MATH REVIEW KIT. Reproduced with permission of the Certified General Accountant Association of Canada.


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1 MATH REVIEW KIT Reproduced with permission of the Certified General Accountant Association of Canada. Copyright 00 by the Certified General Accountant Association of Canada and the UBC Real Estate Division. All rights reserved. No part of this book may be reproduced in any form without written permission from the Certified General Accountant Association of Canada and the UBC Real Estate Division. Published by the UBC Real Estate Division. Printed in Vancouver, Canada. RMATH0
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3 MATH REVIEW KIT Table of Contents Page SECTION : THE NUMBER SYSTEM ARITHMETIC OF SIGNED (NEGATIVE OR POSITIVE) NUMBERS INFINITY PROBLEMS...  SECTION : SIMPLE ALGEBRAIC OPERATIONS PROPERTIES OF ADDITION PROPERTIES OF MULTIPLICATION DISTRIBUTIVE LAW OF ADDITION AND MULTIPLICATION EQUALITY AXIOMS PROBLEMS SECTION : COMPOSITE ALGEBRAIC OPERATIONS ALGEBRAIC EXPRESSIONS GROUPING SYMBOLS CORRECT USE OF PARENTHESES PROBLEMS...  SECTION 4: ALGEBRA OF FRACTIONS THE LOWEST FORM OF FRACTIONS CONVERTING FRACTIONS TO A COMMON DENOMINATOR ALGEBRA OF FRACTIONS PERCENTAGES ROUNDING OFF A DECIMAL COMMON MISTAKES IN THE USE OF FRACTIONS PROBLEMS SECTION : ALGEBRA OF EXPONENTS MEANING OF EXPONENT ALGEBRA OF EXPONENTS COMPOUND INTEREST PROBLEMS SECTION 6: LINEAR FUNCTIONS AND INTERPOLATION LINEAR INTERPOLATION PROBLEMS SECTION 7: SYSTEMS OF LINEAR EQUATIONS AND INEQUALITIES LINEAR EQUATIONS IN ONE VARIABLE PROBLEMS SECTION 8: IDENTITIES AND FACTORIZATION THE IDENTITIES FACTORIZATION PROBLEMS SOLUTIONS... Solutions
4 SECTION THE NUMBER SYSTEM. ARITHMETIC OF SIGNED (NEGATIVE OR POSITIVE) NUMBERS i) Negative of a negative is its absolute value, i.e., () =. This is because () + = 0. To add two numbers of the same sign, we add the absolute values and assign the sign of the two numbers. Example : + 8 =, + ( 9) = 4 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 : + ( 8) =, + 7 = 8, 9 = 7, and + 9 = 8. i 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 : = 7, ( ) ( ) =, ( 0) = 00, ( 8) 6 = 48, & & =, =, =. & &. INFINITY Definition (Infinity). Infinity denoted as is a number larger than any given number. In fact there is no number that would satisfy definition 0. The idea of infinity is to provide some kind of conceptual 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 propositions depict some characteristics: Proposition : There are as many even numbers as there are natural numbers. Consider the natural and even numbers listed as follows:
5 Math Review Kit: Section If we associate to, to 4, to 6, etc., then for every natural number, we are able to produce an even number and vice versa. Proposition : There are infinitely many rational numbers between any two rational numbers. To illustrate this, take two rationals 0 and. Referring to Figure, we can generate numbers, /, /4, /8, /6,... breaking up the lengths to half and moving to the left. There is an infinite number of numbers between 0 and. Figure. PROBLEMS. Indicate whether the following numbers are rational or irrational. &8 B 6,, ,,,, , e, Solve the following: i) + ( 6) 6 + ( 9) i + ( 8) iv) 7 ( ) v) 7 (+6) vi) (8)( ) v + (0) vi ( )( 7) ix) (&6) & (&7) (&)(&) Note that: [(A)(B)] = [A B] = [AB] 
6 SECTION 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.. PROPERTIES OF ADDITION The operation of addition of two arbitrary numbers a and b is the number denoted by a + b. Example : i) If a = 9, b = then a + b = 9 + = 4. Adding to the number a yields + a. Definition (Identity of addition). The number 0 is called the identity of addition with the properties: a + 0 = a () 0 + a = a () Definition (Additive inverse). The additive inverse of a number a is a number b such that a + b = 0 () and b + a = 0 (4) The additive inverse of a is a. Example : Find the additive inverses of and . Since + ( ) = 0 and ( ) + = 0, the additive inverse of is. Similarly, the additive inverse of is. Thus, the process of subtraction can be interpreted in terms of addition. 6 = 4 could be described as the addition: 6 + ( ) = 4. Commutative Law of Addition. If a and b are any two numbers then a + b = b + a () Thus: + = + = 7 and + = + ( ) =. 
