Linear Systems with Two Variables - Pauls Online Math Notes

A linear system of two equations with two variables is any system that can be written in the form. ... where any of the constants can be zero with ... 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Section7-1:LinearSystemswithTwoVariables Alinearsystemoftwoequationswithtwovariablesisanysystemthatcanbewrittenintheform. \[\begin{align*}ax+by&=p\\cx+dy&=q\end{align*}\] whereanyoftheconstantscanbezerowiththeexceptionthateachequationmusthaveatleastonevariableinit. Also,thesystemiscalledlinearifthevariablesareonlytothefirstpower,areonlyinthenumeratorandtherearenoproductsofvariablesinanyoftheequations. Hereisanexampleofasystemwithnumbers. \[\begin{align*}3x-y&=7\\2x+3y&=1\end{align*}\] Beforewediscusshowtosolvesystemsweshouldfirsttalkaboutjustwhatasolutiontoasystemofequationsis.Asolutiontoasystemofequationsisavalueof\(x\)andavalueof\(y\)that,whensubstitutedintotheequations,satisfiesbothequationsatthesametime. Fortheexampleabove\(x=2\)and\(y=-1\)isasolutiontothesystem.Thisiseasyenoughtocheck. \[\begin{align*}3\left(2\right)-\left({-1}\right)&=7\\2\left(2\right)+3\left({-1}\right)&=1\end{align*}\] So,sureenoughthatpairofnumbersisasolutiontothesystem.Donotworryabouthowwegotthesevalues.Thiswillbetheveryfirstsystemthatwesolvewhenwegetintoexamples. Notethatitisimportantthatthepairofnumberssatisfybothequations.Forinstance,\(x=1\)and\(y=-4\)willsatisfythefirstequation,butnotthesecondandsoisn’tasolutiontothesystem.Likewise,\(x=-1\)and\(y=1\)willsatisfythesecondequationbutnotthefirstandsocan’tbeasolutiontothesystem. Now,justwhatdoesasolutiontoasystemoftwoequationsrepresent?Wellifyouthinkaboutitbothoftheequationsinthesystemarelines.So,let’sgraphthemandseewhatweget. Asyoucanseethesolutiontothesystemisthecoordinatesofthepointwherethetwolinesintersect.So,whensolvinglinearsystemswithtwovariableswearereallyaskingwherethetwolineswillintersect. Wewillbelookingattwomethodsforsolvingsystemsinthissection. Thefirstmethodiscalledthemethodofsubstitution.Inthismethodwewillsolveoneoftheequationsforoneofthevariablesandsubstitutethisintotheotherequation.Thiswillyieldoneequationwithonevariablethatwecansolve.Oncethisissolvedwesubstitutethisvaluebackintooneoftheequationstofindthevalueoftheremainingvariable. Inwordsthismethodisnotalwaysveryclear.Let’sworkacoupleofexamplestoseehowthismethodworks. Example1Solveeachofthefollowingsystems. \(\begin{align*}3x-y&=7\\2x+3y&=1\end{align*}\) \(\begin{align*}5x+4y&=1\\3x-6y&=2\end{align*}\) ShowAllSolutions HideAllSolutions a\(\begin{align*}3x-y&=7\\2x+3y&=1\end{align*}\)ShowSolution So,thiswasthefirstsystemthatwelookedatabove.Wealreadyknowthesolution,butthiswillgiveusachancetoverifythevaluesthatwewrotedownforthesolution. Now,themethodsaysthatweneedtosolveoneoftheequationsforoneofthevariables.Whichequationwechooseandwhichvariablethatwechooseisuptoyou,butit’susuallybesttopickanequationandvariablethatwillbeeasytodealwith.Thismeansweshouldtrytoavoidfractionsifatallpossible. Inthiscaseitlookslikeitwillbereallyeasytosolvethefirstequationfor\(y\)solet’sdothat. \[3x-7=y\] Now,substitutethisintothesecondequation. \[2x+3\left({3x-7}\right)=1\] Thisisanequationin\(x\)thatwecansolvesolet’sdothat. \[\begin{align