Math Review qs

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A COMPREHENSIVE OVERVIEW OF BASIC MATHEMATICAL CONCEPTS 
- SET THEORY 
NOTATION 
{ } Brul"e.• indicate the beginning and end ofa set notation; when 
listed e lements or members mllst be separated by commas; 
EXAMPLE: In A= {4 , 8, 16~. the 4, 8, and 16 arc called 
e lement s or members 01' the set; set s are finite (ending, or 
having a last element), unless olherwi se indicated. 
In the middle of a set indi cates c(Jllfilluuti(J1I (JFa patte", ; 
EXAMPLE: 8= {S, 10, 15, ... ,85, 90}. 
At the end of a set indi cates an illfinite set. that is , a set 
with no last element; 
EXAMPLE: C={3, 6, '1, 12, ... }. 
Is a sy mbol which literally means '"such that." 
E 	 Means j ,\' U member (~ror ;~. an elemellt oj; 
EXAMPLE: IfA= {4, 8,12 1, then 12 E A because 12 is 
in set A. 
E 	 Means is 1101 u ",ember ofor i.\' not all element of; 
EXAMPLE: If 8 = {2, 4, 6, X} , then 3 Of 8 because 3 is not 
in set 8. 
o 	Empty set or "ull .'tet: A sct containing no elements or 
members, but which is a subset of all sets ; also written 
as : }. 
e Means is a slibset of; also may be written as ~ . 
(l Means is 11M a.."b.•et ot; also may be w ritten as g:; . 
AeB Indicates that every element of set A is also all elelllellf 
(Jf scI8; 
EX1MPLE: If A= 1). 6 } and 8 = : I, 3. S, 6, 7, 9]. then Ae8 
because the 3 and 6 which are in set A are also in set 8 . 
2" Is 	the IIl1l1/ber of slibsef.,· ".1" a set when II equal s the 
number 01' e lements in that set; 
E."<AMPLE: If A- {4, 5. 6:, then set A has 8 subsets 
because A has 3 clements and 23=8. 
OPERATIONS 
AuB Indicates the /llIi(J1I of 
u UNIONset A with sct 8; eve ry 
clement of this set is EITH ER 
an element of set A, OR an 
' Iement 01' set B; that is, to lorm 
the union of two sets, put all of 
the clements of the two sets 
together into one set, making 
sure not to write any clemcnt more than oncc; 
EXAMPLE: II' A= {2, 4. 6, 8, 10, 12 1 and B= {3, 6 , 9, 12, 
15,18:' then: Au8= {2, 3, 4 , 6, 8, 9,10,12, 15, 18 }. 
Ans~t I~d~~ii::~~e~h~:i~~':.~~i:'~~e~:· n INTERSECTION 
01' this set is al so an clement of 
SCI A AND set 8 ; that is, to form 
the intersection of two scts , list 
only those elements which arc 
found in 80TH of the two sets; 
EXAMPLE: If A= : 2, 4, 6, 8, 
_ 10, 12: and 8 = {3. 6, 'J , 12, 15, 18} , Ihen i\nB= 16, 12}. 
A Indicates the ('oll/plell/em of set 
A; that is, all clements in the COMPLEMENT 
universal set which arc NOT in SET 
set A; 
EXAMPLE: If the universal set 
is the set of Integers and 
 0) . 
A~ { O, 1, 2, 3, ... f, then 
A=:-I , -2 , -3 , -4 , ... f. 
PROPERTIES 
A=B If all of the elements in set i\ are also in set 8 and all 
elements in set 13 are also in set A , although they do not 
have to be in the samc order. 
EXAMPLE: If A= {5, 10 } and 8 ={ 10, 5 j , then i\= B. 
II(A) 	Indicates the IIl1mber ()f e1emellf.. it, .,et A, i.e. , the 
cardinal number of the set. 
,..... 	 EXA,~PLE: Ir A={ 2, 4. 6 }. then n(A)=3. 
A-B Means is eqllivalellt to; that is, set A and set 8 
have thc same number of elements, although the 
clcments thcmselves may or may not be the same. 
EXAMPLE: If A {2.. 4 , 6) and B= : 6, 12, 18 j , then A-8 
because II(A)= 3 and 11(8)= 3. 
AnB=0 Indicates Ji.•j()illf sets which have no elements in 
common. 
EXAMPLE: IfA = : 3, 4 , 5 l and 13; j7, 8, 9), then An 8=0 
because there are no common clements. 
PROPERTIES OF REAL 
NUMBERS 
CLOSURE 
oa + h is a real 	number; when you add 2 real numbers, the 
result is al so a real number. 
EXAMPLE: 3 and 5 are both real numbers; 3 + S = 8 and the 
slim. 8, is also a real number. 
oa -	 b is a rca I number; when you subtrac t 2 real numbers, 
the result is also a real number. 
EXAMPLE: 4 and II are both real numbers; 4- II =-7 and 
the diHerenee, - 7, is also a real number. 
o(a)(b) is a real number; when you multiply 2 real numbers, 
the result is also a real number. 
EXAMPLE: 10 and - 3 are both real numbers; 
(10)(- 3) = -30 and the product , ·-30, is also a real number. 
oalb is a real 	number when b>,O; when you divide 2 real 
numbers, the result is al so a real number unless the denom­
inator (di visor) is zero. 
EXAMPLE: -20 and 5 arc both real numoers ; -20/5 = -4 
and the quotient , -4, is also a real number. 
COMMUTATIVE 
oa + b = b + a ; 	you can add numbers in e ither order and get 
the same answer. 
EXAMPLE: 9 + 15 = 24 and 15 + 9 ; 24. so 9 + 15 = 15 + 9. 
o(a)(b)=(b)(a); you can multiply numbers in either order 
and get the same answer. 
EXAMPLE: (4)(26) = 104 and (26)(4) ~· 104, so (4)(26) = 
(26)(4). 
oa-b .. b-a; you cannot subtract in any order and get the 
same answer. 
EXAMPLE: 8 2 = 6 , but 2 - 8 = -·6. There is no commuta ­
tive property for subtraction. 
oa I b .. b I a; you cannot divide in any order and get the 
same answer. 
EXAMPLE: 812 = 4, but 2 / 8 = .25 , so there is no cOlllmu­
tative property for division. 