7 Math Review Kit: Section 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: + ( + 8) = + = and ( + ) + 8 = =. Example : Consider the following scheme of addition of large numbers: This conventional way of addition involves a combination of commutative and associative laws of addition which can be paraphrased as: = (0 + ) + (80 + ) + (0 + 9) = ( ) + ( + + 9) = = 47.. 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 =, b = Then ab = = 6. If a =, b = 6 then ab = ( 6) =. i times a is a. Definition (Identity of Multiplication). The number is called the identity of multiplication with the properties: a = a (7) a = a (8) 
8 Simple Algebraic Operations Definition 4 (Multiplicative Inverse). The multiplicative inverse of a number a is a number b such that: ab = (9) ba = (0) The multiplicative inverse of a is /a, provided a =/ 0 (a is not zero). Example : Find the multiplicative inverses of 8 and. As 8 /8 = and /8 8 =, /8 is the multiplicative inverse of 8. Similarly, / is the multiplicative inverse of. 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 /b. Proposition : Every real number has a unique additive inverse but only a nonzero number has a (unique) multiplicative inverse. The multiplicative inverse of 0, which is /0 =, is not defined. Commutative Law of Multiplication. If a and b are any two numbers then: ab = ba () Thus: = = 6 ( 9) = ( 9) = 4. Associative Law of Multiplication. If a, b and c are any three numbers, then: a(bc) = (ab)c () Thus: 4 (8 ) = 4 4 = 96 and (4 8) = = 96.. DISTRIBUTIVE LAW OF ADDITION AND MULTIPLICATION Distributive Law. If a, b and c are any three numbers, then: a(b + c) = ab + ac () 
9 Math Review Kit: Section Using the rule (), () can also be written as: (b + c)a = ba + ca (4) Thus: ( + ) = ( ) + ( ) = + 0 =. Example 6: Demonstrate the role of distributive law in the following scheme of multiplication: = (0 + 9) (40 + ) = {(0 + 9) } + {(0 + 9) 40} = {(0 ) + (9 )} + {(0 40) + (9 40)} = = 7.4 EQUALITY AXIOMS When a is equal to b, we write a = b. Following are the selfevident facts (axioms) that the equality relation satisfies: Reflexive: a = a () Symmetric: if a = b, then b = a (6) Transitive: if a = b and b = c, then a = c (7) If a = b, then a +_ c = b +_ c (8) If a = b, then ac = bc (9) If a = b and c = d, then a +_ c = b +_ d (0) 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 (). PROBLEMS. Find the additive inverse of the following numbers: 8, 9, 9, 0, 6,. 4
10 Simple Algebraic Operations. Find the multiplicative inverse of the numbers:,, 9, 9,.. Check whether the following statements are true or false. i) a(b + c) = ab + ac (a +b) = a + b i a(b + ) = ab + b + iv) x(y z) = xy xz v) (y + z w)x = xy + xz xw vi) v (x y)(z w) = xz xw yz + yw (a + b)( c) = a + b ca + cb 
11 SECTION COMPOSITE ALGEBRAIC OPERATIONS In this section, we discuss more complicated algebraic expressions and the use of grouping symbols.. ALGEBRAIC EXPRESSIONS Definition (Constant). A constant is a number or an algebraic symbol that stands for a particular value in a given context. For example, 7,, π, e, are all constants. Sometimes a letter c may denote a constant. Definition (Variable). A variable is an algebraic symbol, say x, that may take on different values from the real number system. Definition (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. For example, x + y + 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, x and 8x; ab and ab; xyz and 4xyz are pairs of like terms. Abbreviation of Some Algebraic Terms i) The sum a + a is written as a. Similarly, a + a + a = a, and so on. The term a also means a. Addition of like terms is obtained by taking the algebraic sum of the coefficients and multiplying by the term. Thus, x + x = 6x, 8ab ab + ab = (8 + l)ab = 7ab. i The product a a means a and a a a means a and so on. iv) Square root of x, denoted by x, is a number y such that y = x. Similarly a cube root of x, x symbolized as, is a number z such that z = x. For example and  are square roots of 4 because = 4 and () = 4. The two square roots are sometimes written as ± 4. Example : i) Find the value of (x x 6) if x =. () () 6 = = 6 = 
12 Math Review Kit: Section Find the value of (x + y) if x = and y =. ( ) = ( ) = 4 i Find the value of (x & x % ) for x =. ( ) & % = 0 = iv) When multiplying two expressions, the multiplication sign (x) may be dispensed with. The distributive law can be used to simplify the product. Example : Simplify the following: i) (x + ) = x + 6 [Using the distributive law] = x + x + ( x) = x + x [Treating + ( x) as ( x), and using the distributive law] = x + Thus, parentheses preceded by a + sign may be removed without changing the expression. i y (x + y ) = y x y + [Treating (x + y ) as (x + y ), and using the distributive law] = y x Thus, parentheses preceded by a  sign may be removed by changing the sign of each term of the expression within the parentheses.. 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 +,,, and " are performed. We now consider the rules of simplifying expressions involving composite algebraic operations. 
