CHAPTER 4 Sstms of Equations 4. Solving Sstms of Linar Equations in Two Variabls 4. Solving Sstms of Linar Equations in Thr Variabls 4. Sstms of Linar Equations and Problm Solving Intgratd Rviw Sstms of Linar Equations Although th numbr of cll phon calls is slowl dcrasing, for th past fw ars, th numbr of tt mssags has incrasd significantl. In fact, tnagrs ag 7 snd about 00 mssags pr month whil thos ag 8 4 snd onl about 600 pr month. In Sction 4., Erciss 57 and 58, w will us th functions graphd blow to solv th sstm of quations. Calls or Tts pr Month 000 800 600 400 00 Avrag Numbr of Monthl Calls vs. Tt Mssags pr Wirlss Subscribr Tt Mssags (snt or rcivd) f() 04.9 7 Calls (snt or rcivd) f() 8.6 75 4.4 Solving Sstms of Equations b Matrics 4.5 Sstms of Linar Inqualitis In this chaptr, two or mor quations in two or mor variabls ar solvd simultanousl. Such a collction of quations is calld a sstm of quations. Sstms of quations ar good mathmatical modls for man ral-world problms bcaus ths problms ma involv svral rlatd pattrns. 0 006 007 008 Yar 009 00 0
0 CHAPTER 4 Sstms of Equations 4. Solving Sstms of Linar Equations in Two Variabls S Dtrmin Whthr an Ordrd Pair Is a Solution of a Sstm of Two Linar Equations. Solv a Sstm b Graphing. Solv a Sstm b Substitution. 4 Solv a Sstm b Elimination. 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. Dollars (in thousands) 0 5 0 5 0 5 Cost Rvnu Brak-vn point (500, 5) 0 0 00 00 00 400 500 600 Numbr of Storag Units Ths lins intrsct at th point 500, 5. This mans that whn 500 storag units ar manufacturd and sold, both cost and rvnu ar $5,000. In businss, this point of intrsction is calld th brak-vn point. Notic that for -valus (units sold) lss than 500, 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 500, 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 - = -7 = 5 b + = 0 + = 9 - = + 6 b = 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 a. b - + = - = - Dtrmin whthr th givn ordrd pair is a solution of th sstm. -, b. b 5 + = - - = Solution a. W rplac with - and with in ach quation. -, - + = First quation - - + Lt = - and =. + = Tru - = - Scond quation - - - Lt = - and =. - - - - = - Tru
Sction 4. Solving Sstms of Linar Equations in Two Variabls 0 Sinc -, maks both quations tru, it is a solution. Using st notation, th solution st is 5 -, 6. b. W rplac with - and with in ach quation. 5 + = - First quation 5 - + - Lt = - and =. -0 + 9 - - = - Tru - = Scond quation - - Lt = - and =. -5 = Fals 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. - - 4 = a. b + = 5, - b. b 4 + = -4 - + = 8 -, 4 Solving a Sstm b Graphing W can stimat th solutions of a sstm b graphing ach quation on th sam coordinat sstm and stimating th coordinats of an point of intrsction. EXAMPLE Solv th sstm b graphing. + = - = - Solution: First w graph th linar quations on th sam rctangular coordinat sstm. Ths lins intrsct at on point as shown. Th coordinats of th point of intrsction appar to b 0,. W chck this stimatd solution b rplacing with 0 and with in both quations. 5 4 5 4 4 5 (0, ) 4 5 Common solution + = First quation 0 + Lt = 0 and =. = Tru - = - Scond quation 0 - - Lt = 0 and =. - = - Tru Th ordrd pair 0, is th solution of th sstm. A sstm that has at last on solution, such as this on, is said to b consistnt. = 5 Solv th sstm b graphing. + = 7 f Hlpful Hint Rading valus from graphs ma not b accurat. Until a proposd solution is chckd in both quations of th sstm, w can onl assum that w hav stimatd a solution.