ASSOCIATIVE 
o(a+b)+c=a+(b+c); you can group numbers in any 
arrangement wh':l1 adding and get the same ans\ver. 
EXAMPLE: (2 + 5) + 9 = 7 + 9 = 16and2 + (5 + 9) = 2 + 14 = 16. 
so (2 -1 5) 1'1 = 2 + (5 + 9). 
o(ab) e = 8 (be) : you can group numbers in any arrangement 
when multiplying and ge t the same an swer. 
EXAMPLE: (4x 5) 8 = (20) 8 = 160 and 4(Sx8) = 4(40) - 11i0, 
so (4xS)8 = 4(5x8 ). 
oThe assuciative property dues 1101 lI 'ork Cor subtraction or 
division. 
EXAMPLES: (10 - 4) - 2 = 6 - 2 = 4, but 10 - (4 - 2) = 10 - 2 - 8; 
fix division, (12 /6) / 2 ~ (2) / 2 = I, but 12 / (6/2) = 12/3 : 4. 
Notice that these answers are not the samc. 
IDENTITIES 
oa + 0 = a ; 7ero is the identity for addition because adding 
Zero does not change the original number. 
EXAMPLE: 9+ 0 = 9 and (H 9 = 9. 
oa(I)=8; one is the identity lor multiplication because multi ­
plying by one does not change the original number. 
EXAMPLE: 23 (I ) = 23 and (I )23 = 23. 
oldentities for suotraction and division become a problem. It 
is true that 45 - 0 = 45 , but 0 - 45 = 45 , not 45. This is also 
the case for divi sion because 4/1 = 4, but 1/4 = .25, so the 
identities do not hold when the numbers are reversed. 
INVERSES 
°a +(-a)=O; a number plus its additive inverse (the number 
with the opposite sign) will always equal zero. 
EXAMPLE: 5 + ( S) = O and (- 5) r 5 = 0. The exception is 
zero because 0 + 0 = 0 already. 
oa(l/a)= I; a number times its multiplicati ve inverse or 
rec iprocal (the number written as a fraction and flipped) 
will always equal one. 
EXAMPLE: S(1 /5) = I. The exception is zero because zero 
cannot be multiplied by any number and result in a product 
of onc. 
DISTRIBUTIVE PROPERTY 
oa(b+c)=ab + ac or a(b - c) = ah - ac; each term in th e 
parentheses mu st be multiplied by Ihe term in Cronl 01' the 
parentheses. 
EXAMPLE: 4( S+ 7) 4(5 )+ 4(7) = 20 + 28 = 48. 
This is a simple example and the distributive property is 
not required 10 Cind the answer. Whe n the problem invo lves 
a variable. however, the distributi ve pro pert y is a necessity. 
E."<AMPLE: 4(5a + 7) 4( 5al + 4(7) = 10a + 28. 
PROPERTIES OF EQUALITY 
oReflexive: a = a ; both s ides of the equation are identical. 
EXAMPLE: 5 + k = 5 +k. 
oSymmetric: If a = h, then b = a. Thi s pro pert y a110\\ s ) OU to 
exchange the two sides of an equation. 
EX4MPLE: 4a - 7 = 9 - 7a + 15 becomes 9 7a + 15 = 4a 7. 
oTransitive: If a=b and b=e, th~n a = e. T hi:; pro pert y 
a llows yo u to connect statemcnts whieh are each equal to 
the same comlllon statement. 
EXAMPLE: Sa 6 = 9k and 9k = a 12; you can eliminate 
the COllllllon term 9k and COllnect the 1()lIowing int o o lle 
equation: Sa (, = a + 2 . 
oAddition Property of Equality: If a=b, then a +c=h+c. 
This property all ows you to add any number or algebraic term 
to anyequation as long as you add it to hoth s ides to keep the 
equation truc . 
EXAMPLE: 5 = 5; if vou acid 3 to one sidc and nut the other. 
the equation become; 8 = 5. which is "li se. bill iI' you add 3 
to botlt sides, yo u get a true equaliun X X. Also. 5a + 4 = 14 
becomes 5a + 4 + ( 4) = 14 + ( 4) ifYllU add 4 to bOlh s ides. 
Thi s results in the cquation Sa 10. 
oMultiplication Pl'Operty of Equality: If a = b, then ac = be 
when c .. O. This property allows you to multiply both sides 
of an equation by any nonzero value. 
EXAMPLE: If 4a = 24 , th en (4a)( .2S )= ( 24)(. 25) and 
a = Ii. Notice that both ,ides of the = were multiplied by .25 . 
SETS OF NUMBERS 
------, 
DEFINITIONS 
• Natural or Counting numhers: : 1. 2. ], 4 , 5, .... I I. 12. ... : 
• \V holc nu mbers: {O, 1,2. 3.... , In, 11. 12. 13. ... : 
oIntegers: {... , -4, -3, -2, -I , n, I. 2. 3. 4, ... } 
oRationa l num bers: : p/ql p and q arc integers. q>'O): the 
sets of Natu ral num bers, Who le numbers, and In tegers. as 
well as numbers which can be w ritten as proper or improper 
fractions, arc all subsets 01" thc set of Rationa l nu m bc rs. 
olrrational 	numbers: { x 1 x is a real lI umber but is not a 
Ration al num be r}: thc sets o f R~l tiona l num bers a nd 
Irrational numbers have no c lement s in common and are. 
there fo re, di sjoint sets . 
oR eal numbers: : x 1 x is Ihe coordinate o Ca po int on a number 
line }; the union of the set llf Rational numhers with the SCI 
of Irrationa l numbers equals the SCI o f Real num bers. 
-Imaginary numbers: 1ai l a is a Rca l Ilum ber Hnd i is the 
number wh ose square is - 11; j 2 I: the sets of' Rea l 
numbers and Imag inary Ilu mbers have no cl ement s in 
COllllllon and arc . therefore. disjoint sets. 
oComplex number s : :a +bil a and b arC Real numbers and 
i is the number whose square is - I:; the SCI of Rea l 
numbers and the se t of Imagina r :v numbers arc bot h 
subsets of the se t of Com plex numbers. 
EXA MPLES: 4 + 7i and J-2i arc Compl e, numbers. 
COMPLEX NUMBERS 
Numbers 
Rational 
Integers 
Whole 
~ ~ 
~/ 
...----.. 