13 Composite Algebraic Operations Expressions with no Grouping Symbol Rule : 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 : Simplify: i) 0 = 6 = = 0 = = = + = = Expressions with more than one Grouping Symbol Rule : First solve the innermost parentheses, then brackets and then outermost braces. Example 4: Simplify: i) [ (0 9)] 6 4 = [ ()] 6 4 = [] 6 4 = 4 = 
14 Math Review Kit: Section x [4x + x(x )] = x [4x + x x] = x [x + x ] = x x x = x x i {8 + [ 9( + 6)] } = {40 + [ 9()] } = {40 + [ 8] } = {40 + [] } = {40 0 } = {40 } = {8} = 76 iv) a 4{a [b + (a ) (b )]} = a 4{a [b + a 6b + 0]} = a 4{a [a 4b + ]} = a 4{a a + b } = a 4{4a + b } = a + 6a 48b + 60 = 8a 48b
15 Composite Algebraic Operations. CORRECT USE OF PARENTHESES The following examples depict some common mistakes in using or not using parentheses. Example : i) From x subtract less than a number y. Answer: x y (incorrect) The correct answer is x (y ). Find the area of a rectangle with sides a and a units. Answer: (a) (a) (incorrect) The correct answer is a(a ). i x (y + ) = x y + (incorrect) The correct solution is x (y + ) = x y. iv) ( x)( y) = xy (incorrect) The correct solution is: ( x)( y) = xy 9 v) (x + l)(y + ) = x + y + (incorrect) The correct solution is: (x + )(y + ) = (x + )y + (x + ) = xy + y + x +.4 PROBLEMS. Find the values of the following expressions at the given variable values. i) x + 7x at x = (x + y) xy + y at x =, y = 
16 Math Review Kit: Section i (x % y) at x =, y = iv) & x % y & at x =, y = v) a bc at a =, b = 8, c = ab. Simplify the following: i) ( 6) i + 8[6 9(8 0)] iv) 0 {7 + [ (9 + )]} v) {( + 9) [ ( + 6 ) + 6( 7)]}. Simplify the following: i) x + x (4x 6x) a (a + b) + (a b) i x + y z (x y + z) (x + y) + (x z) iv) x{x y[z + (x y)]} v) {x + y [x (x + y)]} x y vi) xy + [x y + (x + y)(x y)] 6
17 SECTION 4 ALGEBRA OF FRACTIONS In this section, the algebraic operations discussed in section will be applied to fractions. Percentages and rounding procedure will be discussed as well. 4. THE LOWEST FORM OF FRACTIONS Definition (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 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 (Equality of Fractions). We define a/b = c/d if ad = bc. 4 For example = because 6 = 4. 6 Definition gives rise to the following rule of fractions. Rule : Given a fraction a/b, the numerator (a) and denominator (b) can be multiplied or divided by a nonzero number without changing its value, i.e., a/b = ka/kb, k =/ 0. Example : i) = = = = Definition (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, /, /7, /9, / are fractions in lowest form, whereas, 6/4, 0/0, ab/bd, ab /a c 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 : Reduce the fractions to their lowest form: 0 6a b i),, i, iv) 6 9ab 7a c 8ab c 4
18 Math Review Kit: Section 4 i) = 6 [Writing the numerator and denominator as a product of prime factors] = 4 [Dividing numerator and denominator by common factors ] 0 = = 6a b a a b i = = 9ab a b b a b 7a bc 7 a a b c iv) = = 8ab c 7 a b b c c a 4bc 4. CONVERTING FRACTIONS TO A COMMON DENOMINATOR Rule : 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 : Place the following fractions over a common denominator. i),, i, 6 i) = 6 is a common denominator. The fractions are then: 6 6 ', (i.e.,, 6 ' ) 6 6 = ' The fractions are, (, 8 6 ' ) 8 4
19 Algebra of Fractions i = ' 66 The fractions are,, ( ), ( 4 ' ) 4 Example 4: Place the fractions in example over the lowest common denominator. i), 6 6, 6 have 6 as a common denominator which is lowest. 4 The fractions are,. 6 6 i, have as the lowest common denominator. 6 The fractions are,. 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 : Convert the following into mixed numbers. i) 4 0 So that = 4 + or 4 4
20 Math Review Kit: Section So that = 4. ALGEBRA OF FRACTIONS Rule : i) Sum (Difference) of Fractions. a b c +_ = d ad ± bc bd To find the sum (difference) of two fractions, place them over the lowest common denominator and add (subtract) the numerator. Example : i) + 4 Solve the following: = 8 0 % 0 ' 8 % 0 ' 0 ' 0 & 7 8 ' 6 & 4 6 [Note that 6 is the lowest common denominator] ' & 4 6 ' &9 6 i a b % ac 4bd 44
