Work Sheet 1 Liner Inequlities Rent--Hep, cr rentl compny,chrges $ 15 per week plus $ 0.0 per mile to rent one of their crs. Suppose you re limited y how much money you cn spend for the week : You cn spend t most$ 335. If we let x represent the numer of miles you drive the hep in week, we cn write n inequlity tht models the given conditions : The chrges of must e less The weekly $ 0.0 per mile thn chrge of $ 15 plus for x miles or equl to $ 335 15 + 0.0 x < 335. Plcing n inequlity symol etween polynomil of degree 1 nd constnt results in liner inequlity in one vrile. In this section, we will study how to solve liner inequlities such s the one shown ove. Solving n inequlity is the process of finding the set of numers tht mke the inequlity true sttement. These numers re clled the solutions of the inequlity nd we sy tht they stisfy the inequlity. The set of ll solutions is clled the solution set of the inequlity. Set-uilder nottion nd new nottion, clled intervl nottion, re used to represent solution sets. We egin this section y looking t intervl nottion.
Intervl nottion Susets of rel numers cn e represented using intervl nottion. Suppose tht nd re two rel numers such tht Intervl Nottion Grph The open intervl, represents the set of rel numers etween, ut not including, nd., x / x x is greter thn ( < x ) nd x is less thn ( x < ), The closed intervl, represents the set of rel numers etween, nd including, nd., x / x x is greter thn or equl to ( < x ) nd x is less thn or equl to ( x < ), The infinite intervl, represents the set of rel numers tht re greter thn., x / x The infinity symol does not represent rel numer. It indictes tht the intervl extends indefinitely to the right., The infinite intervl (, ] represents the set of rel numers tht re less thn or equl to. (, ] x / x The negtive infinity symol indictes tht the intervl extends indefinitely to the left. (, ]
Prentheses nd Brckets in Intervl nottion Prentheses indicte endpoints tht re not included in n intervl. Squre rckets indicte endpoints tht re included in n intervl Tle 1 lists nine possile types of intervls used to descrie susets of rel numers. Tle 1 Intervls on the Rel Numer Line Let nd e rel numers such tht Intervl Set-Builder Nottion Nottion, x / x, x / x [, ) x / x (, ] x / x Grph, x / x [, ) x / x (, ) x / x (, ] x / x (, ) x / x is rel numer or (set of ll rel numers ) English Sentence Inequlity x is t lest 5 x 5 greter thn or equl to 5 x is t most 5 x 5 less thn or equl to 5 x is etween 5 nd 7 5 x 7 greter thn 5 nd less thn 7 x is no more thn 5 x 5 Less thn or equl to 5 x is no less thn 5 x 5 greter thn or equl to 5
Exmple 1 Express ech intervl in set-uilder nottion nd grph :. ( 1, 4 ] Solution ( 1, 4 ] = x / 1 x 4-3 - -1 0 1 3 4 5. 4, Solution 4, = x / x 4-6 -5-4 -3 - -1 0 1 c. (, ] Solution (, ] = x / x - 4-3 - -1 0 1 Express ech intervl in set-uilder nottion nd grph : 1. [ 1, ). (, 3 ) 3. [, 5 ] 4. [ 1, 4 ) 5. (, 7 ) Express ech intervl in set-uilder nottion nd grph : 1. [ 3, ). (, ) 3. (, 4. (, 6 ] 5. [, 5 ] 6. ( 5, 0 ) 7. [ 5, ) 8. (, 4 ) )
Work Sheet Solving Liner Inequlities in One Vrile Rent--Hep, cr rentl compny,chrges $ 15 per week plus $ 0.0 per mile to rent one of their crs. You cn spend t most $ 335. We cn write n inequlity : 0.0 x 15 335 Solving Liner Inequlities in One Vrile We know tht liner eqution in x cn e expressed s x 0. A liner inequlity in x cn e written in one of the following forms : x 0, x 0, x 0, x 0. In ech x 0 form, 0. Bck to our question tht opened the section: How mny miles cn you drive your Rent--Hep cr if you cn spend most $ 335 per week? We nswer the question y solving 0.0 x 15 335 for x. The solution procedure is nerly identicl to tht for solving 0.0 x 15 Our gol is to get x y itself on the left side. We do this y sutrcting 15 from oth sides to isolte 0.0 x : 0.0 x 15 335 335 0.0 x 1515 335 15 0.0 x Finlly, we isolte x from 0.0 x y dividing oth sides of the inequlity y 0.0 0.0 x 0.0 x 10 10 0.0 1050 With t most $ 335 per week to spend, you cn trvel t most 1050 miles. We strted with the inequlity 0.0 x 15 335 nd otined the inequlity x 1050 in the finl step.