OPERATIONS OF REAL 
NUMBERS 
V OCABU LARY 
• Total or sum is the answer to an addition problem . The 
numbers added are called addends. 
EXAMPLE: In S + 9 ~ 14, Sand 9 arc addends and 14 is the 
total o r s lim . 
• Differencl' is the answer to a subtraction problem. The number 
subtracted is called the subtrabend. The number t,'om which 
the subtrahend is subtracted is called the minuend. 
EXAMPLE: In 2S -8~ 17, 25 is the m inuend, 8 is the 
subtrahend, and 17 is the diffe rence. 
• Product is the answer to a multiplication problem. T he 
numbers multip lied are each called a facto r. 
EXAMPLE: In 15 x 6 ~ 90, 15 and 6 arc factors and 90 is 
the produc t. 
• Quotient is the answer to a di vision problem. The number 
being divided is call ed the d ividend. The number tha t you 
arc dividing by is called the divisor. If there is a number 
remaining aller the divi s ion process has been completed, 
tha t number is called the remainder. 
EXAMPLE: In 45 : 5~ 9, which may also be written as 5)45 
or 4S15, 45 is the di vidend, 5 is the d ivisor and 9 is the quotient. 
• An 	exponent ind icates t he number of t imes the base is 
multiplied by itsel f; that is, used as a facto r. 
EXAMPLE: In S3, S is the base and 3 is the exponent, or 
power. and 5) (5 ) (S )(5)~ 125; notice tha t the base, 5, was 
multi plicd by itself' 3 ti mes. 
• Prime numbers arc natura l numbers g reate r than I having 
exactly two fact ors, it se lf and one . 
EXAMPLES: 7 is prime because the only two natura l 
numbers th at mu ltiply to equa l 7 are 7 and I ; 13 is pr ime 
because the only two natural numbers that multiply to equal 
I 3 are I 3 and I. 
• Composite numbers are nat ura l numbers that have more 
than two fac tors . 
EXAMPLES: 15 is a composite number because I, 3, 5. 
and 15 a ll multiply in some combination to eq ual 15; 9 is 
composite because I , 3. and 9 a ll mUltiply in some combi­
nation to equa l 9. 
• The greatest common factor (GC F) or greatest common 
divisor (GCD ) of a set o f numbers is the largest natural 
number that is a tactor of each of the numbers in the set; that 
is. the largest natural number that will divide into all of the 
numbers in the set w ithout leav ing a remainder. 
EXAMPLE: The g rea test com mon fac tor (GCF ) of 12, 30 
and 42 is 6 because 6 divides even ly into 12, 30, and 42 
w ithout leaving remai nders. 
• The least comm o n multiple ( LC M ) of a set or numbers is 
the sma llest natu ra l number that can be d iv ided (w ithout 
remainders ) by each of the numbers in the set. 
EXAMPLE:T he least com mon multip le o f2, 3, and 4 is 12 
because a lthough 2, 3, and 4 div ide evenly into Illany 
numbers, including 48,36, 24. and 12, the smallest is 12. 
• The denominato r o f a fract ion is the numbcr in the boltom ; 
that is, the divisor o f the in d icated di v is ion or the li·act io n. 
EX'JMPLE: In 5/8, 8 is the denominator and al so the div i­
sor in the indicated divisi on. 
• The numerator of a frac tion is the number in the top; that 
is, the div idend of indicated division of the fract ion. 
EXAMPLE: In 3/4 . 3 is the numerator and a lso the di vidc nd 
in the indicated d ivision. 
FUNDAMENTAL THEOREM 
OF ARITHMETIC 
The Fundamental Theorem of Arithmetic states that every 
compos ite number can be expressed as a un ique p roduct of 
prime numbers. 
EXAMPLES: 15 ~ (3)(5) , where 15 is composite and both 
3 and 5 are prime; 72 ~ (2 )(2 )(2)(3 )(3). where 72 is compos­
ite and both 2 and 3 arc prime; notice that 72 al so equal s 
(8)(9 ). but thi s does not demonstrate the theorem because 
nei ther 8 nor 9 are prime num bers. 
ORDER OF OPERATIONS 
• Description: T he order in whi ch addition, subtraction, multi ­
plicat ion, and divi sion arc performed determines the answer. 
• Order 
I. 	Parentheses : Any operat ions contained in parentheses 
are done first. if' there are any. Th is also appl ies to these 
cnclosure symbols, : : and [ ]. 
2. E xpo nents: Ex po nent expressions are simpli fie d 
second, if there arc any. 
3. 	 M ultiplication and Division : T hese operations are done 
next in the order in which they are fou nd, go ing le ft to 
right; that is, if divis ion comes ri rst, going lell to right, 
then it is done first. 
4. 	Addi tion and Subtracti on : These operations are done 
next in the orde r in which they arc fo und. go ing left to 
r ight; that is, if subtraction comes fir ·t, go ing left to 
right, then it is do ne firs t. 
DECIMAL NUMBERS 
• The place value of each digit in a base 10 number is dete r­
mined by its positi on with respect to the decima l po int. 
Each position represents multiplication by a power or 10. 
EXAMPLE: In 324, 3 means 300 because it is 3 ti mes 102 
(I 02 ~ 100 ). 2 means 20 because it is 2 times 101(I 0 1~ 10). 
and 4 means 4 times one because it is 4 times I 011 ( IO()~ I). 
The re is an invisible dec imal point to the r ight of the 4. In 
5. 82 . 5 means 5 ti mes one because it is 5 ti mes 
10° (1 OO~ I), X means 8 times one tenth because it is 8 
ti mes 10-1(10-1~.1 ~ 1110). and 2 means 2 tim es one 
hundredth because it is 2 t imes 10-2 (10-2~ .OI ~ 111 00 ). 
WRITING DECIMAL NUMBERS AS 
FRACTIONS 
• W rite the digits that are beh ind the dec imal point as the 
numerato r (top ) o f the li·action . 
• Write the place val ue of the last digit as the denominator 
(bottom ) o f the fracti on . Any dig its in front of the decima l 
point are w hole numbers . 
E)(AMPLE: In 4.068 , the las t digi t behind the decima l 
poin t is 8 and it is in the 1,000ths place ; the refore. 4. 068 
becomes 4~ ' 
t,OOO 
• Notice the number ofze ros in the denominator is equal to the 
number o f digits behind thc decimal point in the o riginalnumber. 