04 CHAPTER 4 Sstms of Equations In Eampl, w hav a consistnt sstm. To rviw, a sstm that has at last on solution is said to b consistnt. Latr, w will talk about dpndnt quations. For now, w dfin an indpndnt quation to b an quation in a sstm of quations that cannot b algbraicall drivd from an othr quation in th sstm. Thus for: Consistnt Sstm: Th sstm has at last on solution. On solution Indpndnt Equations: Each quation in th sstm cannot b algbraicall drivd from th othr. EXAMPLE Solv th sstm b graphing. - = 4 = Solution: W graph ach linar quation. Th lins appar to b paralll. To b sur, lt s writ ach quation in slop intrcpt form, = m + b. To do so, w solv for. - = 4 First quation - = - + 4 Subtract from both sids. = - Divid both sids b - = Scond quation = Divid both sids b. = 5 4 5 4 4 5 4 5 4 If paralll, sstm has no solution Th graphs of ths quations hav th sam slop,, but diffrnt -intrcpts, so ths lins ar paralll. 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. = Solv th sstm b graphing. 4 + - 4 = In Eampl, w hav an inconsistnt sstm. To rviw, a sstm that has no solution is said to b inconsistnt. Lt s now talk about th quations in this sstm. Each quation in this sstm cannot b algbraicall drivd from th othr, so ach quation is indpndnt of th othr. Thus: Inconsistnt Sstm: Th sstm has no solution. No solution Indpndnt Equations: Each quation in th sstm cannot b algbraicall drivd from th othr.
Sction 4. Solving Sstms of Linar Equations in Two Variabls 05 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. Th pairs of quations in Eampls and ar calld indpndnt bcaus thir graphs diffr. In Eampl 4, w s an ampl of dpndnt quations. EXAMPLE 4 Solv th sstm b graphing. + 4 = 0 + = 5 Solution: W graph ach linar quation. W s that th graphs of th quations ar th sam lin. To confirm this, notic that if both sids of th scond quation ar multiplid b, th rsult is th first quation. This mans that th quations hav idntical solutions. An ordrd pair solution of on quation satisfis th othr quation also. Ths quations ar said to b dpndnt quations. Th solution st of th sstm is 5, 0 + = 56 or, quivalntl, 5, 0 + 4 = 06 sinc th lins dscrib idntical ordrd pairs. Writtn th scond wa, th solution st is rad th st of all ordrd pairs,, such that + 4 = 0. Thr is an infinit numbr of solutions to this sstm. - = 4 4 Solv th sstm b graphing. -9 + 6 = - 5 4 4 5 4 0 4 5 6 7 5 In Eampl 4, w hav a consistnt sstm sinc th sstm has at last on solution. In fact, th sstm in Eampl 4 has an infinit numbr of solutions. Lt s now dfin dpndnt quations. W dfin a dpndnt quation to b an quation in a sstm of quations that can b algbraicall drivd from anothr quation in th sstm. Thus: Consistnt Sstm: Th sstm has at last on solution. Infinit numbr of solutions Dpndnt Equations: An quation in a sstm of quations can b algbraicall drivd from anothr. Answr to Concpt Chck: b, c 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. CONCEPT CHECK Th quations in th sstm ar dpndnt and th sstm has an infinit numbr of solutions. Which ordrd pairs blow ar solutions? - + = 4 + 8 = 6 a. 4, 0 b. -4, 0 c. -,
06 CHAPTER 4 Sstms of Equations W can summariz th information discovrd in Eampls through 4 as follows. Possibl Solutions to Sstms of Two Linar Equations On solution: consistnt sstm; indpndnt quations No solution: inconsistnt sstm; indpndnt quations Infinit numbr of solutions: consistnt sstm; dpndnt quations CONCEPT CHECK How can ou tll just b looking at th following sstm that it has no solution? = + 5 = - 7 How can ou tll just b looking at th following sstm that it has infinitl man solutions? + = 5 + = 0 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 + 4 = -6 First quation = - 5 Scond quation EXAMPLE 5 Us th substitution mthod to solv th sstm. + 4 = -6 First quation = - 5 Scond quation Solution In th scond quation, w ar told that is qual to - 5. Sinc th ar qual, w can substitut - 5 for in th first quation. This will giv us an quation in on variabl, which w can solv for. + 4 = -6 First quation $%& b - 5 + 4 = -6 Substitut - 5 for. 4-0 + 4 = -6 8 = 4 Answr to Concpt Chck: answrs ma var = 4 8 = Solv for.