*}2x+9x-21&=1\\11x&=22\\x&=2\end{align*}\] So,thereisthe\(x\)portionofthesolution. Finally,doNOTforgettogobackandfindthe\(y\)portionofthesolution.Thisisoneofthemorecommonmistakesstudentsmakeinsolvingsystems.Tosothiswecaneitherplugthe\(x\)valueintooneoftheoriginalequationsandsolvefor\(y\)orwecanjustplugitintooursubstitutionthatwefoundinthefirststep.Thatwillbeeasiersolet’sdothat. \[y=3x-7=3\left(2\right)-7=-1\] So,thesolutionis\(x=2\)and\(y=-1\)aswenotedabove. b\(\begin{align*}5x+4y&=1\\3x-6y&=2\end{align*}\)ShowSolution Withthissystemwearen’tgoingtobeabletocompletelyavoidfractions.However,itlookslikeifwesolvethesecondequationfor\(x\)wecanminimizethem.Hereisthatwork. \[\begin{align*}3x&=6y+2\\x&=2y+\frac{2}{3}\end{align*}\] Now,substitutethisintothefirstequationandsolvetheresultingequationfor\(y\). \[\begin{align*}5\left({2y+\frac{2}{3}}\right)+4y&=1\\10y+\frac{{10}}{3}+4y&=1\\14y&=1-\frac{{10}}{3}=-\frac{7}{3}\\y&=-\left({\frac{7}{3}}\right)\left({\frac{1}{{14}}}\right)\\y&=-\frac{1}{6}\end{align*}\] Finally,substitutethisintotheoriginalsubstitutiontofind\(x\). \[x=2\left({-\frac{1}{6}}\right)+\frac{2}{3}=-\frac{1}{3}+\frac{2}{3}=\frac{1}{3}\] So,thesolutiontothissystemis\(x=\frac{1}{3}\)and\(y=-\frac{1}{6}\). Aswithsingleequationswecouldalwaysgobackandcheckthissolutionbypluggingitintobothequationsandmakingsurethatitdoessatisfybothequations.Noteaswellthatwereallywouldneedtoplugintobothequations.Itisquitepossiblethatamistakecouldresultinapairofnumbersthatwouldsatisfyoneoftheequationsbutnottheotherone. Let’snowmoveintothenextmethodforsolvingsystemsofequations.Aswesawinthelastpartofthepreviousexamplethemethodofsubstitutionwilloftenforceustodealwithfractions,whichaddstothelikelihoodofmistakes.Thissecondmethodwillnothavethisproblem.Well,that’snotcompletelytrue.Iffractionsaregoingtoshowuptheywillonlyshowupinthefinalstepandtheywillonlyshowupifthesolutioncontainsfractions. Thissecondmethodiscalledthemethodofelimination.Inthismethodwemultiplyoneorbothoftheequationsbyappropriatenumbers(i.e.multiplyeverytermintheequationbythenumber)sothatoneofthevariableswillhavethesamecoefficientwithoppositesigns.Thennextstepistoaddthetwoequationstogether.Becauseoneofthevariableshadthesamecoefficientwithoppositesignsitwillbeeliminatedwhenweaddthetwoequations.Theresultwillbeasingleequationthatwecansolveforoneofthevariables.Oncethisisdonesubstitutethisanswerbackintooneoftheoriginalequations. Aswiththefirstmethodit’smucheasiertoseewhat’sgoingonherewithacoupleofexamples. Example2ProblemStatement. \(\begin{align*}5x+4y&=1\\3x-6y&=2\end{align*}\) \(\begin{align*}2x+4y&=-10\\6x+3y&=6\end{align*}\) ShowAllSolutions HideAllSolutions a\(\begin{align*}5x+4y&=1\\3x-6y&=2\end{align*}\)ShowSolution Thisisthesystemintheprevioussetofexamplesthatmadeusworkwithfractions.Workingitherewillshowthedifferencesbetweenthetwomethodsanditwillalsoshowthateithermethodcanbeusedtogetthesolutiontoasystem. So,weneedtomultiplyoneorbothequationsbyconstantssothatoneofthevariableshasthesamecoefficientwithoppositesigns.So,sincethe\(y\)termsalreadyhaveoppositesignslet’sworkwiththeseterms.Itlookslikeifwemultiplythefirstequationby3andthesecondequationby2the\(y\)termswillhavecoefficientsof12and-12whichiswhatweneedforthismethod. Hereistheworkforthisstep. \[\begin{align*} 