ADDITION 
• Write the decima l n umbers in a verti cal lorm w ith the deci­
mal points lined up one under the other, so digits o f eq ual 
p lace va lue are under each othe r. 
' A dd 
EXAMPLE: 23 .045 + 7.5 + 143 + .034 would become 
23 .045 
7.5 
143,0 
because there is an invisib le deci ma l po int 
beh ind the 143. 
+ .034 
173.579 
SUBTRACTION 
• Write the decimal numbers in a ver tica l [arm wit h the deci­
mal points lined up one under the other. 
• Write add itional zeros aller the last di git beh ind the dec imal 
point in thc minuend (number o n top ) if needed (both the 
minuend and the subtrahend shou ld havc an equal number 
of digits behind the decimal point). 
• EXAMPLE: In 340.06 - 27.3057, 340.06 on ly has 2 digits 
beh ind the decimal po int , so it needs 2 more zeros because 
27 .3057 has 4 digi ts behind the decima l poin t; there fo re, the 
problem becomes: 
340.0600 
-27.3057 
MULTIPLICATION 
• Multiply 
• Coun t the number of digits behind the decimal points in all 
factors . 
' Count the number o f d igits behi nd the decim al po int in the 
answer. T he answer must have the same number of digits 
behind th e decima l po int, as there are digits behind the deci ­
ma l poi nts in all thc factors . It is not necessary to line the 
deci ma l poin ts up in multipl ication. 
EXAMPLE: In (3.05) (.007), mult iply the numbers and count 
the 5 digits behind the decima l points in the problem so you 
can p ut 5 digits behind the decimal poi nt in the product 
(answer); therefore, ( 3 . 05 )(.007) ~ . 02 1 35. This process 
works because .3 ti mes .2 can be written as fract ions, )/10 
ti mes 2/ 10, which equal s 6/ tOO, which equals .06 as a dec imal 
number - two dig its behi nd the dec imal po ints in the prob­
lem and two digits behind the dec imal point in the an swer. 
DIVISION 
• Rule : Always d ivide by a who le number. 
'I fthe d ivisor is a whole number. simply di vide and bri ng the 
decimal point up into the quotien t (answer ). 
2 
70. EXAMPLE: .044)!16 OJ) 3.8 1 
'1 f the divisor is a dec imal number, move the dec ima l po in t 
beh ind the last digit and move the deci ma l point in the 
dividend the same number of p laces. Div ide and bring the 
decima l point up into the quotient (a nswer). 
• Thi s 	process works because both the divisor and the 
di vidend arc actually multipli ed by a po\\ cr of 10; that 
is, 10. 100, 1,000, or 10,000 to move the dec imal point. 
EXAMPLE: .1.5 x~ = 350 ~ 7() 
.05 tOO 5 
ABSOLUTE VALUE 
• Definition: Ixl- x ifx > II or x - () and Ixl ~ x if x < 0; that is. 
the absolute value of a number is always the pOSiti ve \alue 
of that number. 
EXAMPLES: 161 ~ 6 and I 61- 6; the '''lSWer is pos itive 6 in 
both case~. 
ADDITION 
·If the signs of the numbers arc the sUllie. ADD. The 
answer has the same sign as the Ilumbers. 
EXAMPLES: (- 4)+( 9) - 13and 5 + 11 - lt). 
·If the signs of the numbers are diUel'(!III. SUBTRACT. 
The HIlS\Vcr has the sign of the largl'r number (ignoring tlH.' 
signs or taking the absolute va lue of the Ilumhers to dcter­
mine the larger Ilumher). 
EXAMPLES: ( -4 ) + (9'1 ~ 5 and (4) + ( 9) = 5. 
SUBTRACTION 
• Change subtraction to addition of tbe opposite number: 
a - b ~ a I ( b); that is , change the subtraction sign to addi­
tion and also change the sign~()f the number dircc11 y behind 
the subtraction sign to the oppos ite. Theil . to ll o\\' the add i­
tion rules ahovC'. 
EXAMPI.ES: (X) ( 1 2 )~( R) + ( 12) ~ 4 and 
( 8) (12) - ( X) -I ( 1 1 ) ~ 20 and 
(- 8) ( 1 2 ) ~ ( R) 1- (12) - 4. Notice the s ign or thl' number 
in front of the subtraction sign never changes. 
--------------------------------~MULTIPLICATION & DIVISION 
Multiply or divide. then fililow these rules to determine 
the s ig n of the answer. 
·Ifthe numbers have the same signs. the answer is POSITIVE, 
• If the numbers have different Signs. th(, answer is NEGATI\'[, 
-It makes no LlitTcrc ncc which Ilumncr is lanler \\ hen you an~ 
try ing to dL'h.:rmine the sign of the answer.'- ~ 
EXAMPLES: ( 2)( 5) - 10 and ( 7)(3) ~ 2 1 and 
(-2)('1) ~- 18 
DOUBLE NEGATIVE 
' -(-a)~a; that is. the s ign ill ti'on! or the parentheses 
changes the s ign of the contents of the parentheses. 
EXAMPLES: ( 3 )= +3 o r (3) 3; also. (5<1-6) 5a + 6. 
FRACTIONS 
REDUCING 
• Divide 	nUl1lerator (top ) and de nom inato r (bottom) by the 
same number, there by renam ing it to an cqui , alent fraction 
in lower terms. This process may be repeated . 
EXAMPLE: ~O 4 5 
.12 4 
ADDITION 
!.!.. + ~= ~,wherec;tO 
c c 
• Change to equivalent fractions with common d~nomi nator . 
EXAMPLE: To eva luate ~ + ~ + ~. lo llow these steps: 
.1 .j 6 
I. Find the least common d enominator by dete rmi ning the 
sm a ll es t number which can be di vided e \ en ly (no remain­
ders) by all of the num bers in the denominators 
(bottoms). 
EX4 MPLE: 3.4. and 6 divide evenly into 12. 
2. 	 Multiply the n umerator and denomin a tor of each 
fraction so the fradion value has not changed but the 
COllllllon deno minator has been obtai ned. 
EXAMPLE: 
2 4 I 3 5 2 X .> 10 
- x -+- x -+- X -~- I - I -­
.1 4 4 .1 b 2 t2 t2 12 
3. Add the numerators and keep the same denominator 
because the addition of fractions is counting equal parts. 