Sction 4. Solving Sstms of Linar Equations in Two Variabls 07 Th -coordinat of th solution is. To find th -coordinat, w rplac with in th scond quation, = - 5. = - 5 = a b - 5 = - 5 = -4 Th ordrd pair solution is a -4, b. Chck to s that a -4, b satisfis both quations of th sstm. 5 Us th substitution mthod to solv th sstm. = 4 + 7 + = 4 Hlpful Hint If a sstm of quations contains quations with fractions, th first stp is to clar th quations of fractions. Solving a Sstm of Two Equations Using th Substitution Mthod Stp. Solv on of th quations for on of its variabls. Stp. Substitut th prssion for th variabl found in Stp into th othr quation. Stp. Find th valu of on variabl b solving th quation from Stp. Stp 4. Find th valu of th othr variabl b substituting th valu found in Stp into th quation from Stp. Stp 5. Chck th ordrd pair solution in both original quations. EXAMPLE 6 Us th substitution mthod to solv th sstm. - 6 + = μ - 6 = - 4 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 - 6 + b = 6a b μ a - 6 b = a - 4 b - + = simplifis to 4 - = -9 To us th substitution mthod, w now solv th first quation for. - + = First quation - = Solv for. Nt w rplac with - in th scond quation. 4 - = -9 Scond quation $%& b 4 - - = -9 - - = -9 0 = First quation Scond quation = 0 Solv for.
08 CHAPTER 4 Sstms of Equations To find th corrsponding -coordinat, w rplac with in th quation = -. Thn 0 = a 0 b - = 9 0 - = 9 0-0 0 = - 0 Th ordrd pair solution is a - 0, b. Chck to s that this solution satisfis both original quations. 0 6 Us th substitution mthod to solv th sstm. - + 4 = μ 4 - = - 4 4 Solving a Sstm b Elimination 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 7 Us th limination mthod to solv th sstm. - 5 = - First quation - + = 4 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. - 5 = - - + = 4-4 = -8 = First quation Scond quation Add Solv for. 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. - + = 4 Scond quation - + = 4 Lt =. - = = - Th ordrd pair solution is -,. Chck to s that -, satisfis both quations of th sstm. 7 Us th limination mthod to solv th sstm. - = 5 5 + =
Sction 4. Solving Sstms of Linar Equations in Two Variabls 09 Th stps blow summariz th limination mthod. Solving a Sstm of Two Linar Equations Using th Elimination Mthod Stp. Rwrit ach quation in standard form, A + B = C. Stp. If ncssar, multipl on or both quations b som nonzro numbr so that th cofficints of a variabl ar opposits of ach othr. Stp. Add th quations. Stp 4. Find th valu of on variabl b solving th quation from Stp. Stp 5. Find th valu of th scond variabl b substituting th valu found in Stp 4 into ithr original quation. Stp 6. Chck th proposd ordrd pair solution in both original quations. EXAMPLE 8 Us th limination mthod to solv th sstm. - = 0 4 - = 5 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 9-6 = 0 simplifis to -4 - = -5-8 + 6 = -0 Nt w add th lft sids and add th right