We isolted x from 0.0 x y dividing oth sides of 0.0 x 10 y 0.0 positive numer. Let, s see wht hppens if we divide oth sides of the inequlity negtive numer. Consider the inequlity 10 < 14. Divide 10 nd 14 y - 10 14 5 nd 7 Becuse -5 lies to the right of -7 on the numer lie, -5 is greter thn -7 : -5 > - 7 Notice tht the direction of the inequlity symol is reversed : 10 < 14 Dividing y - chnges the direction of The inequlity symol. -5 > - 7 In generl, when we multiply or divide oth sides of n inequlity y negtive numer, the direction of the inequlity symol is reversed. When we reverse the direction of the inequlity symol, we sy tht we chnge the sense of the inequlity.
We cn isolte vrile in liner inequlity the sme wy we cn isolte vrile in liner eqution. The following properties re used to crete equivlent inequlities : Properties of Inequlities Property The Property in Words Exmple The Addition Property of Inequlitiy If <, then +c < +c. If <, then - c < c. The Positive Multipliction Property of Inequlitiy If < nd c is positive, then c < c. If < nd c is positive, then c c. The Negtive Multipliction Property of Inequlitiy If < nd c is negtive, then c > c. If < nd c is negtive, then. c c If the sme quntity is dded to or sutrcted from oth sides of n inequlity, the resulting inequlity is equivlent to the originl one. If we multiply or divide oth sides of n inequlity y the sme positive quntity, the resulting inequlity is equivlent to the originl one. If we multiply or divide oth sides of n inequlity y the sme negtive quntity nd reverse the direction of the inequlity symol, the resulting inequlity is equivlent to the originl one. EX.1.1 Sutrct 3 : Simplify : Divide y : Simplify : x 3 7 x 3 3 7 3 x x x 4 4 4 x. EX.1. 4 x 0 Divide y 4 nd reverse the sense of the inequlity Simplify : 4x 4 0 4 x 5.
Solving Liner Inequlity Exmple Solve nd grph the solution set on numer line: Solution 3 x 11 3 x 11 3 x 3 11 3 x 8 x 8 x 4 The solution set consists of ll rel numers tht re greter thn or equl to 4. The grph of the solution set is shown s follows : Solve nd grph the solution set on numer line. 1. 3 x 5. x 4 x 5 3. 1 4 x
Solve ech liner inequlity nd grph the solution set on numer line. 1. 5x 11 6. 3 x 7 13 3. 9 x 36 4. 8x 11 3x 13 5. 4 x 1 3x 6 6. 8x 3 3 x1 x 5 7. 4 x 3x 0 8. 1 x3 4 x x 3 x 4 4 3 7 x 5 5 x 4 x 5 6 9 18 3 3 x 5 8x 7 5 3 x6 3x5 4 x 3 9. 1 10. 11. 1.
