Supplemental Appendix: Equations of Lines, Compound Inequalities, and Solving Systems of Linear Equations in Two Variables

Transcription

1 _t.pdf /9/ : AM ( ) Supplmntal Appndi: Equations of Lins, Compound Inqualitis, and Solving Sstms of Linar Equations in Two Variabls

2 _t.pdf /9/ : AM ( )

3 _t.pdf /9/ : AM ( ) Equations of Lins F EQUATIONS OF LINES OBJECTIVES Us th slop-intrcpt form to writ th quation of a lin. Graph a lin using its slop and -intrcpt. Us th point-slop form to writ th quation of a lin. Writ quations of vrtical and horizontal lins. Find quations of paralll and prpndicular lins. OBJECTIVE Using slop-intrcpt form to writ quations of lins. In th last sction, w larnd that th slop-intrcpt form of a linar quation is = m + b. Whn a linar quation is writtn in this form, th slop of th lin is th sam as th cofficint m of.also, th -intrcpt of th lin is (0, b). For ampl, th slop of th lin dfind b = + is, and its -intrcpt is (0, ). W ma also us th slop-intrcpt form to writ th quation of a lin givn its slop and -intrcpt. Th quation of a lin is a linar quation in variabls that, if graphd, would produc th lin dscribd. EXAMPLE Writ an quation of th lin with -intrcpt 0, - and slop of. Solution W want to writ a linar quation in variabls that dscribs th lin with -intrcpt 0, - and has a slop of W ar givn th slop and th -intrcpt. Lt m =. and b = -, and writ th quation in slop-intrcpt form, = m + b. = m + b = + - Lt m = and b = -. = - Simplif. Writ an quation of th lin with -intrcpt (0, ) and slop of -. Concpt Chck What is wrong with th following quation of a lin with -intrcpt (0, ) and slop? = + OBJECTIVE Graph a lin using slop and -intrcpt. Givn th slop and -intrcpt of a lin, w ma graph th lin as wll as writ its quation. Lt s graph th lin from Eampl. EXAMPLE Graph = -. Answr to Concpt Chck: -intrcpt and slop wr switchd, should b = + Solution Rcall that th slop of th graph of = is and th -intrcpt is 0, -. To - graph th lin, w first plot th -intrcpt 0, -. To find anothr point on th lin, w rcall that slop is ris Anothr point ma thn b plottd b starting at 0, -, rising unit up, and thn running run =. units to th right. W ar now at th point, -. Th graph is th lin through ths two points. m ~ 6 Run Ris (, ) (0, ) 6

4 _t.pdf /9/ : AM ( ) F Supplmntal Appndi Notic that th lin dos hav a -intrcpt of 0, - Graph =. + and a slop of. EXAMPLE Solution Graph + =. First, w solv th quation for to writ it in slop-intrcpt form. In slop-intrcpt form, th quation is = - Nt w plot th -intrcpt (0, ). +. To find anothr point on th lin, w us th slop - which can b writtn as ris W start at (0, ) and mov down units sinc th numrator of th slop run = -,. is -; thn w mov units to th right sinc th dnominator of th slop is. W arriv at th point (, ). Th lin through ths points is th graph, shown blow to th lft. Down (0, ) Right (, ) m s Lft (, 6) 6 Up (0, ) m s - Th slop can also b writtn as so to find anothr point in Eampl w -, could start at (0, ) and mov up units and thn units to th lft. W would arriv at th point -, 6. Th lin through -, 6 and (0, ) is th sam lin as shown prviousl through (, ) and (0, ). S th graph abov to th right. Graph + = 6. OBJECTIVE Using point-slop form to writ quations of lins. Whn th slop of a lin and a point on th lin ar known, th quation of th lin can also b found.to do this, us th slop formula to writ th slop of a lin that passs through points, and (, ). W hav m = - - Multipl both sids of this quation b - to obtain - = m - This form is calld th point-slop form of th quation of a lin. Point-Slop Form of th Equation of a Lin Th point-slop form of th quation of a lin is slop T - = m( - ) c c point whr m is th slop of th lin and, is a point on th lin.

5 _t.pdf /9/ : AM ( ) Equations of Lins F EXAMPLE Find an quation of th lin with slop - containing th point, -. Writ th quation in slop-intrcpt form, = m + b. Solution Bcaus w know th slop and a point of th lin, w us th point-slop form with m = - and, =, -. - = m - Point-slop form - - = - - Lt m = - and, =, -. + = - + Appl th distributiv proprt. = - - Writ in slop-intrcpt form. In slop-intrcpt form, th quation is = - -. To chck, dfin = - - and graph in an intgr window. From th quation w s that m = -, so th slop is -. Nt, chck on th graph to s that it contains th point, - as in th scrn blow to th lft. 7 7 Th point (, ) is includd in th solution pairs. A scond wa to s that, - is an ordrd pair solution is to amin a tabl of valus as shown abov to th right. Find an quation of th lin with slop - containing th point (-, ). Writ th quation in slop-intrcpt form, = m + b. Hlpful Hint Rmmbr, slop-intrcpt form mans th quation is solvd for. EXAMPLE Find an quation of th lin through points (, 0) and -, -. Writ th quation using function notation. Thn us a graphing utilit to chck. Solution First, find th slop of th lin. m = = - -8 = 8 Nt, mak us of th point-slop form. Rplac, b ithr (, 0) or -, - in th point-slop quation. W will choos th point (, 0). Th lin through (, 0) with slop is 8 - = m - Point-slop form - 0 = - 8 Lt m = and (, ) = (, 0). 8 8 = - 8 = - 0 Multipl both sids b 8. Appl th distributiv proprt. To writ th quation using function notation, w solv for, thn rplac with f(). 8 = - 0 = Divid both sids b 8. f = 8 - Writ using function notation.

6 _t.pdf /9/ : AM ( ) F Supplmntal Appndi To chck, graph = in an intgr window. Trac to find (, 0) and 8 - -, - on th graph of th lin. W can also amin a tabl of ordrd pair solutions to chck Find an quation of th lin through points (-, ) and (, 0). Writ th quation using function notation. Hlpful Hint If two points of a lin ar givn, ithr on ma b usd with th point-slop form to writ an quation of th lin. EXAMPLE 6 standard form. Find an quation of th lin graphd. Writ th quation in Solution First, find th slop of th lin b idntifing th coordinats of th notd points on th graph. Th points hav coordinats (-, ) and (, ). m = Nt, us th point-slop form. W will choos (, ) for (, ), although it maks no diffrnc which point w choos. Th lin through (, ) with slop is - = m( - ) Point-slop form - = ( - ) - - (-) = Lt m = and (, ) = (, ). ( - ) = ( - ) - 0 = - 9 Multipl both sids b. Appl th distributiv proprt. To writ th quation in standard form, mov - and -trms to on sid of th quation and an numbrs (constants) to th othr sid. Th quation of th graphd lin is - + =. 6-0 = = Subtract from both sids and add 0 to both sids. Find an quation of th lin graphd. Writ th quation in standard form. Th point-slop form of an quation is vr usful for solving ral-world problms.

