Algebra Review. How well do you remember your algebra?

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1 Algebr Review How well do you remember your lgebr? 1 The Order of Opertions Wht do we men when we write + 4? If we multiply we get 6 nd dding 4 gives 10. But, if we dd + 4 = 7 first, then multiply by then we get 14 s the nswer. Which nswer do we wnt when we evlute this? The point is tht the order we do these opertions mtters. We must decide convention of which opertions to do first. This is clled order of opertions. Here re the conventions we dopt for the order of opertions: Do things inside prentheses first. Of ll prentheses, work on innermost prentheses first. Net compute ny eponents. Now multiply nd divide. Finlly, do ny ddition or subtrction. So in the cse of the bove problem, we get + 4 = = 10, not 14. Here re couple of other importnt nottions tht we will be using. We often write multipliction with or sometimes without symbol t ll. All of the following men times b. A horizontl line or / mens divide. b = b = b b = /b = b Algebr Hzrd # 1: No Division by Zero. For emple, note tht = = is incorrect. 1

2 Note lso tht in + 1 there is n understood prenthesis on the numertor nd denomintor, so wht we men when we write this is ( + ) or equivlently ( + ) ( 1) ( 1) 1.1 Problems Simplify the following epressions nd show ech individul step in your work: Emple: ( 1) = = = 5 #1. (5 ) + (nswer : 1) # (nswer : 10) # (nswer : 15 ) Associtive, Commuttive, nd Distributive Properties Multipliction nd ddition both hve the properties of ssocitivity nd commuttivity. If the vribles, b, nd c ll represent numbers, the following re the ssocitive properties for ddition nd multipliction: Addition : + (b + c) = ( + b) + c Multipliction : (bc) = (b)c This mens tht we cn dd or multiply numbers in ny order nd you will lwys get the sme thing. Here re couple of emples with rel numbers: + (4 + 1) = 7 = ( + 4) ( ) = 6 = (6 ) The commuttive property is esy, it is + b = b + (ddition) nd b = b (multipliction) This mens tht it doesn t mtter which order we write things in when using ddition or multipliction. (Here s something to strt thinking bout Is the commuttive property true for subtrction nd division?)

3 Some emples wiith numbers: + = 5 = + = 6 = The distributive property is or from the left (b + c) = b + c (b + c) = b + c This property is common one people forget, so mke sure you understnd it. Here it is with some numbers: 4(5 + 1) = 4 = The ssocitive, commuttive, nd distributive properties lso pply to subtrction nd division, but we hve to be creful. To pply the properties to subtrction, we need to first convert the subtrction to ddition. Subtrction is the sme s dding negtive number. Here is n emple of this: = + ( 1) = + ( ) The sme is true for more complicted epressions: ( ) ( 1) = ( 1) + ( ) The distributive property cn further be used to simplify the bove epression. ( 1) + ( ) = ( 1)( 1) + (1)( ) = = 1 Similrly, in order to pply the ssocitive, commuttive nd distributive properties to division, we think of division s multipliction. Dividing is the sme s multiplying by the reciprocl. b = 1 b Here s n emple using the distributive property with division. ( 1) = ( + ( 1)) 1 = 1 + ( 1) 1 The distributive property is lso where the whole business of fctoring comes from. Consider + 4y Both terms in the sum hve fctor of two in them, so we cn fctor it out or use the distributive property to write it different wy. We will do this sort of thing lot in this clss. + 4y = + y = ( + y)

4 Algebr Hzrd # : Fke Distributive Properties. Note tht ( + ) + nd + + You my hve lerned the FOIL (First Outside Inside Lst) method for multiplying things like ( + )( ). It comes from the distributive property. Just tret the stuff in the first set of prentheses s single number nd distribute it into the second set of prentheses. ( + )( ) = ( + ) + ( + )( ) Now use the distributive property gin, this time from the left. Finlly, we simplify by combining like terms.... = + + ( ) + ( ) = + 6 = 6 Notice tht wht we hve done is ectly the FOIL method. We cn use this sme pproch to multiply bigger things. To multiply ( + )( + 1), first distribute ( + ) into ( + 1). ( + )( + 1) = ( + ) + ( + )( ) + ( + )(1) Now distribute gin. = ( + ) + ( )( + ) + (1)( + ) = + + ( ) + ( ) + ( ) + ( )( ) + + = = Problems Use the ssocitive, commuttive nd distributive properties to simplify the following. #4. ( b) [ ( + b)] (ns : ) #5. [( + b) ] [ ( b)] (ns : 0) #6. { [b ( b)]} (ns : 4 + b) Fctor out ny fctors common to ll the terms using the distributive property. Emple : 6 4 = ( ) #7. 1 8y + 0 # y 8 4 z #9. 9bc + b c 4