EXAMPLE: 8 3 10 21 '} 3 
-+ - + -=-= 1- = 1­
12 12 12 12 12 4 
SUBTRACTION 
"-_"-=~ ,wherec;<O 
• Change to equivalent fractions with a common 
denominator. 
Find the least common denominator by determining 
thc smallest number which can be divided evenly by all 
of the numbers in the denominators (boltoms). 
EXAMPLE: '2 _-'­
<) 3 
2. 	 Multiply the nu merator and deno m inator by the samc 
number so the fraction value has not changed, but the 
common denominator has been obtained. 
EXAMPLE: '2 _ -'- x ~=L~ 
9 3 3 <) <) 
3. 	Subtract the numerators and keep the EXAMPLE: 
same denominator because subtraction 3 
of fraction s is finding the dinerenee 
between equal parts. ~9 
MULTIPLICATION 
"- x"- = ~,wherc c;<O and d;<O 
d e x d 
• Common denominators are ]'I;OT needed. 
I. 	 Multiply the numerators (tops) and mUltiply the denomi­
nators (bottoms), then reduce the answer to lowest terms. 
EXAMPLE: 	~ x ~= 12 +~=-'-
3 12 36 12 3 EXAMPLE: 
2. 	OR - reduce any numerator (top) with 
any denominator (boltom) and then 
multip ly thc numerators and multiply the 
denominators. 
DIVISION 
"- + "- = "- x "- = " x d , where c;< 0; d ;< 0; b ;< 0 
d /, eX /, 
'Common denominators are NOT needed. 
I. 	Change division to mul t iplication by the recip rocal; 
that is, Ilip the fraction in back of the division sign and 
change the division sign to a multiplication sign. 
EXAMPLE: LI + ~ becomes LI x~ . EXAMPLE: 
9 3 <) 2 
,..(2 ", I 2 
2. Now, follow the steps for multiplication 
»;ZI= } 
uf fractions, as indicated above. 
PERCENT, RATIO & 
PROPORTION 
PERCENTS 
• Definition: Percent means " out of I00" or " per 100." 
• Percents and equivalent fractions 
I. 	 Percents can be written as fractions by placing the 
number over 100 and simplifying or reducing. 
EXAMPLES: 3IJ% = 	 30 = 2 
100 10 
4.5 45 9 
4,5% = 100 = I.IX)(I = 21KI 
2. Fractions 	can be changed to percents by writing them 
with denominators of 100. The numerator is then the 
percent number. 
EXAMPlE' 2~~x 20 =~ =()O% 
-	 . 5 5 20 100 
• Percents and decimal numbers 
I. To change a percent to a decimal number, move 
".... the decimal point 2 places to the left because percent 
I{ mean s "out of 100" and decimal numbers with two 
I' digit s behind the decimal point also mean "out of 100." 
EXAMPLE: 45'% =~; 125'% = 18; 6% = ~; 
3.5% = ,035 bccause the 5 was already behind the decimal 
point and is not counted as one of the digits inthe " move 
two places." 
2. To change a decimal number to a perccnt, move the 
decimal point two places to the right. 
EXAMPLES: .47=47'Yo; 3.2 = 320.%; .205=~5'% 
• Definition: Comparison between two quantities. 
3 
• Forms : 3 to 5, 3 : 5, 3/5,5' 
PROPORTION 
• Definition: Statement of equality between two ratios or 
tractions. :J 9 
Forms: 3 is to 5 as 9 is to 15,3:5::9:15, 5 15 
SOLVING PROPORTIONS 
• Change the fractions to equivalent fractions with common 
denominators , set numerators (tops) equal to eaeh other, 
and solve the rcsulting statement. 
EXAMPLES: 2 = -'-'- becomes -'2 = -'-'-, so n = 15; 
4 20 20 20 
11+3 10 11+3 5 
-- = - becomes ~- = - , so n + 3 = 5 and n = 2. 
7 14 7 7 
• Cross-multiply and solve the resulting equation. NOTE: 
Cross-multiplication is used to solve proportions illlh' 
and mav ]'I;OT be used in fraction multiplication. Cross­
multiplication may be de scribed as the product of the means 
being equal to the product of the extremes, 
EXAMPLES: 
II 3 	 I -~- , 5n = 21, n = 21 +5, n = 4 ­
755 
3 7 	 I
'4"""';:;:2' 3n + 6 =28, 3n=22, n= 7) 
MIXED NUMBERS & 
IMPROPER FRACTIONS 
GENERAL COMMENTS 
• Description of mixed numbers: W hole numbers followed 
by fracti ons; that is, a whole number added to a frac tion. 
I I I 
EXAMPLE: 	 42 means 41 2 ; 4 /111t! 2 
• Improper fractions a re fracti ons that have a numerator 
(top number) larger tha n the denominator (bollom number) . 
• Conversions 
I. 	 Mixed number to impmper fraction : Multiply the denom­
inator (bottom) by the whole number EXAMPI.E: 
and add the numerator (top) to fi nd 
.,.... 
the numerator of the improper trac­ 2 3x5+2 17 
5- = - -=­tion. T he denominator of the t 3 3 3 
improper fi'3ction is the same as the 
denominator in the mixed number. 
2. 	 Improper fraction to mixed num ber: EXAMPLE: 
Divide the denominator into the 
 3~ 
numerator and write the remainder 17 5 
over the divisor (the divi sor is the 
- nleans 5}T7 
same number as the denominator in 5 15 
thc improper fraction). 2 
ADDITION 
• Add thc whole numbers. 
• Add thc ,,'actions by follow ing the stcps for addition of li'ac­
tions in the fract ion section o f th is study gu ide. 
• I f the answer has an improper fraction, change it to a 
mixed number 
EXAMPLE:and add the 
result ing whole 
number to the 
who le number in 
the answer. 
SUBTRACTION 
• Subtract the fractions first. 
I. 	If the fraction ufthe larger number is larger than the frac­
tion of the smaller number, then follow the steps of 
subtracting fractions in the fraction section of this study 
guide and then subtract the whole nu mbers. 