7 _t.pdf /9/ : AM ( ) Equations of Lins EXAMPLE 7 Prdicting Sals Southrn Star Ralt is an stablishd ral stat compan that has njod constant growth in sals sinc 000. In 00 th compan sold 00 houss, and in 007 th compan sold 7 houss. Us ths figurs to prdict th numbr of houss this compan will sll in th ar 06. Solution. UNDERSTAND. Rad and rrad th problm. Thn lt = th numbr of ars aftr 000 and = th numbr of houss sold in th ar corrsponding to. Th information providd thn givs th ordrd pairs (, 00) and (7, 7). To bttr visualiz th sals of Southrn Star Ralt, w graph th linar quation that passs through th points (, 00) and (7, 7). Numbr of houss sold (7, 7) 0 00 (, 00) Yars aftr 000. TRANSLATE. W writ a linar quation that passs through th points (, 00) and (7, 7). To do so, w first find th slop of th lin. m = = = 7 - Thn, using th point-slop form and th point, 00 to writ th quation, w hav - = m = - Lt m = and (, ) = (, 00) = - 0 Multipl. = + 70 Add 00 to both sids.. SOLVE. To prdict th numbr of houss sold in th ar 06, w us = + 70 and complt th ordrd pair (6, ), sinc = 6. = = 0 Lt = 6.. INTERPRET. Chck: Vrif that th point (6, 0) is a point on th lin graphd in stp. Stat: Southrn Star Ralt should pct to sll 0 houss in th ar 06. Southwst Florida, including Fort Mrs and Cap Coral, has bn a growing 7 ral stat markt in past ars. In 00, thr wr 7 hous sals in th ara, and in 006, thr wr 998 hous sals. Us ths figurs to prdict th numbr of hous sals thr will b in 0. F

8 _t.pdf /9/ : AM ( ) F6 Supplmntal Appndi OBJECTIVE Writing quations of vrtical and horizontal lins. A fw spcial tps of linar quations ar linar quations whos graphs ar vrtical and horizontal lins. EXAMPLE 8 Find an quation of th horizontal lin containing th point (, ). Solution Rcall that a horizontal lin has an quation of th form = b. Sinc th lin contains th point (, ), th quation is =. Blow is a graph of = with th point (, ) indicatd Find th quation of th horizontal lin containing th point (6, -). TECHNOLOGY NOTE Rcall that a vrtical lin dos not rprsnt th graph of a function and cannot b graphd in th Y= ditor in function mod. Vrtical lins ma b drawn using th draw mnu. Blow is an ampl of using a draw vrtical fatur. EXAMPLE 9 undfind slop. Find an quation of th lin containing th point (, ) with Solution Sinc th lin has undfind slop, th lin must b vrtical. A vrtical lin has an quation of th form = c. Sinc th lin contains th point (, ), th quation is =, as shown to th right. 9 slop. Find an quation of th lin containing th point (6, -) (, ) with undfind OBJECTIVE Finding quations of paralll and prpndicular lins. Nt, w find quations of paralll and prpndicular lins. EXAMPLE 0 Find an quation of th lin containing th point (, ) and paralll to th lin + = -6. Writ th quation in standard form. Solution Bcaus th lin w want to find is paralll to th lin + = -6, th two lins must hav qual slops. Find th slop of + = -6 b writing it in th form = m + b. In othr words, solv th quation for. + = -6 = = Subtract from both sids. Divid b. = - - Writ in slop-intrcpt form.

9 _t.pdf /9/ : AM ( ) Equations of Lins F7 Th slop of this lin is - Thus, a lin paralll to this lin will also hav a slop of -.. Th quation w ar askd to find dscribs a lin containing th point (, ) with a slop of - W us th point-slop form.. - = m - Th graphs of paralll lins + = -6 and + = 0. Hlpful Hint Multipl both sids of th quation + = 0 b - and it bcoms - - = -0. Both quations ar in standard form, and thir graphs ar th sam lin. 0 - = = = = 0 Lt m =- and =., =, Multipl both sids b. Appl th distributiv proprt. Writ in standard form. Find an quation of th lin containing th point (8, -) and paralll to th lin + =. Writ th quation in standard form. EXAMPLE Writ a function that dscribs th lin containing th point (, ) and is prpndicular to th lin + = -6. Solution In th prvious ampl, w found that th slop of th lin + = -6 is - A lin prpndicular to this lin will hav a slop that is th ngativ rciprocal of. - or,. From th point-slop quation, w hav - = m - - = - - = = - = - Lt, and m = =, =. Multipl both sids b. Appl th distributiv proprt. Add 8 to both sids. Th graphs of prpndicular lins + = -6 and f = -. = - f = - Divid both sids b. Writ using function notation. Writ a function that dscribs th lin containing th point (8, -) and is prpndicular to th lin + =. Forms of Linar Equations A + B = C = m + b - = m - = c = c Standard form of a linar quation A and B ar not both 0. Slop-intrcpt form of a linar quation Th slop is m, and th -intrcpt is (0, b). Point-slop form of a linar quation Th slop is m, and, is a point on th lin. Horizontal lin Th slop is 0, and th -intrcpt is (0, c). Vrtical lin Th slop is undfind, and th -intrcpt is (c,0). Paralll and Prpndicular Lins Nonvrtical paralll lins hav th sam slop. Th product of th slops of two nonvrtical prpndicular lins is -.

10 _t.pdf /9/ : AM ( ) F8 Supplmntal Appndi VOCABULARY & READINESS CHECK Stat th slop and th -intrcpt of ach lin with th givn quation.. = =.. = = - 7 = - =- + Dcid whthr th lins ar paralll, prpndicular, or nithr = + 6 = - = -9 + = = = = - = - 6 EXERCISE SET Us th slop-intrcpt form of th linar quation to writ th quation of ach lin with th givn slop and -intrcpt. S Eampl.. Slop -; -intrcpt (0, ). Slop -intrcpt 0, -6 ;. Slop ; -intrcpt a0, b. Slop -; -intrcpt a0, - b. Slop -intrcpt (0, 0) 7 ; 6. Slop - -intrcpt (0, 0) ; Graph ach linar quation. S Eampls and. 7. = - 8. = = = =. - + = -6 Find an quation of th lin with th givn slop and containing th givn point.writ th quation in slop-intrcpt form. S Eampl.. Slop ; through (, ). Slop ; through (, ). Slop -; through, - 6. Slop -; through, - 7. Slop through -6, ; 8. Slop through -9, ; 9. Slop - 9 through -, 0 0 ; 0. Slop - through, -6 ; Find an quation of th lin passing through th givn points. Us function notation to writ th quation. S Eampl.. (, 0), (, 6). (, 0), (7, 8). -,, -6,. 7, -, (, 6). -, -, -, , -, -, , -8, -6, , -,, a and a -, 7, 0 b 0 b 0. a and a,, - b b Find an quation of ach lin graphd. Writ th quation in standard form. S Eampl