5x+4y&=1&\underrightarrow{\times\,\,3}\hspace{0.5in}&15x+12y=3\\ 3x-6y&=2&\underrightarrow{\times\,\,2}\hspace{0.5in}&\underline{\,\,6x-12y=4}\\ &&&21x\hspace{0.5in}=7\\ \end{align*}\] So,asthedescriptionofthemethodpromisedwehaveanequationthatcanbesolvedfor\(x\).Doingthisgives,\(x=\frac{1}{3}\)whichisexactlywhatwefoundinthepreviousexample.Noticehowever,thattheonlyfractionthatwehadtodealwithtothispointistheansweritselfwhichisdifferentfromthemethodofsubstitution. Now,againdon’tforgettofind\(y\).Inthiscaseitwillbealittlemoreworkthanthemethodofsubstitution.Tofind\(y\)weneedtosubstitutethevalueof\(x\)intoeitheroftheoriginalequationsandsolvefor\(y\).Since\(x\)isafractionlet’snoticethat,inthiscase,ifweplugthisvalueintothesecondequationwewilllosethefractionsatleasttemporarily.Notethatoftenthiswon’thappenandwe’llbeforcedtodealwithfractionswhetherwewanttoornot. \[\begin{align*}3\left({\frac{1}{3}}\right)-6y&=2\\1-6y&=2\\-6y&=1\\y&=-\frac{1}{6}\end{align*}\] Again,thisisthesamevaluewefoundinthepreviousexample. b\(\begin{align*}2x+4y&=-10\\6x+3y&=6\end{align*}\)ShowSolution Inthispartallthevariablesarepositivesowe’regoingtohavetoforceanoppositesignbymultiplyingbyanegativenumbersomewhere.Let’salsonoticethatinthiscaseifwejustmultiplythefirstequationby-3thenthecoefficientsofthe\(x\)willbe-6and6. Sometimesweonlyneedtomultiplyoneoftheequationsandcanleavetheotheronealone.Hereisthisworkforthispart. \[\begin{align*} 2x+4y&=-10&\underrightarrow{\times\,\,-3}\hspace{0.5in}&-6x-12y=30\\ 6x+3y&=6&\underrightarrow{\text{same}}\hspace{0.5in}&\underline{\hspace{0.35in}6x+3y=6}\\ &&&\hspace{0.5in}-9y=36\\ &&&\hspace{0.85in}y=-4\\ \end{align*}\] Finally,plugthisintoeitheroftheequationsandsolvefor\(x\).Wewillusethefirstequationthistime. \[\begin{align*}2x+4\left({-4}\right)&=-10\\2x-16&=-10\\2x&=6\\x&=3\end{align*}\] So,thesolutiontothissystemis\(x=3\)and\(y=-4\). Thereisathirdmethodthatwe’llbelookingattosolvesystemsoftwoequations,butit’salittlemorecomplicatedandisprobablymoreusefulforsystemswithatleastthreeequationssowe’lllookatitinalatersection. Beforeleavingthissectionweshouldaddressacoupleofspecialcaseinsolvingsystems. Example3Solvethefollowingsystemsofequations. \[\begin{align*}x-y&=6\\-2x+2y&=1\end{align*}\] ShowSolution Wecanuseeithermethodhere,butitlookslikesubstitutionwouldprobablybeslightlyeasier.We’llsolvethefirstequationfor\(x\)andsubstitutethatintothesecondequation. \[\begin{align*}x&=6+y\\&\\-2\left({6+y}\right)+2y&=1\\-12-2y+2y&=1\\-12&=1\,\,\,??\end{align*}\] So,thisisclearlynottrueandtheredoesn’tappeartobeamistakeanywhereinourwork.So,what’stheproblem?Toseelet’sgraphthesetwolinesandseewhatweget. Itappearsthatthesetwolinesareparallel(canyouverifythatwiththeslopes?)andweknowthattwoparallellineswithdifferent\(y\)-intercepts(that’simportant)willnevercross. Aswesawintheopeningdiscussionofthissectionsolutionsrepresentthepointwheretwolinesintersect.Iftwolinesdon’tintersectwecan’thaveasolution. So,whenwegetthiskindofnonsensicalanswerfromourworkwehavetwoparallellinesandthereisnosolutiontothissystemofequations. Thesysteminthepreviousexampleiscalledinconsistent.Noteaswellthatifwe’dusedeliminationonthissystemwewouldhaveendedupwithasimilarnonsensicalanswer. Example4Solvethefollowingsystemofequations. \[\begin{align*}2x+5y&=-1\\-10x-25y&=5\end{align*}\] ShowSolution