sids. 9-6 = 0-8 + 6 = -0 = 0 To find, w lt = 0 in ithr quation of th sstm. - = 0 First quation 0 - = 0 Lt = 0. - = 0 = -5 Th ordrd pair solution is 0, -5. Chck to s that 0, -5 satisfis both quations of th sstm. 8 Us th limination mthod to solv th sstm. - = -6 4 + 5 = -8 EXAMPLE 9 Us th limination mthod to solv th sstm. + = 6 + = 5 Solution If w multipl both sids of th first quation b -, th cofficints of in th two quations will b opposits. Thn -a + b = - 6 + = 5 simplifis to -6 - = -4 6 + = 5
0 CHAPTER 4 Sstms of Equations Now w can add th lft sids and add th right sids. -6 - = -4 6 + = 5 0 = Fals Th rsulting quation, 0 =, is fals for all valus of or. Thus, th sstm has no solution. Th solution st is 5 6 or. This sstm is inconsistnt, and th graphs of th quations ar paralll lins. 9 Us th limination mthod to solv th sstm. 8 + = 6 + 4 = - EXAMPLE 0 Us th limination mthod to solv th sstm. -5 - = 9 0 + 6 = -8 Solution To liminat whn th quations ar addd, w multipl both sids of th first quation b. Thn -5 - = 9 0 + 6 = -8-0 - 6 = 8 simplifis to 0 + 6 = -8 Nt w add th quations. -0-6 = 8 0 + 6 = -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 is an infinit numbr of solutions. Thus, th quations of this sstm ar dpndnt, and th solution st of th sstm is 5, 0-5 - = 96 or, quivalntl, 5, 0 0 + 6 = -86. 0 Us th limination mthod to solv th sstm. - + = - 9-6 = Hlpful Hint Rmmbr that not all ordrd pairs ar solutions of th sstm in Eampl 0, onl th infinit numbr of ordrd pairs that satisf -5 - = 9 or quivalntl 0 + 6 = -8.
Sction 4. Solving Sstms of Linar Equations in Two Variabls Graphing Calculator Eplorations 0 4. 4.9 0 0 0.6 5.6 A graphing calculator ma b usd to approimat solutions of sstms of quations b graphing ach quation on th sam st of as and approimating an points of intrsction. For ampl, approimat th solution of th sstm = -.6 + 5.6 = 4. - 4.9 First us a standard window and graph both quations on a singl scrn. 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 or a Zoom In fatur. Using ithr mthod, w find that th approimat point of intrsction is (.5,.64). Solv ach sstm of quations. Approimat th solutions to two dcimal placs.. = -.65 +.65 = 4.56-9.44.. - 4.7 = 0.6 5.86 + 6. = -8.89. = 7.6 +.48 = -.6-6.4 4. -7.89-5.68 =.6 -.65 + 4.98 =.77 Vocabular, Radinss & Vido Chck Match ach graph with th solution of th corrsponding sstm. A B C D. no solution. Infinit numbr of solutions., - 4. -, 0 Martin-Ga Intractiv Vidos S Vido 4. Watch th sction lctur vido and answr th following qustions. 4 5. In Eampl, th ordrd pair is a solution of th first quation of th sstm. Wh is this not nough to dtrmin if th ordrd pair is a solution of th sstm? 6. From Eampl, what potntial drawbacks dos th graphing mthod hav? 7. Th sstm in Eampl nds on of its quations solvd for a variabl as a first stp. What important point is thn mphasizd? 8. Wh is Stp in Solving a Sstm b Elimination skippd in Eampl 5?