5 I 4 2 
EXAMPLE: 	7--2-=5-=5­() 6 6 3 EXAMPLE:2. 	 If that is not the case, then 
2 7 2 9
borrow ONE from 	 the 6 -= 5 +-+ -=5­
whole number and add it to 7 7 7 7 
5the fracti on 	 (must have 
-3­
common denominators) 7 
before subtracting. 	 42­
7 
3 
• Short Cut For Borrowing: EXAMPL E: 
Reduce the whole nwnber by 2 ' 2+ 7, l)
6-= 5+-- = 5 ­
one, replace the numerator by 7 7 7 
the sum (add) of the numera­
 -3~ = -3~ 
tor and denom inator of the 7 7 
fraction, and keep the same 
 4 
? ­denominator. 
-7 
MULTIPLICATION & DIVISION 
Change each In ixcd nu mber to an improper frae! io n and 
fo llow the steps for multip lyi ng and di vid ing frac tions. 
GEOMETRIC FORMULAS 
P E R 1M ETE R: The perimeter. P, o f a two-dimens io nal 
shape is the SUIll of all side le ngths. 
AREA: The area, A. 01' a two-di me nsional shape i, the 
number of square units that can be put in the region 
cncloscd hy the s ides. NOTE: Area is obtaincd through 
some combination of multiplying he ights and bases . which 
always form 90u angles with each othe r, except ill circl es . 
VOLUME: Thc volullle, Y, of a threc-d imens iona l shapc is 
the numbe r of cubic units that can be put ill the region 
enclosed by all the sides. 
Square Area: A = hb 
If h=g, then b=H: also. as all sides 
arc equal in a square, the-n: 
A = 64 sq uare: units 
Rectangle Arca: A = hb. or A = Iw 
If h=4 and b = 12, then : A =( 4){ 121 
A = 48 st.J Lmre un ilS 
Triangle A rea: A = II., bh 
If h= 8 and b= 12. th-en : A = '/,IK I( 12) 
A = 48 square unit s -
ParallclognllTl Area: A = hh 
Il'h=6 and b=l). then: A=(6){YI 
A = 54 square units 
Cin.:le Area: A = rcr:!' 
If1[= 3. 14 and 1'=5, th~n: A=13.1 4 )(:i)', 
A;;; 78.5 square units 
Cin:ulllference: C = 2rcr 
C=(2)(3.14)(5)=31.4 units 
Pythagorean Theorem 
If a right triangle has hypote nuse c. and 
:,ides a and b. then c 2 ~a2 +b2 
Rectangular Prism Ytllull1c 
V=lwh ; ifl= 12. w=:1, h=4. then: hc:r7V=( 12 1( 31(4). V= 144 cubic units 
w 
Cubc Volume. Y= e'\ 
Each edge le ngth . c. j:-. equal to the 
other edges in a cube. 
If e =8. then: v= (XIIX)iS). 
V= 512 cubic units 
Cy linder Volume. V= 1[r' h 
If radius, 1' =9, h =8 . the n: 
Y= 1[(9 1' 18i, Y= 3.14(8 1\(SI. 
Y=2034.72 cubic unit s 
Cone Vo lume. V= l !3nr1h 
If r=6 and h = 8, thell: 
V= 1/1 1((6)1(8 ). 
Y= 'I; (3. 14)(36)(81 
V=30 1.44 cu bic uni ts 
TriangUI.<Ir Prism VOIUIl1.l'": V=(area 0 1" Irianglc)hc2#_ 
If ~ IHis an area equal (() 
12 S h 
'/2 (5)1121. then: V=30h alld if h= 8. thell: \1 ­
Y=(30j(XI, V=240 cubi~ units 
Rectangular Pyramid V()lulllc: 
V= 1/.1, (:.trca of r~L'l a nglc)h 
I f I =5 and w = 4. the rCdanglc ha~ all 
arca of 20, then: 
V='/,(20Ih and irh=9. 
thcn: 'V= '/ ,(20)191. V= nll cubic units 
Sphere VUIUIllI.!, V= 4/[,J Irradius, 1'= 5, 
4(3.1.~ )(5)~ • then: 
V= 1.:70. V= 523.J cuhic unih 
PERCENT APPLICATIONS 
% INCREASE 
_~(_!i_"c_r_t'l_ls_ = amount oj'incI't:(lseC:' or 
IO!) origi/la/l',Jilie 
FORMULAS: 
(original "alue) x (% increase) ~ amount ofincrease 
If the amount of increase is not given, it may be fOllnd 
through this subtraction: 
(new value) -(original value) ~ amount of increase. 
EXAMPLE: The Smyth Company had 10,000 employees in 
1992 and 12.000 in 1993. Find the % increase. Amount of 
increase ~ 12,000 - 10,000 = 2,OO() 
'% increase: ~ 2,O()() 
I(X) 10,000 
SO,11 = 20 .Uld the (!~. incrc:-L';C = 2fY~) becatlse 1Y;) means "out of 100:' 
% DISCOUNT 
FOR~1ULAS: ';" diS ( UIIII{ = WI1011llIo(dis("()f1ll1 or 
100 originol \',dlle 
(original price) x (%1 discount) = $ discount 
If not given, ($ discount) = (original price) - (new price) 
EXAMPLE: The Smyth Company put suits thai usually sell 
for $250 on sale for $ISO. Find the percent disCLlunt. 
II $llXI 
% discount: 100 ~ $250 
so, II = 40 and the (;0 discount '---- 4U1%. 
SIMPLE INTEREST 
FORMULAS: i~prt. or (total amount) ~ 
(principal) + interest 
Vv'hcrc 	 i = interest 
p = principal: money borrowed or lent 
r = rate: percent ratl' 
t .":::: time: expressed in the same period as the rate, 
i.e., ifrak is per year. then time is in years or part ora year. 
Ii'rate is per month. then time is in months. 
EXAMPLE: Carolyn borrowed S5,000 Irom the bank at 6% 
~ simple inkrest per year. I f she borrowed the money for only 
3 months, find the total amount that she paid the bank. 
I S interest = prl ~ ($5,000) (6"1.,) (.2S) " $75 
I Notice that the 3 momhs was changed to .25 of a year. l ~T__ I __n____ ____ ~ $'5 ,_______ ~ S:5 ,_7 5. ..._()Ia_A ,()u I't_= I) ' i____()( )t) ' $ 75____O____________ 
a: % COMMISSION 
FORl\lULAS: '~-(~ cOf1lmissiun _ S cOlI/missioll 
I(}I) - S .';alt.'s 
= ($ sales) x (% commission) ~ S commission 
EXAMPLE: Missy earned 4'1, on a hOLlse she suld for 
S125,000. Find her dollar commission. 