21 Algebra of Fractions Though Rule (i) can be used, we may also see that bd is the lowest common denominator. a b 4d 4d % ac 4bd a 4d bd % ac bd ' 8ad % ac bd ' a(8d % c) bd Rule 4: Multiplication of Fractions a b c d ' ac bd Example 7: Solve the following: i) 9 ' ' 0 ' 76 a b b a ' a b ' a b b b a b a a ' b a (Note here the use of Section 4. [definition ] in reducing to a lowestform fraction.) Rule : Division of Fractions a b c d ' a b d c ' ad bc 4
22 Math Review Kit: Section 4 Example 8: Solve the following. i) 8 ' 8 ' 6 ' a b c a b ' a b a b c ' a b 6bc ' a c 4.4 PERCENTAGES Definition (Percentage). Percent, denoted by %, is a fraction with 00 as the denominator. Example 9: Convert the following percentages into fractions and decimals. i) 8% 8 8% = = 00 8 Also, = /% % ' 00 ' 00 ' 00 Also, 00'.0 00 '.00 Example 0: Convert the following fractions to percentages: i) 46
23 Algebra of Fractions ' 0 0 ' ' 60% 6 In this case, it is difficult to convert the denominator to 00. Therefore convert the fraction to a decimal then multiply and divide by ' ' ' 8 % _ (Note that above the last figure of decimal indicates that figure repeats to infinity; thus.8 is really.8... ). 4. 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 or more, or add nothing if the next decimal digit to the right is less than. For example: Rounding off.769 to two decimal places gives.77. Rounding off.874 to three decimal places gives.87. Rounding off to two decimal places gives 8.8. Rounding off to nearest cents Example : Round off to nearest cents. i) of $ Now 6 '.8 Multiplication by 00 will shift the decimal two places to the right. Therefore, round off at the fourth decimal place. Hence 00 =.8 00 = $
24 Math Review Kit: Section 4 of $ Now, = Multiplying by 000 will shift the decimal three places to the right. Therefore, round off at the fifth place ' ' $4.86 (Note: In Mortgage Finance calculations, payments are always rounded up, even if the decimal digit is less than.) 4.6 COMMON MISTAKES IN THE USE OF FRACTIONS x % y i) ' x % y (wrong) The correction solution is: x % y ' x % y a b c d ' a c bd (wrong) In fact, a b c d ' a b c d ' a b d c ' ad bc i 7 8 ' 7 7 ' 0 4 ' 0 (wrong) 7 8 ' 7 7 ' 4 Since 7 7 ' 48
25 Algebra of Fractions 4.7 PROBLEMS. Reduce the following to lowest form: 4 6 i) i iv) abc bcd. Place the fractions over their lowest common denominator: 7,, 7, 4 i) i 6 9, 6 a b, c iv) v) bd x yx, y xz. Insert one of <, = or > in the following: i) i & & iv) & 9 & 0 8 Hint: Note: Convert to common denominator and compare numerators. > means greater than ; < means less than. 4. Solve the following and reduce to lowest form: % 7 7 % i) i 4 8 & 6 iv) 4 % 6 & a b % c x % y v) vi) & x & y v bd z z a vi ix) bc x y b x) d 9 /6 a x xi) x xi a xiv) 8 /7 b bx xy zu x u. Convert the following percentages to fractions: i) % % i % iv) %
26 Math Review Kit: Section 4 6. Convert the following fractions to percentages: i) i iv) 8 7. Round off the following to the nearest cent: i) i of $000 over $0 8 of a million dollars iv) of $0, Simplify the following: [ 8 9 & ( & )] i) { 4 % [( & ) & ( & 6 )]} i { x % [x % y & (x & y)]} iv) ( a & b )( 4 & a 6 ) 40
27 SECTION ALGEBRA OF EXPONENTS In section, we introduced terms like x 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.. MEANING OF EXPONENT n An expression, symbolized by a, is called an exponential number or expression. The nonzero 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 : n 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] For example, = = 8. Similarly ( ) = ; ( ) = /8; ( ) = ; and 8 6 =. n Definition : If the exponent is a negative integer, then we define a = a n For example, = = 8 ( )  Similarly = ; = = = Definition : a =. Definition is a consequence of definitions and, as we shall see later in this section. /n n Definition 4: If n is a positive integer, then we define a as a number b such that b = a. In other n a /n th words, a is n root of a, and is denoted by. 4 Thus, a, which is also written as a, is called a square root of a, a is called a cube root of a and a is n a th th called a fourth root of a and so on. Note that we called as a n root and not the n root because there th may be more than one n root. For example 4 has two roots, = 4 and = 4. 