Inthisexampleitlookslikeeliminationwouldbetheeasiestmethod. \[\begin{align*} 2x+5y&=-1&\underrightarrow{\times\,\,5}\hspace{0.5in}&\,\,\,\,10x+25y=-5\\ -10x-25y&=5&\underrightarrow{\text{same}}\hspace{0.5in}&\underline{-10x-25y=5}\\ &&&\hspace{0.9in}0=0\\ \end{align*}\] Onfirstglancethismightappeartobethesameproblemasthepreviousexample.However,inthatcaseweendedupwithanequalitythatsimplywasn’ttrue.Inthiscasewehave0=0andthatisatrueequalityandsointhatsensethereisnothingwrongwiththis. However,thisisclearlynotwhatwewereexpectingforananswerhereandsoweneedtodeterminejustwhatisgoingon. We’llleaveittoyoutoverifythis,butifyoufindtheslopeand\(y\)-interceptsforthesetwolinesyouwillfindthatbothlineshaveexactlythesameslopeandbothlineshaveexactlythesame\(y\)-intercept.So,whatdoesthismeanforus?Welliftwolineshavethesameslopeandthesame\(y\)-interceptthenthegraphsofthetwolinesarethesamegraph.Inotherwords,thegraphsofthesetwolinesarethesamegraph.Inthesecasesanysetofpointsthatsatisfiesoneoftheequationswillalsosatisfytheotherequation. Also,recallthatthegraphofanequationisnothingmorethanthesetofallpointsthatsatisfiestheequation.Inotherwords,thereisaninfinitesetofpointsthatwillsatisfythissetofequations. Inthesecaseswedowanttowritedownsomethingforasolution.So,whatwe’lldoissolveoneoftheequationsforoneofthevariables(itdoesn’tmatterwhichyouchoose).We’llsolvethefirstfor\(y\). \[\begin{align*}2x+5y&=-1\\5y&=-2x-1\\y&=-\frac{2}{5}x-\frac{1}{5}\end{align*}\] Then,givenany\(x\)wecanfinda\(y\)andthesetwonumberswillformasolutiontothesystemofequations.Weusuallydenotethisbywritingthesolutionasfollows, \[\begin{array}{*{20}{c}}\begin{aligned}x&=t\\y&=-\frac{2}{5}t-\frac{1}{5}\end{aligned}&{\hspace{0.25in}{\mbox{where}}\,t{\mbox{isanyrealnumber}}}\end{array}\] Toshowthatthesegivesolutionslet’sworkthroughacoupleofvaluesof\(t\). \(t=0\) \[x=0\hspace{0.25in}y=-\frac{1}{5}\] Toshowthatthisisasolutionweneedtoplugitintobothequationsinthesystem. \[\begin{align*}2\left(0\right)+5\left({-\frac{1}{5}}\right)&\mathop=\limits^?-1&\hspace{0.25in}-10\left(0\right)-25\left({-\frac{1}{5}}\right)&\mathop=\limits^?5\\-1&=-1&\hspace{0.25in}5&=5\end{align*}\] So,\(x=0\)and\(y=-\frac{1}{5}\)isasolutiontothesystem.Let’sdoanotheronerealquick. \(t=-3\) \[x=-3\hspace{0.25in}y=-\frac{2}{5}\left({-3}\right)-\frac{1}{5}=\frac{6}{5}-\frac{1}{5}=1\] Againweneedtoplugitintobothequationsinthesystemtoshowthatit’sasolution. \[\begin{align*}2\left({-3}\right)+5\left(1\right)&\mathop=\limits^?-1&\hspace{0.25in}-10\left({-3}\right)-25\left(1\right)&\mathop=\limits^?5\\-1&=-1&\hspace{0.25in}5&=5\end{align*}\] Sureenough\(x=-3\)and\(y=1\)isasolution. So,sincethereareaninfinitenumberofpossible\(t\)’stheremustbeaninfinitenumberofsolutionstothissystemandtheyaregivenby, \[\begin{array}{*{20}{c}}\begin{aligned}x&=t\\y&=-\frac{2}{5}t-\frac{1}{5}\end{aligned}&{\hspace{0.25in}{\mbox{where}}\,t{\mbox{isanyrealnumber}}}\end{array}\] Systemssuchasthoseinthepreviousexamplesarecalleddependent. We’venowseenallthreepossibilitiesforthesolutiontoasystemofequations.Asystemofequationwillhaveeithernosolution,exactlyonesolutionorinfinitelymanysolutions.