\/~) commission: ~:::::: '£ C()!1I111is,~ or 
ItXI $1 co. (X~I 
($ 12S,()(XlI x (4%) = $ commission, so S cOl1lmission ~ $5,000. 
% MARKUP 
FORl\lULAS: ~'I/ Il/wJ.lljJ = S lIIurkllP or 
100 origil/a! IJI"ic(' 
(original price)x('Yo markup)~ $ markup 
lfnot given, ($ markup) ~ (new price) - (original price) 
EXAMPLE: The Smyth Company bought blouses lor S20 
each and sold them lor $44 each. Find the percent markup. 
S markup ~ $44-$20~$24 '1;' markup: ~~ 24, 
lOCI 20 
so, n = 120 and the f% markup = 120(%. 
% PROFIT 
FORMULAS: " (,	 Cmlit _ S prulil 
J()() lolu/ S illCOlllt' 
or (total $ income) x (% profit) ~ $ profit 
if not givcn, S profit ~ (total $ income) - ($ expensesl. 
I: EXAMPLE: The Smyth Company had expenses of 
$ISO,OOO and a profit of $1 0,000. Find the 'Yr, profit.
Z total S incomc - $150,000+ $1 UJ)OO = $160,()OO 
% profit: ~~ 10.000 or 
~ 100 16lJ,(JIlO 
($ I60,(JOO) x (n)=S I 0.000. In either case, the 0;., profit ~ 6.2S%. 
I ~~~--------------~ U S $ 4 95 free downloads & 
"" •• • t U" '~
"IS" & "OF" 
Any problems that are or can be stated with percent and the 
words "is" and "of' can be solved using these formulas: 
FORMULAS: Vi; _ "is" numher 
i7io - "of' number 
or "of' means multiply and "is" means equals. 
EXAMPLE I: What percent of 12S is 50" -"- ~ ~ or 
I!~) 125 
II x 125 ~ 50: in either case, the percent ~ 40%. 
EXAMPLE 2: What number is 125% of 80" 	 125 ~ -"- or 
HXI KO 
( 1.2S)(80) = II. In either ease, the number ~ 100. 
% DECREASE 
FOR~,tULAS: % decrease = Oil/Olin! o(deCfCUSt' or 
IfJf) originol mille 
(original value)x(% decrease) ~ amount of decrease 
Irnot given, amount ordecrease~(original value) · (new value) 
EXAMPLE: The Smyth Company had 12,OO() employees in 
1993 and 9 ,000 in 19l)4. Find the percent decrease. 
Amount of decrease = 12,000 - 9,000 = 3,000 
II ~.(XX) 
~/o decrease: 10() = 12.()(IO' so, n=25 and the (% del'fcase 25(~/~1. 
% EXPENSES OR COSTS 
FORMULAS: % expellses , S C.\{'('llses7: 
I ()() rOfU/ S illnJ/11(' 
or (tolal $ income)x(% expenses) ~ $ expenses 
EXAMPLE: The Smyth Company had a total income or 
5250,000 and $7,500 profit last month. Find the percent 
expenses. $ expenses: = $250,000 - $7 ,SOO = $242,5()O 
(% expenses:~:::::: 242.~ 
IIXI 250.t)(X) 
so. II = 97 and the (!/(I expenses = 971!-r(1. 
COMPOUND INTEREST 
FORMULA: A = 1'( I +~r 
Whcre: 	 A = total amount 
p=principal; money saved or invested 
r=rate of interest: llsually a 1% per year 
t= time: expressed in years 
II =total number of periods fit 
EXAMPLE: John put SI00 A- P (I +.':.) 
.. 	 II (4 8II1to a savings account a~ 4(% .. ( .(M, ) ~ x~ 1 
compounded quarterly for R A -- 100 I+ 4 
years. How much was in the A = I OO( 1.0 I ),1 
account at the end of 8 years? 
A ~ 100( 1.3749) 
A ~ 137.49 
ALGEBRA 
VOCABULARY 
oVariables are letters used ill represent numbers. 
oConstants arc specific numbers that arc not multiplied by 
anv variables. 
oC,iefficients are numbers that are multiplied by one or more 
variables. 
EXAMPLES: -4xy has a coefficient of-4: 9m' has a coef­
ficient of 9: x has -an invisible coefficient of I. 
• Terms are constants or variable expressions. 
EXAMPLES: .la: -Sc4d: 25 mpJ r,; 7 are all terms. 
• Like 	or similar ternlS arc terms that have the same vari­
ables to the same degree or exponent value. Coelficients do 
not,- matter, they n~ay be eq;lal or. nut. 
ExAMPLES: 3rn- and 7m~ are like terms because they bOlh 
have the same variable to the same power or cx~onent ·valuc. 
- ISa"b and 6a"b are like terms, but 2,4 and 6x are not like 
terms because although they have the same variable, x, it is to 
the power of 4 in one term and to the puwer uf 3 in the other. 
oAlgebraic expressions arc terms that are connected by 
either addition or subtraction. 
EXAMPLES: 2s , 4a1 - 5 is an algebraic expression with 3 
terms: 2s, 4a1 and 5. 
oAlgebraic equations arc statements of equality between at 
least two terms. 
EXAMPLES: 4z ~ 28 is an algebraic equation. 
3(a-4) 'c 6a= IIJ-a is also an algebraic equation. Notice 
that both statements have equal signs in them. 
oAlgebraic inequalities are statements that have either > or 
< between at least two terms. 
EXAMPLES: 50 <-2x is an algebraic inequality. 
3(2n + 7) > - lOis an algebraic inequality. 
