from both sides]�2x = 0x = 0


1. i) 81m � 25n2 2

= (9m) � (5n)2 2

= (9m + 5n)(9m � 5n) [using identity (3)]

12

14

14

14


Solutions-11

ii) a + 276

= (a ) + 32 3 3

Using identify (6) we get:(a + 3)[(a ) � a � 3 + 3 ]2 2 2 2 2

= (a + 3)(a � 3a + 9)2 4 2

iii) 8a � 125b3 2

= (2a) � (5b)3 3

= (2a � 5b)[(2a) + 2a � 5b + (5b) ] [using identity (7)]2 2

= (2a � 5b)(4a + 10ab + 25b )2 2

iv) 25abc + 5a b + 10abc2 2

Factoring out 5ab, we get:

5ab(5c + a + 2c )2

v) x y + x y3 2 2 3

Factoring out x y , we get:2 2

x y (2x + y)2 2

MATH REVIEW KIT - UBC Sauder School of Business · Math Review Kit: Section 1 1-2 Figure 3 If we...

Documents

Transcript of MATH REVIEW KIT - UBC Sauder School of Business · Math Review Kit: Section 1 1-2 Figure 3 If we...