5 Multiply nd simplify by using the distributive property. #10. ( )( + ) #11. ( 1)( + ) Frctions Multiplying frctions. This is the esiest, just multiply the numertors(top) together nd multiply the denomintors(bottom) together. b c d = c bd Here s n emple with numbers. 4 1 = 8 And nother. 1 1 = = 1 = 4 1 = 4 1 = 1 4 = 4 This is good, since hlf of three qurters is three eighths nd third of twelve is four. Also notice tht we hve used the fct tht 1 = 1 1 nd tht = 1. These re importnt observtions tht we use often. To dd frctions, we need common denomintor. This mounts to sying tht if we wnt to dd things up, it is esier if ll the pieces re the sme size. Sy we wnt to dd three qurters to one hlf. The pieces re different sizes, but we know tht one hlf is the sme s two qurters. Therefore we cn dd three qurters plus two qurters. It is esy now, the nswer is five qurters. This is common denomintors. Now in more mthemticl terms Generlly, the esiest common denomintor to work with is the lest common multiple of the denomintors in question. In this cse, the lest common multiple of 4 nd is 4. Our gol is to write both frctions with denomintor 4. The first is done, the second we do s follows, using the fct tht = 1. 1 = 1 1 = 1 = 1 = 4 Now we hve = = 5 4 It works the sme if the frctions re more complicted. Try

6 The lest common multiple of the denomintor is ( + 1)( + ). So this time we need to fi both denomintors. First, to fi + +1 we use the fct tht + = 1. Now to fi + Becuse we hve + 1 = = + + we use the fct tht = = ( + ) ( + 1)( + ) + = 1 + = = ( + 1) ( + 1)( + ) + 1 = ( + ) ( + 1)( + ) nd + = ( + 1) ( + 1)( + ) = ( + ) ( + 1)( + ) + ( + 1) ( + 1)( + ) Now we hve written the sum in frctions with common denomintor so we cn dd nd simplify to get = ( + ) + ( + 1) ( + 1)( + ) = ( + 1)( + ) = Wht if you hve frction divided by frction? As you my lredy know, you multiply the top frction by the reciprocl of the bottom frction. (Recll tht you get the reciprocl of frction by moving the top to the bottom nd the bottom to the top.) Here s why it works. First notice wht hppens if you multiply frction by its reciprocl. b b = b b = 1 Now let s see wht hppens if frction is divided by frction. We ll use our observtion tht b b = 1. c/d b/ = 1 c/d b/ = /b /b c/d b/ = (/b)(c/d) (/b)(b/) = (/b)(c/d) 1 = (/b)(c/d) We ve done ectly wht I sid we would do. We multiplied the top frction by the reciprocl of the bottom frction. Now we see tht dividing four by hlf is the sme s multiplying by two. 4 1 = 4/1 1/ = = 4 = 8 6

7 One more thing bout frctions, we lwys try to write them in lowest terms. This mens tht we write the frction so tht the numertor nd denomintor hve no common fctors. Here re couple of emples. The frction 4 is not in lowest terms. Both the numertor nd the denomintor hve fctor of in them. We fctor it out nd cncel s follows. This one is little more difficult. 4 = 1 = 1 = 1 1 = We begin by noticing tht the numertor nd denomintor hve common fctor of +. Now we my proceed s in the previous emple ( + )( + ) = ( + )( ) = = + Algebr Hzrd # : Mgic Cncelltion in Frctions. Note tht +b b + nd Problems Use the rules of bsic lgebr to write the following epressions s frctions in lowest terms. 1. b b (nswer : b b ) (nswer : ) (nswer : 1 ) 16. y + y y b + b 4 Skim bck through this worksheet nd circle nything you cme cross tht ws prticulrly useful nd ny ides tht seem prticulrly importnt to you, especilly nything tht ws unfmilir. We will use this worksheet s reference during the rest of the course. Being consistent with lgebr will mke everything else go lot smoother. 7

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