11 _t.pdf /9/ : AM ( ) Equations of Lins F9 Us th graph of th following function f() to find ach valu. 68. Undfind slop; through 0, Through (6, ); paralll to th lin 8 - = Through (, ); prpndicular to th lin - = 8 7. Through, -6; prpndicular to = 9 7. Through -, -; paralll to = 9 7. Through, -8 and -6, -; us function notation. 7. Through -, - and -6, ; us function notation.. f(0) f() 8. f() 9. Find such that f = Find such that f =. Writ an quation of ach lin. S Eampls 8 and 9.. Slop 0; through -, -. Horizontal; through -,. Vrtical; through (, 7). Vrtical; through (, 6). Horizontal; through (0, ) 6. Undfind slop; through (0, ) Find an quation of ach lin. Writ th quation using function notation. S Eampls 0 and. 7. Through (, 8); paralll to f = - 8. Through (, ); paralll to f = - 9. Through, -; prpndicular to = Through -, 8; prpndicular to - =. Through -, -; paralll to + =. Through -, -; prpndicular to + = MIXED Find th quation of ach lin. Writ th quation in standard form unlss indicatd othrwis. S Eampls,, and 8 through.. Slop ; through -,. Slop ; through -,. Through (, 6) and (, ); us function notation. 6. Through (, 9) and (8, 6) 7. With slop - -intrcpt ; 8. With slop -; -intrcpt us function notation. 9 ; 9. Through -7, - and 0, Through, -8 and -, - 6. Slop - through -, 0 ; f- 6. Slop - through, - ; 6. Vrtical lin; through -, Horizontal lin; through (, 0) 6. Through 6, -; paralll to th lin + = Through 8, -; paralll to th lin 6 + = 67. Slop 0; through -9, Solv. S Eampl Dl Mont Fruit Compan rcntl rlasd a nw applsauc. B th nd of its first ar, profits on this product amountd to $0,000. Th anticipatd profit for th nd of th fourth ar is $66,000. Th ratio of chang in tim to chang in profit is constant. Lt b ars and P b profit. a. Writ a linar function P() that prsss profit as a function of tim. b. Us this function to prdict th compan s profit at th nd of th svnth ar. c. Prdict whn th profit should rach $6, Th valu of a computr bought in 00 dprciats, or dcrass, as tim passs. Two ars aftr th computr was bought, it was worth $000; ars aftr it was bought, it was worth $800. a. If this rlationship btwn numbr of ars past 00 and valu of computr is linar, writ an quation dscribing this rlationship. [Us ordrd pairs of th form (ars past 00, valu of computr).] b. Us this quation to stimat th valu of th computr in th ar Th Pool Fun Compan has larnd that, b pricing a nwl rlasd Fun Noodl at $, sals will rach 0,000 Fun Noodls pr da during th summr. Raising th pric to $ will caus th sals to fall to 8000 Fun Noodls pr da. a. Assum that th rlationship btwn sals pric and numbr of Fun Noodls sold is linar and writ an quation dscribing this rlationship. b. Prdict th dail sals of Fun Noodls if th pric is $ Th valu of a building bought in 990 apprciats, or incrass, as tim passs. Svn ars aftr th building was bought, it was worth $6,000; ars aftr it was bought, it was worth $80,000. a. If this rlationship btwn numbr of ars past 990 and valu of building is linar, writ an quation dscribing this rlationship. [Us ordrd pairs of th form (ars past 990, valu of building).] b. Us this quation to stimat th valu of th building in th ar In 006, th mdian pric of an isting hom in th Unitd Stats was approimatl $,000. In 00, th mdian pric of an isting hom was $0,900. Lt b th mdian pric of an isting hom in th ar, whr = 0 rprsnts 00. (Sourc: National Association of REALTORS ) a. Writ a linar quation that modls th mdian isting hom pric in trms of th ar. [Hint: Th lin must pass through th points (0, 0,900) and (,,000).]

12 _t.pdf /9/ : AM ( ) F0 Supplmntal Appndi b. Us this quation to prdict th mdian isting hom pric for th ar = = - 0 c. Intrprt th slop of th quation found in part a. 80. Th numbr of births (in thousands) in th Unitd Stats in 000 was 060. Th numbr of births (in thousands) in th Unitd Stats in 00 was 6. Lt b th numbr of births (in thousands) in th ar, whr = 0 rprsnts 000. (Sourc: National Cntr for Halth Statistics) CONCEPT EXTENSIONS Answr tru or fals. 89. A vrtical lin is alwas prpndicular to a horizontal lin. a. Writ a linar quation that modls th numbr of births (in thousands) in trms of th ar. (S hint for Ercis 79a.) 90. A vrtical lin is alwas paralll to a vrtical lin. b. Us this quation to prdict th numbr of births in th Unitd Stats for th ar 0. Find an quation of th prpndicular bisctor of th lin sgmnt whos ndpoints ar (, 6) and (0, ). Eampl: c. Intrprt th slop of th quation in part a. 8. Th numbr of popl mplod in th Unitd Stats as mdical assistants was 87 thousand in 00. B th ar 0, this numbr is pctd to ris to 89 thousand. Lt b th numbr of mdical assistants (in thousands) mplod in th Unitd Stats in th ar, whr = 0 rprsnts 00. (Sourc: Burau of Labor Statistics) Prpndicular bisctor 7 6 (, 6) Lin sgmnt (0, ) 9 Solution: A prpndicular bisctor is a lin that contains th midpoint of th givn sgmnt and is prpndicular to th sgmnt. Stp : Th midpoint of th sgmnt with nd points (, 6) and 0, - is (, ). Stp : Th slop of th sgmnt containing points (, 6) and a. Writ a linar quation that modls th numbr of popl (in thousands) mplod as mdical assistants in th ar. (S hint for Ercis 79a.) b. Us this quation to stimat th numbr of popl who will b mplod as mdical assistants in th ar Th numbr of popl mplod in th Unitd Stats as sstms analsts was 87 thousand in 00. B th ar 0, this numbr is pctd to ris to 60 thousand. Lt b th numbr of sstms analsts (in thousands) mplod in th Unitd Stats in th ar, whr = 0 rprsnts 00. (Sourc: Burau of Labor Statistics) a. Writ a linar quation that modls th numbr of popl (in thousands) mplod as sstms analsts in th ar. (S hint for Ercis 79a.) b. Us this quation to stimat th numbr of popl who will b mplod as sstms analsts in th ar 0. 0, - is. Stp : A lin prpndicular to this lin sgmnt will hav slop of -. Stp : Th quation of th lin through th midpoint (, ) with a slop of - will b th quation of th prpndicular bisctor. This quation in standard form is + = 9. Find an quation of th prpndicular bisctor of th lin sgmnt whos nd points ar givn. S th prvious ampl. 9., - ; -, 9. -6, - ; - 8, , 6; -, - 9. (, 8); (7, ) 9. (, ); -, , 8; -, - REVIEW AND PREVIEW Solv. S Sction = = = = Dscrib how to chck to s if th graph of - = 7 passs through th points., -.0 and 0, -.7. Thn follow our dirctions and chck ths points.