28 Math Review Kit: Section Example : Simplify the following: i) 8 8 i iv) / /4 / / / i) 8 = because = 8 /4 4 8 = because = 8 / i = = because = / / iv) = = because =. / As in parts i and iv), the same rule applies to negative fractional exponents as to negative integral exponents. (Refer to definition.) /n Care has to be taken when, in a, a is negative. Then the roots are not defined when n is even. However, / when n is odd, a root may exist. For example, () or & is not defined in the context of real numbers as there is no number b such that b =. But & =  as () = . For the sake of simplifying /n matters, we shall assume that a > 0 whenever finding a root of a is involved. Also, a would mean the th th positive n root of a. Talking of n roots, the following theorem, which generates many irrational numbers, will not be out of place. Theorem : If a is a prime number, then for every n where n =,,, 4,..., is an irrational number. n a Thus,,, 7, 4, are all irrational numbers. Example : Use your calculator, if possible, to check the following: i) ' ' i 4 ' iv) &9 ' & &.07 p/q th Definition : Let p/q be a fraction, then a is defined as the q root of a raised to the power p. q p/q a p /q p Symbolically, a = ( ) or (a ) 
29 Algebra of Exponents Example : / i) 4 = ( 4) = = 8 i 4/ = ( ) = = 8 / = = = = / ( ) iv) / (0.09) = ( 0.09) = (0.) =.07 n The exponential expression, a, 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 a n. Example 4: i) = = 9.7 [Using a calculator and rounding off at third decimal] B = = 8.87 [Using a calculator and rounding off at third decimal]. ALGEBRA OF EXPONENTS n Now that a 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 nonzero numbers. m n m+n Rule : aa = a m n mn Rule : (a) = a Rule : (ab) = ab n n n a n Rule 4: ( ) = b a n b n a m Rule : = a m n a n 
30 Math Review Kit: Section Example : Simplify the following: i) ' 4 4 ' 8 [using rule ] 8 = [using rule ] 4 = = ' 4 ' 7 7 [using rule ] ' = 0 This example gives us an opportunity to check that a = for a =. 7 7 ' 7&7 ' 0 [using rule ] 0 Hence = = i (a b ) (a ) (b) (ab) 4 ' a a bb 4 a 4 b 4 [using rule ] ' ' a b 6a 4 b 4 6 ab & [using rule ] [using rule ] 4
31 Algebra of Exponents ' a 6b 4 iv) (x yz ) (x yz ) ' (x yz ) x yz 4 ' (x ) y (z ) x yz 4 [using rule ] ' 4x 4 y z 6 x yz 4 [using rule ] = 4xyz [using rule ] v) ab c x y ' (ab c ) (x y ) [using rule 4] ' a b 4 c 6 x 6 y 4 [using rules and ]. COMPOUND INTEREST The compound interest formula is an application of exponential expressions. Example 6: A person borrows $000 at 8% interest, compounded annually. What is the amount due after years? Figure 
32 Math Review Kit: Section The interest rate is $0.08 per dollar per year. Interest during first year = 000(.08) Amount due after one year = Principal + interest = (.08) = 000( +.08) Now, this amount becomes the principal at the beginning of the second year, then, interest during second year = 000( +.08)(.08). Amount due after two years = 000( +.08) + 000( +.08)(.08) = 000( +.08)[ +.08] = 000( +.08) This amount is the principal for the third year. Then, interest during the third year = 000( +.08) (.08). Amount due after years = 000( +.08) + 000( +.08) (.08) = 000( +.08) [ +.08] = 000( +.08) = 000(.08) = (using a calculator) = $9.7 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( +i) n This is the compound interest formula. Example 7: Solve example 6 using the compound interest formula. Here P = 000, i =.08, n = Therefore, A = 000( +.08) = = $9.7.4 PROBLEMS. Compute the following (do not give decimal answers): i) 4 7 i iv) ( v) (0.06) /4 49 )&/ / 4/ 6
