2.4 Linear Inequalities and Interval Notation


 Kimberly Stokes
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1 .4 Liner Inequlities nd Intervl Nottion We wnt to solve equtions tht hve n inequlity symol insted of n equl sign. There re four inequlity symols tht we will look t: Less thn <, Greter thn >, Less thn or equl to nd Greter thn or equl to. The or equl to types of inequlities re like comining the inequlity symol nd n equl sign into the sme symol. As usul, we re interested in finding the solutions to these inequlities. As it turns out, these inequlities hve n infinite numer of solutions. So, for this reson, we wnt to grph these inequlities ecuse it s the only wy to see ll of the solutions t one time. We need to strt with the following fcts. Grphing Inequlities When grphing n inequlity of type or we use [ nd ] to represent the vlue we re working with is ctully included in the solutions hence the or equl to prt). When grphing n inequlity of type < or > we use nd ) to represent the vlue we re working with is not included in the solutions Let s tke look t some grphing to get the ide. Emple 1: Grph.. > To grph, we need to first recognize which inequlity symol we re deling with. In this cse we re working with the greter thn symol. This mens, the solution is ll of the vlues greter or lrger) thn . Also, since we hve regulr greter thn symol, we use n open circle. The reson for this is ecuse the  is not ctully included in the solutions. So we get the grph This time we wnt to grph ll of the vlues less thn or equl to three. This mens we wnt ll of the vlues less or smller) thn 3 s well s the 3 itself. So we use the closed circle on the three nd get the grph s follows ] 3 4
2 In order to simplify mtters, we wnt to define simple nottion for inequlities. This new nottion is clled using intervls. There re two types of intervls on the rel numer line; ounded nd unounded. Definitions: Bounded intervl An intervl with finite length, i.e. if we sutrct the endpoints of the intervl we get rel numer. Unounded intervl Any intervl which is not of finite length is unounded. Let us proceed to define these intervls y relting them to inequlities. We strt with the unounded intervls. Unounded Intervls Inequlity Intervl Type Nottion Grph Hlfopen [ Open,, Hlfopen Open Entire Line,,, ] ) Notice tht when writing in intervl nottion, we lwys write our intervls in incresing order. Tht is, we lwys hve the smller numers on the left. There re other types of inequlities tht we will look t in more depth lter in the section, ut for the ske of orgniztion, we will include the intervls here. These relte to the compound inequlities tht we tret t the end of the section. Bounded Intervls Inequlity Intervl Type Nottion Grph Closed, Open, Hlfopen, Hlfopen, [ ] ) [ ) ] Note tht the lengths of ll the intervls ove re ove intervls re ounded y definition.. Which is rel numer nd thus ll the Emple : Write the following in intervl nottion c. 3 d.. This is ounded intervl. It my prove helpful to grph the inequlity first. [ )
3 3 1 So, s n intervl we get.. Agin this is ounded intervl. The grph is , So, we get. c. This is n unounded intervl. Grphing we get 43 Hence our intervl is   3, [ ) d. Finlly, we hve Thus, our intervl is,. 1 ] So, now tht we cn grph nd del with intervls, how do we solve n inequlity? We need the following properties. Properties of Inequlities 1. If, then c c nd c c.. If, nd c is positive, then c c nd. c c 3. If, nd c is negtive, then c c nd. c c The rules re similr for <, nd. The ide is tht we cn dd or sutrct ny vlue on oth sides of n inequlity symol nd nothing chnges nd we cn multiply or divide ny positive vlue on oth sides of n inequlity symol nd nothing chnges. However, if we multiply or divide ny negtive vlue on oth sides of n inequlity symol, we must flip the inequlity symol. We use these properties to solve inequlities. Bsiclly, ll we hve to rememer is tht we solve them just s we solved equtions with the dded restriction tht ny time we multiply or divide y negtive, we hve to chnge the inequlity symol. Emple 3: Solve nd grph c So we cn simply solve this inequlity s we solved equtions. We just get the lone on one side like we did with equtions. The only thing we hve to rememer is tht when