13 _t.pdf /9/ : AM ( ) Compound Inqualitis F COMPOUND INEQUALITIES OBJECTIVES Find th intrsction of two sts. Solv compound inqualitis containing and. Find th union of two sts. Solv compound inqualitis containing or. Two inqualitis joind b th words and or or ar calld compound inqualitis. Compound Inqualitis and 7 Ú or OBJECTIVE Finding th intrsction of two sts. Th solution st of a compound inqualit formd b th word and is th intrsction of th solution sts of th two inqualitis. W us th smbol to rprsnt intrsction. Intrsction of Two Sts Th intrsction of two sts, A and B, is th st of all lmnts common to both sts. A intrsct B is dnotd b A B A B EXAMPLE If A = ƒ is an vn numbr gratr than 0 and lss than 06 and B =,,, 66, find A B. Solution Lt s list th lmnts in st A. Th numbrs and 6 ar in sts A and B. Th intrsction is, 66. A =,, 6, 86 If A = ƒ is an odd numbr gratr than 0 and lss than 06 and B =,,, 6, find A B. OBJECTIVE Solving compound inqualitis containing and. A valu is a solution of a compound inqualit formd b th word and if it is a solution of both inqualitis. For ampl, th solution st of th compound inqualit and Ú contains all valus of that mak th inqualit a tru statmnt and th inqualit Ú a tru statmnt. Th first graph shown blow is th graph of, th scond graph is th graph of Ú, and th third graph shows th intrsction of th two graphs.th third graph is th graph of and Ú. ƒ q, ] ƒ Ú 6 ƒ and Ú 6 also ƒ 6 (s blow) [, q [, ] Sinc Ú is th sam as, th compound inqualit and can b writtn in a mor compact form as. Th solution st ƒ 6 includs all numbrs that ar gratr than or qual to and at th sam tim lss than or qual to. In intrval notation, th st ƒ and Ú 6 or ƒ 6 is writtn as,.

14 _t.pdf /9/ : AM ( ) F Supplmntal Appndi Hlpful Hint Don t forgt that som compound inqualitis containing and can b writtn in a mor compact form. Compound Inqualit Compact Form Intrval Notation and 6 Graph: , 6 EXAMPLE Solv: and Solution First w solv ach inqualit sparatl. Now w can graph th two intrvals on two numbr lins and find thir intrsction. Thir intrsction is shown on th third numbr lin. ƒ 6 96 ƒ 6 6 ƒ 6 9 and 6 6 = ƒ 6 6 Th solution st is - q, and and and q, 9 - q, - q, Solv: and - 6. Writ th solution st in intrval notation. EXAMPLE Solv: Ú 0 and Solution First w solv ach inqualit sparatl. Now w can graph th two intrvals on numbr lins and find thir intrsction. ƒ Ú 06 ƒ -6 ƒ Ú 0 and -6 = Ú 0 and - -9 Ú 0 and -8 Ú 0 and [0, q - q, -] Thr is no numbr that is gratr than or qual to 0 and lss than or qual to -. Th solution st is. Solv: 0 and Writ th solution st in intrval notation. Hlpful Hint Eampl shows that som compound inqualitis hav no solution. Also, som hav all ral numbrs as solutions.

15 _t.pdf /9/ : AM ( ) Compound Inqualitis F To solv a compound inqualit writtn in a compact form, such as 6-6 7, w gt alon in th middl part. Sinc a compound inqualit is rall two inqualitis in on statmnt, w must prform th sam oprations on all thr parts of th inqualit. EXAMPLE Solv: Hlpful Hint Don t forgt to rvrs both inqualit smbols. Solution To gt alon, w first subtract from all thr parts Subtract from all thr parts. Simplif. Divid all thr parts b - and rvrs th inqualit smbols. This is quivalnt to Th solution st in intrval notation is -,, and its numbr-lin graph is shown. 0 Solv: Writ th solution st in intrval notation. DISCOVER THE CONCEPT For th compound inqualit 6-6 7, in Eampl, graph =, = -, and = 7. With this notation, w can think of our inqualit as 6 6. a. Find th point of intrsction of and. b. Find th point of intrsction of and. c. Dtrmin whr th graph of is btwn th graphs of and. d. Find th -valus for which Writ th solution st of 6 6. In th discovr abov, w find th points of intrsction as shown Th solution of 6 6, or contains th -valus whr th graph of = - is btwn = and = 7. Ths -valus in intrval notation ar -,. (Parnthss ar usd bcaus of th inqualit smbols 6.)

16 _t.pdf /9/ : AM ( ) F Supplmntal Appndi EXAMPLE Solv algbraicall: - Chck graphicall. +. Solution First, clar th inqualit of fractions b multipling all thr parts b th LCD of a + b Multipl all thr parts b th LCD of. Us th distributiv proprt and multipl. Subtract from all thr parts. Simplif. Divid all thr parts b. Simplif. Th numbr-lin graph of th solution is shown. t Th solution st in intrval notation is c -9, - 9 d. To chck, graph = -, = and = in a [-,, ] b [-,, ] window and solv +,. Th solution of consists of th -valus whr th graph of is btwn or qual to th graphs of and. Th solution st in intrval notation is [-9, -.] or c -9, - 9 d. -. Chck graphicall and writ th solu- - Solv algbraicall: tion st in intrval notation. OBJECTIVE Finding th union of two sts. Th solution st of a compound inqualit formd b th word or is th union of th solution sts of th two inqualitis. W us th smbol to dnot union. Hlpful Hint Th word ithr in this dfinition mans on or th othr or both. Union of Two Sts Th union of two sts, A and B, is th st of lmnts that blong to ithr of th sts. A union B is dnotd b A B A B

17 _t.pdf /9/ : AM ( ) Compound Inqualitis F EXAMPLE 6 If A = ƒ is an vn numbr gratr than 0 and lss than 06 and B =,,, 66, find A B. Solution Rcall from Eampl that A =,, 6, 86. Th numbrs that ar in ithr st or both sts ar,,,, 6, 86. This st is th union. 6 If A = ƒ is an odd numbr gratr than 0 and lss than 06 and B =,,,, 66. Find A B. OBJECTIVE Solving compound inqualitis containing or. A valu is a solution of a compound inqualit formd b th word or if it is a solution of ithr inqualit. For ampl, th solution st of th compound inqualit or Ú contains all numbrs that mak th inqualit a tru statmnt or th inqualit Ú a tru statmnt. ƒ 6 ƒ Ú 6 ƒ or Ú q, ] [, q - q, ] [, q In intrval notation, th st ƒ or Ú 6 is writtn as - q, ] [, q. EXAMPLE 7 Solv: - 0 or + Ú. Solution First w solv ach inqualit sparatl. - 0 or + Ú or Ú or Ú Now w can graph ach intrval on a numbr lin and find thir union. ƒ f ƒ Ú 6 ƒ or Ú f Th solution st is { a - q, d [, q. { 0 6 a - q, d [, q a - q, d [, q 7 Solv: or - Ú. Writ th solution st in intrval notation. EXAMPLE 8 Solv: or Solution First w solv ach inqualit sparatl or or or 6 0