33 Algebra of Exponents. Evaluate the following: 7 9 i) ( ) ( ) i / &/ / / (&8) / (&7) 4/ & iv) v) 4 / (&) / &/ &/. Approximate the irrational exponent by a threedigit decimal and compute using a calculator: i) B i e 4. Simplify: (x ) (x ) x yz i) ( ) i (x) (x ) axb (a ) (b ) (ab) iv) x[x + (x x)] v) (x y) vi) (x y) (xyz) / / / /. If $,000 is invested for years at 8% compounded yearly, what will this investment be worth at the end of this time? 6. On January, 97, a man incurs a debt of $,000 at % compounded annually. If the lender demands payment on January, 979, 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 0 years when he becomes 6? 7
34 SECTION 6 LINEAR FUNCTIONS AND INTERPOLATION 6. 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 x between x and x such that the point (x,y ) lies on the straight line. This process is called linear interpolation., If x is outside x and x, then it is called intrapolation. In both the cases, the two point formula as indicated in the following example is used. Example : An investment of $000 for years at 6% interest compounded annually yields $8. and at 8% compounded annually yields $469.. Use linear interpolation to find the amount at 7% compounded annually. Consider the points (6, 8.) and (8, 469.). Substituting these points for (x,y ) and (x,y ) into the two point formula and letting x = 7, we get: y & y y y = (x & x x & x ) [The two point formula] 469. & 8. y 8. = (7 6) 8 & 6.0 = () = 6. Hence y = = Therefore, at 7%, the interpolated amount is $ PROBLEMS. On a piece of land, the application of gallon of fertilizer yields bushels of a certain crop and the application of gallons of fertilizer yields bushels of the crop. Estimate the yield if gallons of fertilizer were used, using linear (twopoint) interpolation. 6
35 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. LINEAR EQUATIONS IN ONE VARIABLE A linear equation in one variable, x, is of the form: ax + b = 0, a =/ 0. (A) Definition (Solution). A value of x is said to be the solution of equation (A) if it satisfies the equation. Example : i) Solve x = 0 and check to see that the solution satisfies the equation. Using the laws of equality, we have: x + = (adding on both sides) x = This is also known as transposing on the right hand side of the equation by changing its sign. Simplifying further, we get: x = (dividing both sides by ) Substituting x = in the original equation, we get: ( ) = 0 = 0 0 = 0 Hence, x = is the solution of the given equation. 7
36 Math Review Kit: Section 7 Solve x + = x + and check the solution. Adding (x ) on both sides we get: x + x = x + x x = 0 x = (dividing the equation by ) Substituting x = in the original equation, we get: () + = + + = 0 0 = 0 Hence, x = is the solution of the given equation. 7. PROBLEMS. Solve for x: i) (x +) = x + x + = x + 0 i (x + ) +( x) = 7
37 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. THE IDENTITIES Definition (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, x = x + x; x + x = x(x + ). 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) = x + xy + y () (x y) = x xy + y () (x + y)(x y) = x y () (x + y) = x + x y + xy + y (4) (x y) = x x y + xy y () (x + y)(x xy + y ) = x + y (6) (x  y)(x + xy + y ) = x y (7) As an illustration, we check identity () L.H.S. = (x + y) = (x + y)(x + y) = x(x + y) + y(x + y) [using the distributive law] = x + xy + yx + y = x + xy + y = R.H.S. Example : Simplify the following: i) (x + ), (a ), i (m + )(m ), iv) (x + ) 8