CHAPTER 4 Sstms of Equations 4. Ercis St Dtrmin whthr ach givn ordrd pair is a solution of ach sstm. S Eampl. - =, - - = -4 -,.. - 4 = 8 + 0 = 4 - = -9, 5. 4 + = - 5. = -5 = - -, 0 6. = 6 = - - 5 = - 4, 4. + 4 = 4 -, 6 + 7 = -9 7. a -6 = 5 + 8, -b 4 + 5 = -7 8. -8 = - a 4, -b Solv ach sstm b graphing. S Eampls, through 4. 9. + = - = 4-4 = 0. + = 5. - = 4 6 - = 4 0. - = 8 + =. 4 - = 6 - = 0 - + = 6 4. - 9 = 9 Solv ach sstm of quations b th substitution mthod. S Eampls 5 and 6. 5. + = 0 = 4 7. 4 - = 9 + = -7 9. μ. + 4 = - 4 4-4 = + = 4 - + = 5 + = -7 6. = 8. - = 6-4 + = -8 0. μ. μ 5 + 5 = - + 5 = - 8 5 8 - = - = Solv ach sstm of quations b th limination mthod. S Eampls 7 through 0. - + = 0. + = 5 5 + = 5. - = 7 7. μ 4 + 5 = 6-4 = - - 5 = 9. - 6 = - + = 0 4. + 6 = 6 - = -5 6. 4 - = 6 8. μ + 4 = - - 4 = - 6 - = - 0. 4 + 5 = -9. - = 4-4 = 4. + = = - 6 MIXED Solv ach sstm of quations. + 5 = 8 5. 6 + = 0 + = 7. - = 4 + = 4 9. μ - 4 - = - 4 + 6 = 8 4. + 9 = 4 + = 5 4. + = - 0 - = 45. 5 = 4-6 5 - = 7 47. - + 5 = 8 = + 49. 5-5 = 0 - = - 5. = - 5. = 6 = 5 - + 5 55. μ 5 = 6-4 = - 7 0 4-7 = 7 57. - = 4 - + = 6. - 9 = 9 = - 5 4. 8-4 = 0-4 = -5 6. - - 8 = 0 - + = -8 8. + = 4 - = - 40. μ + = - = - 4. - 6 = - + 6 = 5 44. + 4 = + 4 = 0 46. 7 = + 4 = 48. + 5 = - = 7 + 50. - 7 = - = 5 5. - = -4 54. = + 4 - = 5 5 = 8-56. μ = - 8-5 = 58. -4 + 0 = 0-4 = - - = - 59. μ - 6 + 60. μ 8 = 8 + 6 = 0 0.7-0. = -.6 6. 0. - = -.4 4 -.5 = 0. 6. + 7.8 = -5.68-0.7 + 0.6 =. 6. 0.5-0. = -0.8 - = -5. 64. 6. + 6 =.96
Sction 4. Solving Sstms of Linar Equations in Two Variabls REVIEW AND PREVIEW Dtrmin whthr th givn rplacmnt valus mak ach quation tru or fals. S Sction.. 65. - 4 + z = 5; =, =, and z = 5 66. + - z = 7; =, = -, and z = 67. - - 5 + z = 5; = 0, = -, and z = 5 68. -4 + - 8z = 4; =, = 0, and z = - Add th quations. S Sction 4.. 69. 7. + - 5z = 0 - + 4 + z = 5 0 + 5 + 6z = 4-9 + 5-6z = - CONCEPT EXTENSIONS 70. 7. + 4-5z = 0-4 - z = -7-9 - 8 - z = 9 + 4 - z = Without graphing, dtrmin whthr ach sstm has on solution, no solution, or an infinit numbr of solutions. S th scond Concpt Chck in this sction. 7. = - 5 = + + = 75. 5 + 5 = 5 = - 74. μ = - + 5 76. = 5 - = - 5-77. Can a sstm consisting of two linar quations hav actl two solutions? Eplain wh or wh not. 78. 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. 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 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 79 through 8. Cost of a Prviousl Rntd DVD (in dollars) 6 4 0 8 6 Dmand Suppl Equilibrium point 79. Find th numbr of DVDs and th pric pr DVD whn suppl quals dmand. 80. Whn is btwn and 4, is suppl gratr than dmand or is dmand gratr than suppl? 8. Whn is gratr than 6, is suppl gratr than dmand or is dmand gratr than suppl? 8. 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 =.5, whr is th numbr of tubs of toothpast sold and is th total incom for slling tubs. Th cost quation is = 0.9 + 000, 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 88. Dollars 6000 5000 4000 000 000 000 Cost Rvnu 0 0 500 000 500 000 500 Tubs of Toothpast 000 8. Find th coordinats of th point of intrsction, or brak-vn point, b solving th sstm =.5 = 0.9 + 000 84. Eplain th maning of th -valu of th point of intrsction. 