18 _t.pdf /9/ : AM ( ) F6 Supplmntal Appndi Now w can graph ach intrval on a numbr lin and find thir union. ƒ 7-6 -, q 0 ƒ q, 0 0 ƒ 7 - or q, q 0 = all ral numbrs Th solution st is - q, q. 8 Solv: or 7 0. Writ th solution st in intrval notation. Answr to Concpt Chck: b is not corrct Concpt Chck Which of th following is not a corrct wa to rprsnt th st of all numbrs btwn - and? a. ƒ b. - 6 or 6 c. (-, ) d. 7 - and 6 VOCABULARY & READINESS CHECK Us th choics blow to fill in ach blank. Som choics ma b usd mor than onc. or and compound. Two inqualitis joind b th words and or or ar calld inqualitis.. Th word mans intrsction.. Th word mans union.. Th smbol rprsnts intrsction.. Th smbol rprsnts union. 6. Th smbol is th mpt st. 7. Th inqualit - 6 mans ƒ 6 0 and 7 06 = EXERCISE SET MIXED If A = ƒ is an vn intgr6, B = ƒ is an odd intgr6, C =,,, 6, and D =,, 6, 76, list th lmnts of ach st. S Eampls and 6.. C D. C D. A D. A D. A B 6. A B 7. B D 8. B D 9. B C 0. B C. A C. A C Solv ach compound inqualit. Graph th solution st on a numbr lin and writ it in intrval notation. S Eampls and.. 6 and and Ú -. - and Ú and and 6 8. Ú- and 7 Solv ach compound inqualit. Writ solutions in intrval notation. S Eampls and Ú 7 and - Ú 0. + Ú and - Ú and and 7 0

19 _t.pdf /9/ : AM ( ) Compound Inqualitis F and or and and + Ú Solv ach compound inqualit. S Eampls and or Solv ach compound inqualit. Graph th solution st on a numbr lin and writ it in intrval notation. S Eampls 7 and or 6. Ú - or. - or Ú or or 6 8. Ú - or - Solv ach compound inqualit. Writ solutions in intrval notation. S Eampls 7 and or - 0 Ú or - Ú or or 7 -. ( - ) 6 or ( - ) Ú - or - Solv ach compound inqualit using th graphing utilit scrns. 7. a. b. 0 (, ) 0 (., 0) 0 7. a. b or or Ú (9, 8) (, 6) 7. a. (., ) 0 0 (., ) 76. a b. 6 or 7 b. or Ú 0 ( 6., ) (0, 8) MIXED Solv ach compound inqualit. Writ solutions in intrval notation. S Eampls through and and or or Ú and +. - Ú and - 7. Ú or or ( - ) ( + ) and and or REVIEW AND PREVIEW ƒ ƒ ƒ ƒ ƒ Evaluat th following. S Sctions. and ƒ - ƒ9ƒ ƒ ƒ 80. -ƒ ƒ Find b inspction all valus for that mak ach quation tru. 8. ƒƒ = 7 8. ƒƒ = 8. ƒƒ = 0 8. ƒƒ = - CONCEPT EXTENSIONS Us th graph to answr Erciss 8 and 86. Gallons U.S. Annual Consumption of Bottld Watr or Dit Soda pr Prson Bottld watr Dit soda Yar

20 _t.pdf /9/ : AM ( ) F8 Supplmntal Appndi 8. For what ars was th consumption of bottld watr gratr than 0 gallons pr prson and th consumption of dit soda gratr than gallons pr prson? 86. For what ars was th consumption of bottld watr lss than gallons pr prson or th consumption of dit soda gratr than gallons pr prson? Th formula for convrting Fahrnhit tmpraturs to Clsius tmpraturs is C = F -. Us this formula for Erciss 87 9 and During a rcnt ar, th tmpraturs in Chicago rangd from -9 to C. Us a compound inqualit to convrt ths tmpraturs to Fahrnhit tmpraturs. 88. In Oslo, th avrag tmpratur rangs from -0 to 8 Clsius. Us a compound inqualit to convrt ths tmpraturs to th Fahrnhit scal. Solv. 89. Christian D Anglo has scors of 68, 6, 7, and 78 on his algbra tsts. Us a compound inqualit to find th scors h can mak on his final am to rciv a C in th cours. Th final am counts as two tsts, and a C is rcivd if th final cours avrag is from 70 to Wnd Wood has scors of 80, 90, 8, and 7 on hr chmistr tsts. Us a compound inqualit to find th rang of scors sh can mak on hr final am to rciv a B in th cours. Th final am counts as two tsts, and a B is rcivd if th final cours avrag is from 80 to 89. Solv ach compound inqualit for. S th ampl blow. To solv , notic that this inqualit contains a variabl not onl in th middl, but also on th lft and th right. Whn this occurs, w solv b rwriting th inqualit using th word and and and and , or -, THE BIGGER PICTURE SOLVING EQUATIONS AND INEQUALITIES W now continu th outlin from Sctions. and.. Solv. Writ inqualit solutions in intrval notation. Although suggstions will b givn, this outlin should b in. - and - Ú - our own words. Onc ou complt this nw portion, tr th rciss blow Solving Equations and Inqualitis = I. Equations A. Linar quations (Sctions. and.). - or or - 7 II. Inqualitis t 7. A. Linar Inqualitis (Sction.) - t = 7 B. Compound Inqualitis: Two inqualit signs or 8. ( - ) + + Ú ( + ) + inqualitis sparatd b and or or. Or mans union and and mans intrsction. (Sction.) and 6-7 or (- q, -7) and (intrsction) 0 (- q, or (union)

21 _t.pdf /9/ : AM ( ) Solving Sstms of Linar Equations in Two Variabls F9 SOLVING SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES OBJECTIVES Dtrmin whthr an ordrd pair is a solution of a sstm of two linar quations. Solv a sstm b graphing. An important problm that oftn occurs in th filds of businss and conomics concrns th concpts of rvnu and cost. For ampl, suppos that a small manufacturing compan bgins to manufactur and sll compact disc storag units. Th rvnu of a compan is th compan s incom from slling ths units, and th cost is th amount of mon that a compan spnds to manufactur ths units. Th following coordinat sstm shows th graphs of rvnu and cost for th storag units. Solv a sstm b substitution. 0 Solv a sstm b limination. Dollars (in thousands) Brak-vn point (00,,000) 0 Cost 0 Rvnu Numbr of Storag Units Ths lins intrsct at th point (00,,000). This mans that whn 00 storag units ar manufacturd and sold, both cost and rvnu ar $,000. In businss, this point of intrsction is calld th brak-vn point. Notic that for -valus (units sold) lss than 00, th cost graph is abov th rvnu graph, maning that cost of manufacturing is gratr than rvnu, and so th compan is losing mon. For -valus (units sold) gratr than 00, th rvnu graph is abov th cost graph, maning that rvnu is gratr than cost, and so th compan is making mon. Rcall from Chaptr that ach lin is a graph of som linar quation in two variabls. Both quations togthr form a sstm of quations. Th common point of intrsction is calld th solution of th sstm. Som ampls of sstms of linar quations in two variabls ar Sstms of Linar Equations in Two Variabls - = = 0 c = + = 9 - = + 6 = OBJECTIVE Dtrmining whthr an ordrd pair is a solution. Rcall that a solution of an quation in two variabls is an ordrd pair (, ) that maks th quation tru. A solution of a sstm of two quations in two variabls is an ordrd pair (, ) that maks both quations tru. EXAMPLE Dtrmin whthr th givn ordrd pair is a solution of th sstm. a. - + = - = - -, b. -, + = - - = Solution a. W rplac with - and with in ach quation. - + =? =? + = = - = - Scond quation? Lt = - and =. - - = - Lt = - and =. First quation? Tru - - = - - = - Tru