38 Math Review Kit: Section 8 i) Using identity (), we have: (x + ) = (x) + [(x) ()] + = 4x + x + 9 Using identity (), we get: (a ) = (a) [(a) ()] + = 9a 0a + i In this case, identity () applies. Thus: (m + )(m ) = m = m iv) Using identity (4), we have: (x + ) = x + (x ) + (9x) + = x + 9x + 7x FACTORIZATION Definition (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 and the algebraic expression itself are not used in factorization. For example: x + 0 = (x + ) and x y = (x + y)(x y) Note that these two examples are also identities. The identities and the distributive law are extensively used in factorization. Example : Factorize the following expressions: i) xy + xz + x i a b + ab + ab x y + 6x y + 7x y + x y 8
39 Identities and Factorization iv) 7m 8n i) We note that x is a common factor in each term. Invoking the distributive law, we factor out x: xy + xz + x = x(y + z + ) Factoring out ab and using the laws of exponents: a b + ab + ab = ab(a + + b) i Factoring out x y and using the laws of exponents: x y + 6x y + 7x y + x y = x y(x + y + 9y + ) iv) 7m 8n = (m) (n) = (m n)[(m) + (m n) + (n) ] [using the identity (7)] = (m n)(9m + 6mn + 4n ) 8. PROBLEMS. Factorize the following: i) 8m n 6 a + 7 i iv) v) 8a b abc + a b + 0abc x y % 4 x y 8
40 Solutions
41
42 SOLUTIONS Solutions to the Problems in Section 8. The rational numbers are: 6, ,,, , i) + ( 6) = 7 i iv) 6 + ( 9) = + ( 8) = 7 ( ) = v) 7 (+6) = vi) v (8)( ) = 04 + (0) = vi ( )( 7) = ix) (&6) & (&7) (&)(&) ' & & (&) ' 9 ' Solutions to the Problems in Section. The additive inverse (a.i.) of 8 is 8 because 8 + ( 8) = 0 and = 0. Similarly the a.i. of 9 is 9. The a.i. of 0 is 0. The a.i. of 6 is 6. The a.i. of is. The a.i. of 9 is 9.. The multiplicative inverse (m.i.) of is because = and =. Similarly the m.i. of is, the m.i. of 9 is, the m.i. of 9 is and the m.i. of is. 9 9 Solutions
43 Math Review Kit: Solutions. i) True True i False because a(b + ) = ab + a iv) True v) True vi) True v False because (a + b)( c) = a( c) + b( c) = a ac + b bc Solutions to the Problems in Section. i) ( ) + 7( ) = ( )( ) + ( 4) = 4 4 = 8 6 = 8 ( + ) () + = = = = i iv) [ % (&)] ' (9&) ' 8 ' 64 ' & % & '& % 9 & '&8 v) 8 (&) 8 ' 4 8(&) 6 ' 4 &8 6 ' &96 4 '&4. i) = 9 + = = = ( 6) = 4 ( ) = + = 4 i + 8[6 9(8 0)] = + 8[6 9( )] = + 8[6 ( 8)] = + 8[6 + 8] = + 8[4] = + 9 = 94 Solutions
44 Math Review Kit: Solutions iv) 0 {7 + [ (9 + )]} = 0 {7 + [ 6]} = 0 {7 8} = 0 { } = 0 + = v) {( + 9 ( [ ( + 6 ) + 6( 7)]} = {( + 9) [ (9 ) + 6( 4)]} = {8 [ ( ) 4]} = {8 [ 9]} = {8 + 8} = {46} = 9. i) x + x (4x 6x) = 4x 4x + 6x = 6x a (a + b) + (a b) = a a b + a b = b i x + y z (x y + z) (x + y) + (x z) = x + y z x + y z x y + x z = x x x + x + y + y y z z z = x + y z iv) x{x y[z + (x y)]} = x{x y[z + x y]} = x{x yz yx + y } = x xyz x y + 6xy v) {x + y [x (x + y]} x y = {x + y [x x y]} x y = {x + y [x y]} x y = {x + y x + y} x y = 4y x y = y x vi) xy + [x y + (x + y)(x y)] = xy + [x y + x(x y) + y(x y)] = xy + [x y + x xy + xy y ] = xy + x y + x xy + xy y = x y + xy xy + xy + x y = x y xy + x y Solutions
45 Math Review Kit: Solutions Solutions to the Problems in Section 4. i) i 4 ' ' 49 ' ' ' ' 9 6 iv) abc bcd ' a b c c b c d ' ac d. i) Lowest common demoninator (l.c.d.) is 7 = 9, hence the fractions are: 7 and 7 7 or 9 and 4 9 The l.c.d. is 7. The fractions are:, 7, 4 0 or 7, 7, 7 i The l.c.d. is 9. The fractions are: 6 9, 6 or 80 9, 6 9 iv) The l.c.d. is bd. The fractions are: a b b d, x x yz x, c bd or ad bd, c bd v) The l.c.d. is xyz. The fractions are: y y xz y or x xyz, y xyz. i) < 8 as ' > 7 as 0 ' 70 and 7 ' 0 70 Solutions4