85. If th compan slls 000 tubs of toothpast, dos th compan mak mon or los mon? 86. If th compan slls 000 tubs of toothpast, dos th compan mak mon or los mon? 87. For what -valus will th compan mak a profit? (Hint: For what -valus is th rvnu graph highr than th cost graph?) 88. For what -valus will th compan los mon? (Hint: For what -valus is th rvnu graph lowr than th cost graph?) 89. Writ a sstm of two linar quations in and that has th ordrd pair solution (, 5). 90. Which mthod would ou us to solv th sstm? 5 - = 6 + = 5 Eplain our choic. 0 4 5 6 Numbr of DVDs (in thousands) 7
4 CHAPTER 4 Sstms of Equations 9. Th amount of bottld watr consumd pr prson in th Unitd Stats (in gallons) in th ar can b modld b th linar quation =.47 + 9.6. Th amount of carbonatd dit soft drinks consumd pr prson in th Unitd Stats (in gallons) in th ar can b modld b th linar quation = 0. +.55. In both modls, = 0 rprsnts th ar 995. (Sourc: Basd on data from th Economic Rsarch Srvic, U.S. Dpartmnt of Agricultur) a. What dos th slop of ach quation tll ou about th pattrns of bottld watr and carbonatd dit soft drink 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). 9. Th amount of U.S. fdral govrnmnt incom (in billions of dollars) for fiscal ar, from 006 through 009 ( = 0 rprsnts 006), can b modld b th linar quation = -95 + 406. Th amount of U.S. fdral govrnmnt pnditurs (in billions of dollars) for th sam priod can b modld b th linar quation = 85 + 655. (Sourc: Basd on data from Financial Managmnt Srvic, U.S. Dpartmnt of th Trasur, 006 009) a. What dos th slop of ach quation tll ou about th pattrns of U.S. fdral govrnmnt incom and pnditurs? b. Solv this sstm of quations. (Round our final rsults to th narst whol numbrs.) c. Did pnss vr qual incom during th priod from 006 through 009? 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. μ 95. μ 97. μ 99. μ 00. μ + = - = 4 + = 5 - = - 4 = 5 - = - = -8 + = 5 + 7 = - 0-4 = 0 + = 7 94. μ + = 6 96. μ 98. μ + = 5 5 - = + = - - = 8 4. Solving Sstms of Linar Equations in Thr Variabls Solv a Sstm of Thr Linar Equations in Thr Variabls. In this sction, th algbraic mthods of solving sstms of two linar quations in two variabls ar tndd to sstms of thr linar quations in thr variabls. W call th quation - + z = -5, for ampl, a linar quation in thr variabls sinc thr ar thr variabls and ach variabl is raisd onl to th powr. A solution of this quation is an ordrd tripl (,, z) that maks th quation a tru statmnt. For ampl, th ordrd tripl, 0, - is a solution of - + z = -5 sinc rplacing with, with 0, and z with - ilds th tru statmnt - 0 + - = -5. Th graph of this quation is a plan in thrdimnsional spac, just as th graph of a linar quation in two variabls is a lin in two-dimnsional spac. Although w will not discuss th tchniqus for graphing quations in thr variabls, visualizing th possibl pattrns of intrscting plans givs us insight into th possibl pattrns of solutions of a sstm of thr thr-variabl linar quations. Thr ar four possibl pattrns.. Thr plans hav a singl point in common. This point rprsnts th singl solution of th sstm. This sstm is consistnt.