Solving Liner Inequlity in form: x 0 Solving Liner Inequlity in One Vrile in x 0 form. We know tht liner eqution in x cn e expressed s x 0. We cn solve the Inequlity in form: x 0 like solving x 0. Exmple 3 Solve nd grph the solution set on numer line: x - 3 4x - 5 4 6 x - 3 4x - 5 1 ( ) 1 ( ) 4 6 3 ( x 3 ) ( 4 x 5 ) 6 x - 9 8 x - 10 - x - 1 The solution set consists of ll rel numers except 1 The grph of the solution set is shown s follows : x 1-3 - -1 0 1 1 3 Solve nd grph the solution set on numer line. 3x + 5x - 5 1. 3 Solve nd grph the solution set on numer line. 1. 3x + 9 5x - 10
Solving Inequlity with Asolute Vlue We know tht x descries the distnce of x from zero on rel numer line. We cn use this geometric interprettion to n inequlity such s x This mens tht the distnce of x from 0 is less thn, s shown in Figure 1. Figure 1 x, so x -3 - -1 0 1 3 The intervl shows vlue of x tht lie less thn units from 0. Thus, x cn lie etween nd. Tht is, x is greter thn nd less thn. We write ( -, ) or { x / x } Some solute vlue Inequlities use the greter thn symol. For exmple, x mens tht the distnce of x from 0 is greter thn, s shown in Figure Figure x, so x or x -3 - -1 0 1 3 Thus, x cn e less thn or greter thn. We write x or x. These oservtions suggest the following principles for solving Inequlities with solute vlue. Solving n Asolute Vlue Inequlity If x is n lgeric expression nd c is positive numer, 1. The solutions of x c re the numers tht stisfy c x c. The solutions of x c re the numers tht stisfy x c or x c These rules re vlid if < is replce y < nd > is replce y > Exmple 4 Solve nd grph the solution set on numer line: x 4 3 Solution We rewrite the inequlity without solute vlue rs. x 4 3, mens 3 3 4 x 4 4 3 4 Add 4 to ll three prts 1 x The solution set consists of ll rel numers greter thn 1 nd less thn 7. The grph of the solution set is shown s follows : 7 0 1 3 4 5 6 7 8
In the In the x c cse, we hve one compound inequlity to solve. x c cse, we hve two seprte inequlities to solve. Solve nd grph the solution set on numer line: 1. x 5 Solve nd grph the solution set on numer line: 1. x 5. x 3 3. x 3 4 4. x 1
Work Sheet 3 Applictions Our next exmple shows how to use n inequlity to select the etter del etween two pricing options. We use our strtegy for solving word prolems, trnslting from the verl conditions of the prolem to liner Inequlity. Exmple 5 Selecting the Better Del Acme Cr rentl gency chrges $ 4 dy plus $ 0.15 per mile, wheres Interstte rentl gency chrges $ 0 dy plus $ 0.05 per mile. How mny miles must e driven to mke the dily cost of n Acme Cr rentl gency etter del thn n Interstte rentl? Solution Step 1 Let x represent one of the quntities. We re looking for the numer of miles tht must e driven in dy to mke Acme the etter del. Thus Let x = the numer of miles driven in dy Step Represent other quntities in terms of x. We re not sked to find nother quntity, so we cn skip this step. Step 3 Write n inequlity in x tht models the conditions. Acme is etter del thn Interstte if the dily cost of Acme is less thn the dily cost of Interstte. The dily cost of Acme is less thn The dily cost of Interstte 15 cents the numers of 5 cents the numers of $ 4 plus times miles driven $ 0 plus times miles driven 4 + 0.15. x < 0 + 0.05. x
Step 4 Solve the inequlity nd nswer the questions. 4 0.15 x 0 0. 05 x 4 0.15 x 0.05 x 0 0.05 x 0. 05 x 4 0.1x 0 4 0.1x 4 0 4 0.1x 0.1 x 0.1 x 16 16 0.1 160 Thus,driving fewer thn 160 miles per dy mke Acme the etter del. Step 5 Check the proposed solution in the originl wording of the prolem. One wy to do this is to tke milege less thn 160 miles per dy to see if Acme is the etter del. Suppose tht 150 miles re driven in dy. Cost for Acme = 4 0.15(150) 6. 50 Cost for Interstte = 0 0.05(150) 7. 50 Acme hs lower dily cost, mking Acme the etter del. Suppose tht 160 miles re driven in dy. Cost for Acme = 4 0.15(160) 8 Cost for Interstte = 0 0.05(160) 8 Suppose tht 170 miles re driven in dy. Cost for Acme = 4 0.15(170) 9. 50 Cost for Interstte = 0 0.05(150) 8. 50 Acme hs higher dily cost.
Use the strtegy for solving word prolems, trnslting from the verl conditions of the prolem to liner inequlity. 1. A locl nk chrges 8 per month plus 0.05 per check. The credit union chrges per month plus 0.08 per check. How mny checks should e written ech month to mke the credit union etter del? Use the strtegy for solving word prolems, trnslting from the verl conditions of the prolem to liner inequlity. 1. You re choosing etween two-long distnce telephone plns. Pln A hs monthly fee of $ 15 with chrge of $ 0.08 per minute for ll long distnce clls. Pln B hs monthly fee of $ 3 with chrge of $ 0.1 per minute for ll long distnce clls. How mny minutes of long distnce clls in month mke pln A the etter del?.