REVIEW Chapter 1 The Real Number System


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1 Mth 7 REVIEW Chpter The Rel Number System In clss work: Solve ll exercises. (Sections. &. Definition A set is collection of objects (elements. The Set of Nturl Numbers N N = {,,,, 5, } The Set of Whole Numbers W W = { 0,,,,, 5, } N W The Set of Integers Z Z = {, , , , ,0,,,,, 5, } N W Z The Set of Rtionl Numbers Q Q = b, Z, b 0 N W Z Q b The Set of Irrtionl Numbers Exmples:, 5,π The Set of Rel Numbers R R = { x xisrtionl or xis irrtionl } N W Z Q R Mthemticl Symbols SYMBOL MEANING EXAMPLES = is equl to is not equl to belongs to ( bout n element it doesn t belong to < is less thn is less thn or equl to > is greter thn is greter thn or equl to
2 Properties of Rel Numbers PROPERTIES ADDITION + MULTIPLICATION COMMUTATIVITY + b= b+, b, R b = b b, R ASSOCIATIVITY ( + b + c = + ( b + c, bc,, R ( b c = bc (, bc,, R IDENTITY ELEMENT = 0 + =, R = =, R INVERSE ELEMENT R, there is R such tht + ( = ( + = 0 R, 0, there is R such tht = = DISTRIBUTIVITY ( b+ c = b+ c multiply out (remove prentheses fctor out the common fctor Exercise # Find the opposite nd the reciprocl (if ny of ech number: The Number Its Opposite Its Reciprocl The Double Negtive Rule ( =
3 (Section. The Absolute Vlue of Number Definition ( The bsolute vlue of number is the distnce between the number nd 0 (the origin on the number line. = dist(,0 Property 0, R Definition (, if 0 =, if < 0 Properties ( b = b, b, R ( =, b, R, b 0 b b Note: + b + b Exmple: b b Exmple: Exercise # Simplify the following: 7 = c 7 = b ( 7 = d ( 7 = Exercise # Simplify the following: ( (A: 0 d b c ( ( + ( 5 9[ (+ 6] ( (A: 7/ 6 (A: 9 7 e (A: 7 (A: /0 Exercise # Evlute the following expressions if x =, y =, z = : y x + yz (A: 8 b yz ( xy (A: 9
4 (Section.6 Properties of Integrl Exponents Definition If n N, then n =... n times is clled bse nd n is clled power (exponent. PROPERTY EXAMPLES The Product Rule m n m n = + The Quotient Rule m n = m n The ZeroExponent Rule 0 =, 0 The NegtiveExponent Rule n = n The Power Rule m ( n = mn Products to Power ( n n n b = b Quotients to Power n n = n b b
5 5 Exercise #5 Simplify the following expressions: x 5[ x 5( x 5] (A: x 5 b xyxy ( x 7 xy( xy x ( A: 6xy + 6xy 5 xy xy xy ( A:x 7 y 6 + ( ( A:x c ( 8 ( ( d x x x x ( x e ( x ( x+ 5 ( x( x+ f 7x x 5( x+ ( A :9 x+ 6 g x+ y x+ y x 6 x+ y Exercise #6 Simplify ech expression. Write nswers without using prentheses or negtive exponents. y yy b b b b c + b ( d 5 b c 0 x y x y e xy xy f 0 bc xy xy 5 g ( 5 x y ( xy 5 6 ( A: y 6 b ( A : b c ( A:xy 9 8 y x x 8y 5 6 Exercise #7 Find the set A { xx Z, x } = <. b Find the set B = { xx N, 0 < x 5} Exercise #8 Find x such tht N, x Z x. b Find x such tht 5 Z, x Z. x +
6 6 (Sections. &.5 Liner Equtions Definition An eqution is mthemticl sttement tht two lgebric expressions re equl. Exmples: Types of Equtions ( IDENTITY = n eqution which is lwys true regrdless of the vlue of the vrible. Exmples: = x+ = x + ( CONTRADICTION = n eqution which is lwys flse regrdless of the (INCONSISTENT vlue of the vrible. Exmples: 5= 7 x+ = x+ ( CONDITIONAL = n eqution whose truth or flsehood depends on the vlue of the vrible. Exmples: x + = 5 Exercise #0 Determine the type of ech of the following equtions: x = x ( b ( x c ( w 5 + = 5x+ 0 + = w+ Definition Definition A solution of n eqution is the vlue of the vrible tht stisfies the eqution. The process of finding the vlues tht stisfy n eqution is clled solving the eqution.