46 Math Review Kit: Solutions i iv) & > & 7 as & '&4 and & 7 '& & 9 '&0 8 as & 9 '& 9 '& i) % 7 ' 7 % ' 7 % 7 % 4 ' 4 4 % 4 ' 7 4 ' ' i 8 & 6 ' 88 & ' & ' 7 88 iv) v) 4 % 6 & ' 0 68 % 9 68 & 8 68 ' 0 % 9 & 8 68 a b % c bd ' ad bd % c bd ' ad % c bd ' 68 ' 8 vi) x % y z & x & y z ' x % y & (x & y) z ' x % y & x % y z ' y z v vi ix) x) xi) x 7 8 ' 4 40 ' ' 0 ' a bc x y b d ' axb bcyd ' ax cyd 9 % ' 9 ' 6 9 ' /6 8 ' 6 8' 6 8 ' 48 /7 ' 7 ' 7 ' ' xi a x b a bx ' a x b bx a ' a x b ab ' ax ' ax Solutions
47 Math Review Kit: Solutions xiv) xy zu x u ' xy zu u x ' xyu zux ' yu zx. i) i iv) % ' 00 ' 0 4 % ' /4 00 ' 4 00 ' % ' /4 00 ' 4 00 ' 400 % ' 00 ' ' 4 ' 4 6. i) i iv) = 00% = 4% = 00% = 40% = 00% = 66.6% = 66 % 8 8 = 00% =.% = % 7. i) i iv) of $000 =. 000 = $.00 8 of $0 = = $8.4 of $,000,000 =.,000,000 = $,. 9 of $0,000 = ,000 = $, i) [ 8 9 & ( & )] = [ 8 9 & 6 ]as & ' 6 & 6 ' 6 Solutions6
48 Math Review Kit: Solutions = = = [ 6 8 & 8 ] [ 8 ] ' { 4 % [( & ) & ( & 6 )]} = { 4 % [ & 6 ]} as ' and ' 6 = { 4 & 7 6 }as ' 4 6 { & 4 } ' = (& ) = & 4 i = = = = = = { x % [x % y & (x & y)]} { x % [x % y & x % y]} { x % [ x % 4 y]} { x % x % y} { 6 x % y} x % 4 y iv) ( a & b )( 4 & a 6 ) = = = a ( 4 & a 6 ) & b ( 4 & a 6 ) a 8 & a & b % ab 8 8 a & a & b % 8 ab Solutions to the Problems in Section. i) 4 ' (4) ' ( ) ' ' ' 9 Solutions7
49 Math Review Kit: Solutions 6 7 ' ( ) ' ' ' 9 i & 4 ' ' 4/ ( ) ' 4/ ' 4 6 iv) 49 &/ ' 49 / ' 49/ ' (7 ) / / ( ) 76/ 7 ' ' / 6/ ' 4 v) (0.06) ' [(.) 4 ] '. '.. i) ' 4 ' 6 ' & ' 9 4 ' ( ) ( ) ' 4 4 ' & ' i iv) / &/ / / ' ' & ' (&8) / (&7) 4/ ' (& ) / (& ) 4/ ' &6/ & / 4 / (&) / ( ) / (& ) / 6/ & 6/ ' & & 4 ' 4 8 & 8 ' 4 00 ' 8 0 ' 0 v) & &/ &/ ' / / / ' / / / ' ( % 8 % % ) ' / ' ' * Alternatively: ( ) / (& ) 4/ ( ) / & (& ) ' (&) (&) 4 / ( )(& ) i) = = = 4.78 (rounded value) B ' '.4 ' Solutions8
50 Math Review Kit: Solutions i e ' '.78 ' (i) (x )(x ) (x)(x ) ' x 7 x 6 ' x ' x ( x y axb ' x 6 y z a x b ' x y z ' xyz a b ab (i (iv) (v) (a ) (b ) ' a 6 b 6 ' a b (ab) a b x[x % (x & x)] ' x[x % x & x] ' x % x & x (x &/ y) ' [x &/ y / ] ' [x &/ ] [y / ] ' x & y / ' y / x (vi) (x y) / (xyz) &/ ' [(x ) / y / ][x &/ y &/ z &/ ] ' [x 4/ y / ][x &/ y &/ z &/ ] ' xy / z &/ ' xy / z /. P = 00, i =.08, n = A = 000( +.08) = = $ P = 000, i =., n = 4 A = 000( +.) 4 = = $ P = 8000, i =.09, n = 0 = 8000( +.09) 0 = = $8,98.9 Solutions to the Problems in Section 6. Let x denote the amount of fertilizer and y, the crop yield. Then the two points are (,) and (,). The question of the line passing through these points is: Solutions9
51 Math Review Kit: Solutions y & y ' y & y x & x (x & x ) [The Two Point Formula] y & ' & & (x & ) ' (x & ) y ' x & % [adding to both sides] y ' x % for x =, y ' () % ' 6 % ' 7 ' Hence, the estimated yield corresponding to gallons of fertilizer is / bushels. Solutions to the Problems in Section 7. i) (x + ) = x + = x + = x = 0 [adding ( ) to both sides] x = x + x + = x + 0 x + 7 = x + 0 Adding 7 and x both sides, x x = x x x = x = i (x + ) + ( x) = x x = x + = [subtracting from both sides] x = 0 x = 0 Solutions to the Problems in Section 8. i) 8m n = (9m) (n) = (9m + n)(9m n) [using identity ()] Solutions0
52 Math Review Kit: Solutions 6 a + 7 = (a ) + Using identify (6) we get: (a + )[(a ) a + ] 4 = (a + )(a a + 9) i iv) 8a b = (a) (b) = (a b)[(a) + a b + (b) ] [using identity (7)] = (a b)(4a + 0ab + b ) abc + a b + 0abc Factoring out ab, we get: ab(c + a + c ) v) xy + xy 4 Factoring out 4 x y, we get: 4 x y (x + y) Solutions
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