'tles att !df . ISBN-13: 978-157222506-0~ Customer Hotline # 1,800,230,9522 qUlc 5 U y.com 
ISBN - l0: 157222506 - 8 
!\OTICI:: 1'0 STUDE'T Thi s Oui<.:kSllId y · gllid..: j , a liall!!), 
a!1nohll<.'d ~.'U id~ to baSIC m;r tli <,:Olll'>CS 
\ 11 righl~~ rclit' rH'd. Nil pan oj' lhi~ 
duco.!d or Ir,lIls mi a ....d in 1"(l1"l11. ell 
or llh.:dla l1 ic:d 
 911~ 11I)llII ~III!1!IJ~ll~llllr l llllillll lll 
1X'[(\lI~ ~ ((l!l 
6 3 
4 
DISTRIBUTIVE PROPERTY 
FOR POLYNOMIALS 
oType I: a(c+d) ~ ae+ad 
EXAMPLE: 4x-'(2xy -t y' ) ~ 8x4y + 4x3y' 
o'lype 2: (a + b)(e+d)~ a(c+d) + b(c+d) ~ ae + ad + be + bd 
EX1MPLE: (2x + y)(3x - 5v) 2xt:lx -5y ) + y(3x-Sy ) ~ 6x1-
I Oxy + 3xy - 5y1 ~ Iix2 - 7xy: 5y' . T hi s may als,) be clone b) 
using the FOIL Method j'.r Products of Binomials 
(sec Algebra I chart). This is a pupular method "". 
multiplying 2 terms by 2 terms only. FO IL means firs! 
term timcs first term. outer term times outer term. inn!.!!" 
term times inner tcnn. and last tcrm tirncs last term. 
COMBINING LIKE TERMS 
(adding or subtracting) 
o RU LE: Co mbine (acid or subtrael) only the eoellie ient ,; of 
like terms and never change the c,\poncnt~ during addition 
or subtraction: a + a = 2a. 
EX4MPLES: 4",3 and _7v3, arc like tcrl1l ~ . even thouch the 
x and yl arc 1l9t iii th~' S~l11lt: or9cr. and may be comhined 111 Ihi" 
mantleI': 4~y;l + 7y-'x =. ] xv" ; notice only the cn('trici~nt" 
\vere combined and no CXpO~1Cl1t chan~ed: I)a2lx' and 3bca:' 
arc not likt..' terms bt..'l"[\usl.' the CXplllll.'nts o f Ihl.' a art..' no! tht..' 
sallie in both terms. ~o they Illay not he added or ~tlbtrat..'tl.'d. 
MULTIPLYING TERMS 
oDefinition: 35 (3)(3)(3)(3)(31: that is, 3 is called the bali<' and 
it is multiplied by itself 5 times because the exponenl is 5. 
am~(a)(a)(a) ... (a): that is, the a is multiplied by itselfm time,. 
o Product Rule for Exponents: (am)(II") ~ a""": that i,. 
when Illultiplying the sam ..... hase, a in this cas...... simply add 
the exponents. 
• Any terms may he Illultiplied. not just like terllls. 
oRULE: Multiply the coelri,'icnts and multiply the \1IriahJc, 
(this m~aJ1S you have tn add the cxpon ..... llt ~ ()ftlle same \ariable). 
EXAMPLE: (4a~c)(-12a2h.1c)=-4!1a·b·lc2: notice that 4 
times - 12 became - 4X. a4 times a 2 bel·am.: at" c tim~~ .." 
became c2. and thL' bJ \vas written 10 intiiGltc Illultiplication 
by b, but the expunent did not cltange un the h heeause thcre 
was only one h in the problem. 
SOLVING A FIRST·DEGREE 
EQUATION WITH ONE VARIABLE 
oEliminale any fractions by l"inl' the Multiplicatiun PfIlpr:11) 
or Equality (could be tric~y if not handled properly) . ~ 
EXAMPl.E: 1/2( 3a + S) ~ 2!3 (7a 5) + ~ "ould be lTlulti­
plied on both sides of the sign by thl? IO\\'l?st cornl11oll S'. j 
denominator of 1/2 [Ind 2/3• \vhil"h is 6: the result \\Quld 
be 3(3a + 5) ~ 4(7a 5) + 54: notice that only 1,, 2'1. and 
9 were multiplied by 6 and not the contcilts-ofti le par,'ntheses: 
the parentheses will be handled in Ihe next ,tep, di"tribution. 
• [)istribute to remove any J1arl'nthe~l" s . ifthcre arC' all )'. 
EXAMPLE: 3(3a + 5) 4(7a 5) + 5-1 becomes 
9a + 15 ~ 28a - 2() + 54. 
• Comhine allY like h:Tll1S that arc 011 the sanll' s ide l.lfthe l'qual..; sign. 
EXAMPLE: 9a + 15 = na 10 + ~4 becomcs 
9a + 15 ~ 28a + 3-1 because the only like terms on the same 
side of the equals sign were the 20 and the +S4. 
• Use the addition property (If equality to add the same terms 
on both sides of the equals sign. This may be done mOI"l: than 
oncc. The objective here is to get all terms with th ..... ~arnc vari­
able on one side of the equals Sign and all constants \\ ithulI! 
the variable on the other side of the equals s ign. 
EXAMPLE:9a + 15 ~ na T 34 becollles 'Ia + 15 - 28a ­
15 =2~a 1-34-288-15. Notice that both - 2Sa and 15 wcre 
added to both ~ ides of the equal s " ign at the same time. This 
results in - llJa = 19 after lik..: terms arc added or subtracted. 
oUse the lIlultiplication property of "'qualit~ to make the 
coellicient or the variable a I. 
EXAMPLE: 19a ~ 19 would be multiplied (In both s ides b) 
'l,y (or divided by 10) til get" I in front of the a, so the 
equation would become la 19( _ I/ ,y )or s implya- - I. 
oCheck the answer by " ubstituting it Il)r the variable in the 
original equation to sec if it works. 
SOLVING A FIRST·DEGREE 
INEQUALITY WITH ONE VARIABLE 
• Follow the same steps tor solving a first -degree equality as 
described above, except for one step in the process. This 
exception follows. 
oException: When applying the multiplication property. the 
inequality sign must turn artlund if ~'ou multiplied b~ a 
negative number. 
EXAMPLES: In 4m > -48, you need to multiply both ~ 
sides or the > symbol by II.. Therefore, J 
4m( 1/4»-48( 1/4), This results in Ill > 12 . otiee the '> 
did not turn around hecause you multipli ed hy a posi ­
tive 1/4. However, in 5x > 65. you need to multiply both 
sides by _ 1/5 ' Therefore, 5x( 1/, )< 65( 1/5 ). Thi s re>ldts in 
x < 13. Notice the > did turn around and bC'L·0111C < becall se 
you multiplied by a negative number, _ 1/5' 
·Check the solution by substituting some Ilumerical \alu..: s 
of the variable in the original inequalit y.