7 Exercise # Determine which of the listed vlues stisfies the given eqution: x + = 6, x= 0, x= b 6 w= 0 w, w=, w= Properties of Equlity If = b, then + c= b+ c, c R c = b c, c R c = bc, c R b =, c 0 c c Exercise # Solve the following equtions. 7 x = 8 g p = 5 n x y+ y 8 b + 5= 5 h 9x = 8 5 o z+ = z 7 7 u 5x+ 8= x+ 8 c 6= x + j x = q 5.6t+ =.6t v x = x d 6x = 5 q k t = p 5x+ x= 0 x + = 5 7 e 0t = 6 l + = r 6x 5 7x x 5 t = 0 + = + t ( = + y ( f = 0 m x = 5 s ( p+ 5 ( 9 + p = z ( ( x w 7 5( = ( 5 α x ( x+ 8 = 5x ( x x Exercise # Solve the following equtions. 5 = 0 x+ x+ z = b y = c + = w+ w+ d = e ( 6 v = f ( 6 u = g 5 ( t = (t+ + h 5 ( n n + 6 ( s+ = + ( s+ i = j x 6 + = k m m = + l k k = ( k 5
8 8 x+ + 6 x 6 = x+ x + x+ = x+ m ( x x+ 5= n ( ( 6 5 q q q 80 6 o ( = ( + p ( ( Exercise # Solve the following equtions. 0.8q.=.6 b.r.7 =.5 c.s+.7s =.9 d 5.m+.m=.96 e 0.(0.n 0. = 0.0 f 0.8(0.p 0.5 = 0.8 g 0.8q 0.(0 q = 80 h 0.r.(0 0.8( r 0 x x = = + i ( ( j ( y 0.87 y=.98 k 0.y+ 0.( 0 y = 0.y+ 6 l 0.x+ 0.05( x 00 = 05 Exercise #5 Solve the following equtions. 5% q = 6 b 0% r = 9 c50% s+ s = d 75% t + t = 05 e 0% u + 5% u = 8 f 50% t + 0%(90 t = 0 g 0% + 0%(5 s = 9 Answers #: /; b ; c ; d /;e 7; f 6/; g 7; h ; i ; m 5/; o  Answers #: 6; b ; c.7; d.; e /8; f 5; g 0; h 6 Answers #5: 0; b 0; c 8; d 60; e 0; f 0; g Exercise #6 Evlute x ( xy y for x stisfying ( x + = x + 6 nd y stisfying y 0 = 5y Exercise #7 Solve ech formul for the specified vrible: v k v = k+ gt, for t t = g S π b S=πd+ π, for d d = π A p c A= P( + rt, for r r = pt d A= w + lw, for l = + for e A h( b f A= lw+ lh+ wh for l A w l = w A = b h A wh l = ( w+ h
9 9 Chpter 5 Review of Fctoring Expressions Note: Fctoring n expression chnges it from sum into product. I Fctoring The Gretest Common Fctor (5. This is direct ppliction of the distributive property: b+ c= b ( + c Exercise # 8 Fctor s completely s possible:. 8x + 0x xy + 5xy 60xy 9 ( ( x x 6 9m n+ 8p 0 xy xy. b 6c + 8b 7 t 8t. 5xy 5xy 8 x + z( z 5( z x d d 8x y + y. cb ( ( b+ 9. s( r t( r 5. ( ( 6. z( b+ ( b+ II Fctoring by Grouping (5. We use this method when we usully hve four or more terms. Exercise # 9 Fctor s completely s possible: x + x+ xy+ y b + b 5 x xy x y + 7 6m m p mp+ p y y y z + z z m + mn+ 9m+ 9n 6 + b b 9 0x 5x x+ 0 8pq+ p+ 0q+ 5 b b+ b 8uv + 6uv+ 0uv + 0uv III Specil Products (5.5 Two Terms Three Terms b = b + b Perfect Squre Trinomils Difference of Squres: ( ( Sum of Squres: + b  not fctorble + b+ b = ( + b Difference of Cubes: b ( b( b b Sum of Cubes: + b = ( + b( b+ b = + + b+ b = ( b
10 Exercise # 0 Fctor s complete ly s possible: 0 x x x 0x+ 5 p 6 9 m 8 0 x w + w+ 7 6y 7 8 y x + 6 x + y 6 5 b n c p 7 9 ( x IV Fctoring Trinomils x + bx+ c (5. Cse Leding coefficient is: Exercise # Fctor s completely s possible: x x x 6 x 8 ( d + ( d 9 9 = Fctor x + bx+ c= ( x ( x product = c sum = b + 5x x 5x 6 5x 6 6 x + x 5 0 m + 6m 8 7x t 8t+ x xy+ y x x 6 8 z 7z+ 0 t tz 6z + d + d x 7x 5 5 w w 96 6 w w+ Cse Leding coefficient is not : Fctor x + bx+ c= x + x+ x+ c by splitting the middle term bx then using grouping product = c sum = b Fctor s completely s possible: 7 6x + 7x 0 0 x x 0 7x x 0 8 t 7t+ 5+ 6b 9b 6p + 8p y + 5y b b 6 r + 9r u + 8u 8 0b b x + 7x 8 5q + 57q+ 8 6k + 8k 6k k + 7k 70k
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