4 we hve to divide y negtive, we will need to flip the inequlity symol. We proceed s follows + 3 < Sutrct 3 from oth sides < > > 1 Divide y , flip the inequlity symol Now we simply grph. We since we hve > symol So in intervl nottion our solution is 1, ).. Agin, we solve s we did with equtions nd flip the inequlity symol if needed. We get Sutrct on oth sides Sutrct 4 on oth sides So we grph. Here we need ] ecuse we re working with symol. So our solution is, ]. ] c. Agin, we proceed s we did ove. Just follow the steps tht we did when we lerned how to solve equtions ) ) Use the distriutive property Comine like terms Sutrct 8 on oth sides Add 5 on oth sides Divide y 1 on oth sides Reduce
5 5 4 Now we grph. When doing so with frctionl vlue, we hve to do our est to plot the point roughly where it should go. We hve So the solution is [ 5 4, ). [ Compound Inequlities There is nother type of inequlity tht we need to e le to solve clled in compound inequlity. Compound inequlities come in severl forms. We will only concentrte on three primry types: doule inequlities, inequlities contining nd, inequlities contining or. As we will see, the doule inequlities will relte directly to our ounded intervls tht we discussed t the eginning of the section. Also, wht we need to keep in mind is tht when working with the nd nd or prolems, nd mens, or mens We will illustrte how to solve ll of these types in the following emples. Emple 4: Solve nd grph In this emple we hve the so clled doule inequlities. In order to solve these, we wnt to get the lone in the middle of the inequlity symols. We do this the sme wy we did in the previous emple. The only difference is whtever we do one prt of the inequlity, we need to do to ll three prts of the inequlity. So we proceed s follows < < < 1 3 Sutrct 4 everywhere Divide y 3 everywhere 3 < 4 For the grph nd solution to this inequlity, ll we need know is s long s the endpoints re set up in consistent wy s they re for this one, since 3 is less thn 4), then the solution is everything etween the endpoints. So we hve grph of So the solution is [3, 4). [ )
6 . Agin, we proceed s we did ove. Rememer, whenever multiply or divide y negtive, we must flip the inequlity symol. We get < 5 4 < Sutrct 5 everywhere 7 < 4 < > 4 4 > 4 4 Divide y 4 everywhere Don t forget to flip the inequlity symols) 7 4 > > 1 Now gin, the endpoints re consistent ecuse everything etween them. This gives This gives us n intervl of 1, ) ) is lrger thn 1 so the grph is Notice tht whenever deling with doule inequlity, s long s the endpoints re in the correct order, the inequlity will lwys e ll vlues in etween the endpoints. Now we need to del with the compound inequlities tht hve n nd or n or. We simply need to rememer tht when deling with n nd we wnt only the overlpping portion of the grph, when deling with n or we get to keep everything we grph. Emple 5: Solve nd grph or nd c. 9 7 nd d or First, we cn strt y solving ech of the inequlities seprtely nd del with the or lter. We get < 10 or 1 > < 3 or > < 3 3 or > 6 < 1 or > 3
7 Now, to grph, we grph ech of the inequlities nd s stted ove, since we hve n or we get to keep everything we grph. This gives 1 ) This gives us solution of, 1) 3, ).. Agin, we strt y solving ech inequlity seprtely nd tke cre of the nd lter. 3 5 nd 3 1 > nd 3 > 1 8 nd 3 > nd > 4 Since this time we hve n nd we need to just grph the overlpping section. So we strt y grphing ech inequlity individully, ove or elow the grph, then put the overlp onto the finished grph. 4 [ 3 4 So we cn see clerly tht the grphs overlp 4 onwrd. However, t the vlue of 4, they do not overlp since the ottom piece hs prenthesis, nd thus does not contin the vlue of 4. So our grph is This mens our solution is ctully just the 4, ). We need to e wre tht when we re working with n nd prolem, the grph in the end will usully give us the ctul solution. Keep this in mind s we continue to work on the compound inequlities. c. We will gin proceed s efore < 7 nd 3 5 > < 9 nd 3 > 15 9 < 9 nd 3 > < 1 nd > 5
8 Since we hve n nd we wnt only the overlpping section. Grphing individully we get ) 1 Since there is no overlpping section, there must e no solution. d. Lstly, we will solve s we did ove or or or or Since we hve n or we grph oth of the inequlities on the sme line nd we get to keep everything tht we grph. We get So since the entire numer line hs een shded, ll rel numers must e the solution. We write this in intervl nottion s, )..4 Eercises Grph [ ] >. > < 1 8. < 4 9. > < Solve nd grph ) ) 5)
9 or or nd nd nd or or nd or nd or or nd or or or nd nd nd or or nd 1 4 5