22 _t.pdf /9/ : AM ( ) F0 Supplmntal Appndi Lft sids of th quations of th sstm Right sids of th quations of th sstm A calculator chck for Eampl a. Sinc -, maks both quations tru, it is a solution. Using st notation, th solution st is {-, }. b. W rplac with - and with in ach quation. + = - First quation - + =? - Lt = - and = =? - - = - Tru - = Scond quation - - =? Lt = - and =. - = Fals Lft sids of th quations of th sstm Right sid of first quation of th sstm Not right sid of scond quation of th sstm Sinc th ordrd pair -, dos not mak both quations tru, it is not a solution of th sstm. Dtrmin whthr th givn ordrd pair is a solution of th sstm. - - = a. (, -) + = A calculator chck for Eampl b. + = - b. (-, ) - + = 8 Eampl abov shows how to dtrmin that an ordrd pair is a solution of a sstm of quations, but how do w find such a solution? Actuall, thr ar various mthods to find th solution. W will invstigat svral in this chaptr: graphing, substitution, limination, matrics, and dtrminants. OBJECTIVE Solving a sstm b graphing. To solv b graphing, w graph ach quation in an appropriat window and find th coordinats of an points of intrsction. EXAMPLE Solv th sstm b graphing. + = - = - Solution Sinc th graph of a linar quation in two variabls is a lin, graphing two such quations ilds two lins in a plan.to us a graphing utilit, solv ach quation for. First quation = - + = + Scond quation Graph = - + and = + and find th point of intrsction.

23 _t.pdf /9/ : AM ( ) Solving Sstms of Linar Equations in Two Variabls F TECHNOLOGY NOTE Whn solving a sstm of quations b graphing, tr th standard or intgr window first and s if th intrsction appars. If it dos not appar in on of ths windows, ou ma b abl to s nough of th graph to stimat whr th intrsction will occur and adjust th window stting accordingl A chck of th solution: (0, ). Vrif th ordrd pair solution (0, ) b rplacing with 0 and with in both original quations and sing that tru statmnts rsult ach tim. Th scrn on th right abov shows that th ordrd pair (0, ) dos satisf both quations. W conclud thrfor that (0, ) is th solution of th sstm. A sstm that has at last on solution, such as this on, is said to b consistnt. Solv ach sstm b graphing. If th sstm has just on solution, find th solution. - = = = a. b. c = - + = 7 L - = DISCOVER THE CONCEPT Us our graphing utilit to solv th sstm - = = In th discovr abov, w s that solving ach quation for producs th following: - = - = - + = - First quation Subtract from both sids. Divid both sids b -. = = = Scond quation Divid both sids b. 0 q q Notic that ach quation is now in th form = m + b. From this form, w s that both lins hav th sam slop,, but diffrnt -intrcpts, so th ar paralll as shown to th lft. Thrfor, th sstm has no solution sinc th quations hav no common solution (thr ar no intrsction points). A sstm that has no solution is said to b inconsistnt. DISCOVER THE CONCEPT Us our graphing utilit to solv th sstm + = 0 + = In th discovr abov, w s that solving ach quation for producs th following: + = 0 First quation + = Scond quation = - + = - +

24 _t.pdf /9/ : AM ( ) F Supplmntal Appndi Hlpful Hint If a sstm of quations has at last on solution, th sstm is consistnt. If a sstm of quations has no solution, th sstm is inconsistnt. Notic that both lins hav th sam slop, - and th sam -intrcpt, This,. mans that th graph of ach quation is th sam lin Hlpful Hint If th graphs of two quations diffr, th ar indpndnt quations. If th graphs of two quations ar th sam, th ar dpndnt quations. To confirm this, notic that th ntris for and ar th sam in th tabl shown on th right abov. Th quations hav idntical solutions and an ordrd pair solution of on quation satisfis th othr quation also. Thus, ths quations ar said to b dpndnt quations. Th solution st of th sstm is, ƒ + = 6 or, quivalntl,, ƒ + = 06 sinc th quations dscrib idntical ordrd pairs. Writtn this wa, th solution st is rad th st of all ordrd pairs (, ), such that + = 0. Thr ar thrfor an infinit numbr of solutions to th sstm. Concpt Chck Th quations in th sstm ar dpndnt and th sstm has an infinit numbr of solutions. Which ordrd pairs blow ar solutions? - + = + 8 = 6 a. (, 0) b. -, 0 c. -, W can summariz th information discovrd in Eampl as follows. On solution: consistnt sstm; indpndnt quations No solution: inconsistnt sstm; indpndnt quations Infinit numbr of solutions: consistnt sstm; dpndnt quations Concpt Chck How can ou tll just b looking at th following sstm that it has no solution? = + = - 7 How can ou tll just b looking at th following sstm that it has infinitl man solutions? + = + = 0 Answr to Concpt Chck: b, c; answrs ma var A graphing calculator is a vr usful tool for approimating solutions to a sstm of quations in two variabls. S th nt ampl.

25 _t.pdf /9/ : AM ( ) Solving Sstms of Linar Equations in Two Variabls F..9 EXAMPLE dcimal placs. Solv th sstm b graphing. Approimat th solution to two +.6 =.6 -. = -.9 Solution First us a standard window and graph both quations on a singl scrn. Th scrn in th margin shows that th two lins intrsct. To approimat th point of intrsction, trac to th point of intrsction and us an Intrsct fatur of th graphing calculator. W find that th approimat point of intrsction is (.,.6). Bcaus th solution is an approimation, notic that th numrical chck with ths approimations dos not show quivalnt prssions. For ampl, instad of +.6 =.6, w hav +.6 =.9. Th numbr.9 is clos to.6, but not qual to.6. Kp this in mind whn chcking approimations..6.6 Solving graphicall. - and -valus ar automaticall stord to dcimal placs. Solv b graphing. Approimat th solution to two dcimal placs = = -. Th numrical chck with dcimal approimations. OBJECTIVE Solving a sstm b substitution. Graphing th quations of a sstm b hand is oftn a good mthod of finding approimat solutions of a sstm, but it is not a rliabl mthod of finding act solutions of a sstm. W turn instad to two algbraic mthods of solving sstms. W us th first mthod, th substitution mthod, to solv th sstm + = -6 = - First quation Scond quation EXAMPLE Us th substitution mthod to solv th sstm. + = -6 = - First quation Scond quation Solution In th scond quation, w ar told that is qual to -. Sinc th ar qual, w can substitut - for in th first quation. This will giv us an quation in on variabl, which w can solv for. + = = = -6 8 = First quation Substitut - for. = 8 = Solv for. A numric chck of th solution a -, for b Eampl. Th -coordinat of th solution is To find th -coordinat, w rplac with in. th scond quation, = -. = - = a b - = - = - Th ordrd pair solution is a -, Chck to s that a -, satisfis both quations of th sstm. b. b

26 _t.pdf /9/ : AM ( ) F Supplmntal Appndi Us th substitution mthod to solv th sstm. = = Th stps blow summariz th substitution mthod. Solving a Sstm of Two Equations Using th Substitution Mthod STEP. Solv on of th quations for on of its variabls. STEP. Substitut th prssion for th variabl found in Stp into th othr quation. STEP. Find th valu of on variabl b solving th quation from Stp. STEP. Find th valu of th othr variabl b substituting th valu found in Stp into th quation from Stp. STEP. Chck th ordrd pair solution in both original quations. Hlpful Hint If a sstm of quations contains quations with fractions, first clar th quations of fractions. EXAMPLE Us th substitution mthod to solv th sstm = d - = - 6 Solution First w multipl ach quation b its last common dnominator to clar th sstm of fractions. W multipl th first quation b 6 and th scond quation b. Hlpful Hint To avoid tdious fractions, solv for a variabl whos cofficint is or -, if possibl. 6a b d a - 6 b = 6a b = a - b simplifis to To us th substitution mthod, w now solv th first quation for. - + = - = First quation Solv for. Nt w rplac with - in th scond quation. - = = = -9 0 = = = - = -9 Scond quation Solv for. First quation Scond quation

27 _t.pdf /9/ : AM ( ) Solving Sstms of Linar Equations in Two Variabls F To find th corrsponding -coordinat, w rplac with -. Thn 0 = a 0 b - = = =- 0 in th quation Th ordrd pair solution is a- or quivalntl (-., 0.). W chck this 0, 0 b solution graphicall to th lft. = Us th substitution mthod to solv th sstm. - + = μ - = - OBJECTIVE Solving a sstm b limination. Th limination mthod, or addition mthod, is a scond algbraic tchniqu for solving sstms of quations. For this mthod, w rl on a vrsion of th addition proprt of qualit, which stats that quals addd to quals ar qual. If A = B and C = D thn A + C = B + D. EXAMPLE 6 Us th limination mthod to solv th sstm. - = = First quation Scond quation Solution Sinc th lft sid of ach quation is qual to th right sid, w add qual quantitis b adding th lft sids of th quations and th right sids of th quations. This sum givs us an quation in on variabl,, which w can solv for. - = = - = -8 = First quation Scond quation Add. Solv for. Satisfis Th -coordinat of th solution is. To find th corrsponding -coordinat, w rplac with in ithr original quation of th sstm. Lt s us th scond quation. - + = - + = - = = - Scond quation Lt =. Th ordrd pair solution is -,. W chck numricall (to th lft) to s that -, satisfis both quations of th sstm. Satisfis 6 Us th limination mthod to solv th sstm. - = + =

28 _t.pdf /9/ : AM ( ) F6 Supplmntal Appndi Th stps blow summariz th limination mthod. Solving a Sstm of Two Linar Equations Using th Elimination Mthod STEP. Rwrit ach quation in standard form, A + B = C. STEP. If ncssar, multipl on or both quations b som nonzro numbr so that th cofficints of a variabl ar opposits of ach othr. STEP. Add th quations. STEP. Find th valu of on variabl b solving th quation from Stp. STEP. Find th valu of th scond variabl b substituting th valu found in Stp into ithr original quation. STEP 6. Chck th proposd ordrd pair solution in both original quations. EXAMPLE 7 Us th limination mthod to solv th sstm. - = 0 - = Solution If w add th two quations, th sum will still b an quation in two variabls. Notic, howvr, that w can liminat whn th quations ar addd if w multipl both sids of th first quation b and both sids of th scond quation b -. Thn - = = 0 simplifis to - - = = -0 Nt w add th lft sids and add th right sids. 9-6 = = -0 = 0 To find, w lt = 0 in ithr quation of th sstm. - = = 0 - = 0 = - First quation Lt = 0. Th ordrd pair solution is 0, -. Chck to s that 0, - satisfis both quations of th sstm. 7 Us th limination mthod to solv th sstm. - = -6 + = -8

29 _t.pdf /9/ : AM ( ) Solving Sstms of Linar Equations in Two Variabls F7 EXAMPLE 8 Us th limination mthod to solv th sstm. + = L 6 + = Solution If w multipl both sids of th first quation b -, th cofficints of in th two quations will b opposits. Thn 6 ( ) Th two graphs appar to b paralll lins, supporting no solution to th sstm of Eampl 8. Now w can add th lft sids and add th right sids = = 0 = Fals Th rsulting quation, 0 =, is fals for all valus of or. Thus, th sstm has no solution. Th solution st is 6 or. This sstm is inconsistnt, and th graphs of th quations ar paralll lins. 8 -a + b = - L simplifis to 6 + = Us th limination mthod to solv th sstm. 8 + = = = L + = - EXAMPLE 9 Us th limination mthod to solv th sstm. - - = = -8 Solution To liminat whn th quations ar addd, w multipl both sids of th first quation b. Thn - - = = 8 simplifis to = = -8 Nt w add th quations = = -8 0 = 0 Th rsulting quation, 0 = 0, is tru for all possibl valus of or. Notic in th original sstm that if both sids of th first quation ar multiplid b -, th rsult is th scond quation. This mans that th two quations ar quivalnt. Th hav th sam solution st and thr ar an infinit numbr of solutions. Thus, th quations of this sstm ar dpndnt, and th solution st of th sstm is, ƒ - - = 96 or, quivalntl,, ƒ0 + 6 = -86. Th graph (shown on a standard window) and tabl indicat that th graph of both quations is th sam lin. This supports th solution abov for Eampl 9.

30 _t.pdf /9/ : AM ( ) F8 Supplmntal Appndi 9 Us th limination mthod to solv th sstm. - + = = Hlpful Hint Rmmbr that not all ordrd pairs ar solutions of th sstm in Eampl 9, onl th infinit numbr of ordrd pairs that satisf - - = 9 or quivalntl = -8. EXAMPLE 0 Finding th Brak-Evn Point A small manufacturing compan manufacturs and slls compact disc storag units. Th rvnu quation for ths units is = 0 whr is th numbr of units sold and is th rvnu, or incom, in dollars for slling units. Th cost quation for ths units is = 0 + 0,000 whr is th numbr of units manufacturd and is th total cost in dollars for manufacturing units. Us ths quations to find th numbr of units to b sold for th compan to brak vn. Solution Th brak-vn point is found b solving th sstm = 0 = 0 + 0,000 First quation Scond quation To solv th sstm, graph = 0 and = 0 + 0,000 and find th point of intrsction, th brak-vn point. 0, ,000 0, Th ordrd pair solution is (00,,000). This mans that th businss must sll 00 compact disc storag units to brak vn. A hand-drawn graph of th quations in this sstm can b found at th bginning of this sction. 0 Th rvnu quation for a crtain product is = 7, whr is th numbr of units sold and is th rvnu in dollars. Th cost quation for th product is = , whr is th numbr of units manufacturd and is th cost in dollars for manufacturing units. Find th numbr of units for th compan to brak vn.

31 _t.pdf /9/ : AM ( ) Solving Sstms of Linar Equations in Two Variabls F9 VOCABULARY & READINESS CHECK Match ach graph with th solution of th corrsponding sstm. A B C D. no solution. Infinit numbr of solutions. (, -). (-, 0) EXERCISE SET Dtrmin whthr ach givn ordrd pair is a solution of ach sstm. S Eampl A sstm of quations and th graph of ach quation of th sstm is givn blow. Find th solution of th sstm and vrif that it is th solution. S Eampl =, - - = 8 - = - -, + 0 = - = -9, + = - - = -, + = + 7 = -9-6 = = -7-8 = - + = = 0 a, -b a, -b = 8 - = Solv ach sstm b graphing. S Eampls and.. + =. - = 8 - = + = - = =. - = = - = 6 - = = 6-9 = 9 7. Can a sstm consisting of two linar quations hav actl two solutions? Eplain wh or wh not. 8. Suppos th graph of th quations in a sstm of two quations in two variabls consists of a circl and a lin. Discuss th possibl numbr of solutions for this sstm = - = Solv ach sstm of quations b th substitution mthod. S Eampls and. + = = 0 0. = = 9. - = = = 9. + = -7 - = = =- = -. d. d + - = =-8 0

32 _t.pdf /9/ : AM ( ) F0 Supplmntal Appndi = =. d 6. d - + = - = Solv ach sstm of quations b th limination mthod. S Eampls 6 through = = = + 6 = + = = = =-. d. d = - = - - = =. - = 6. - = 6 - = - - = = - + = = 6-9 = 9 - = = - 7. = 6 8. = - + = 6-9. d = 0-7 = = - = - 6. d = = =-. = L - = - = + - = d d = 8 - = = = 0 - = = = = = 8. = - 6 = = 0 -. = = = =.96 MIXED Solv ach sstm of quations = = = - =.. + = d - - = = = = + = = = = = 0 - = 8 + = - =- d + =- = = = + = + = 0 7 = REVIEW AND PREVIEW Dtrmin whthr th givn rplacmnt valus mak ach quation tru or fals. S Sction z = ; =, =, and z = z = 7; =, = -, and z = z = ; = 0, = -, and z = z = ; =, = 0, and z = - Add th quations. S Sction z = z = z = z = z = z = z = z = - = = 8 + = + = - = = L - = 0-7 = - CONCEPT EXTENSIONS Th concpt of suppl and dmand is usd oftn in businss. In gnral, as th unit pric of a commodit incrass, th dmand for that commodit dcrass. Also, as a commodit s unit pric incrass, th manufacturr normall incrass th suppl. Th point whr suppl is qual to dmand is calld th quilibrium point. Th following shows th graph of a dmand quation and th

33 _t.pdf /9/ : AM ( ) Solving Sstms of Linar Equations in Two Variabls F graph of a suppl quation for prviousl rntd DVDs. Th -ais rprsnts th numbr of DVDs in thousands, and th -ais rprsnts th cost of a DVD. Us this graph to answr Erciss 77 through 80. S Eampl 0. Cost of a Prviousl Rntd DVD (in dollars) 77. Find th numbr of DVDs and th pric pr DVD whn suppl quals dmand. 78. Whn is btwn and, is suppl gratr than dmand or is dmand gratr than suppl? 79. Whn is gratr than 7, is suppl gratr than dmand or is dmand gratr than suppl? 80. For what -valus ar th -valus corrsponding to th suppl quation gratr than th -valus corrsponding to th dmand quation? Th rvnu quation for a crtain brand of toothpast is =., whr is th numbr of tubs of toothpast sold and is th total incom for slling tubs. Th cost quation is = , whr is th numbr of tubs of toothpast manufacturd and is th cost of producing tubs.th following st of as shows th graph of th cost and rvnu quations. Us this graph for Erciss 8 through 86. S Eampl 0. Dollars Cost Dmand Suppl 6 Numbr of DVDs (in thousands) Rvnu Tubs of Toothpast 8. Find th coordinats of th point of intrsction, or brakvn point, b solving th sstm =. = Equilibrium point Eplain th maning of th ordrd pair point of intrsction If th compan slls 000 tubs of toothpast, dos th compan mak mon or los mon? 8. If th compan slls 000 tubs of toothpast, dos th compan mak mon or los mon? 8. For what -valus will th compan mak a profit? (Hint: For what -valus is th rvnu graph highr than th cost graph?) 86. For what -valus will th compan los mon? (Hint: For what -valus is th rvnu graph lowr than th cost graph?) 87. Writ a sstm of two linar quations in and that has th ordrd pair solution (, ). 88. Which mthod would ou us to solv th sstm? Eplain our choic. - = 6 + = 89. Th amount of rd mat consumd pr prson in th Unitd Stats (in pounds) in th ar can b modld b th linar quation = Th amount of all poultr consumd pr prson in th Unitd Stats (in pounds) in th ar can b modld b th linar quation = In both modls, = 0 rprsnts th ar 000. (Sourc: Basd on data and forcasts from th Economic Rsarch Srvic, U.S. Dpartmnt of Agricultur) a. What dos th slop of ach quation tll ou about th pattrns of rd mat and poultr consumption in th Unitd Stats? b. Solv this sstm of quations. (Round our final rsults to th narst whol numbrs.) c. Eplain th maning of our answr to part b. 90. Th numbr of books (in thousands) in th Univrsit of Tas libraris for th ars 00 through 00 can b modld b th linar quation = For th sam tim priod, th numbr of books (in thousands) in th Columbia Univrsit libraris can b modld b = , whr is th numbr of ars sinc 00.(Sourc: Association of Rsarch Libraris) a. What dos th slop of ach quation tll ou about th pattrn of books in ths two univrsit libraris? b. Solv this sstm of quations. (Round our rsults to th narst whol numbr.) c. Eplain th maning of our answr to part b. Solv ach sstm. To do so ou ma want to lt a = (if is in th dnominator) and lt b = (if is in th dnominator). 9. d + = - = + = 7 9. d + = 6