CHAPTER 7 Systems Of Equations And Inequalities

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C H A P T E R 7Systems of Equations and InequalitiesSection 7.1Linear and Nonlinear Systems of Equations . . . . . . . . 611Section 7.2Two-Variable Linear Systems. . . . . . . . . . . . . . . 625Section 7.3Multivariable Linear Systems. . . . . . . . . . . . . . . 638Section 7.4Partial FractionsSection 7.5Systems of Inequalities . . . . . . . . . . . . . . . . . . . 674Section 7.6Linear Programming . . . . . . . . . . . . . . . . . . . . 685Review Exercises. . . . . . . . . . . . . . . . . . . . . . 661. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 696Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 710Practice Test. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 716

C H A P T E R 7Systems of Equations and InequalitiesSection 7.1 Linear and Nonlinear Systems of EquationsYou should be able to solve systems of equations by the method of substitution.1. Solve one of the equations for one of the variables.2. Substitute this expression into the other equation and solve.3. Back-substitute into the first equation to find the value of the other variable.4. Check your answer in each of the original equations. You should be able to find solutions graphically. (See Example 5 in textbook.)Vocabulary Check1. system of equations2. solution3. solving4. substitution5. point of intersection6. break-even1. 4x y 6x y 61(a) 4 0 3 1 0, 3 is not a solution.(b) 4 1 4 1 1, 4 is not a solution.(c) 4 2 1 2 is not a solution.4 12 3 16 12 3 6 12, 3 is a solution. 32 32,(d)2.4x x y 112 y 3?(a) 4 2 2 13 316 13 3? 2 13 11 2 13 11 2, 13 is a solution.?(b) 4 2 2 9 316 9 3 2, 9 is not a solution.3 231 ?(c) 4 2 3 3364 313 3 32, 313 is not a solution.7 237 ?(d) 4 4 4 3494 374 3737 ? 4 4 1174 374 11 74, 374 is a solution.611

6123.Chapter 7Systems of Equations and Inequalitiesy 2ex 3x y 4. log x 3 y1289x y 92(a) 0 2e 237(a) log 9 3 9 9, 379 is not a solution. 2, 0 is not a solution. 2 2e0(b)(b) log 10 3 23 0 2 219 10 0, 2 is a solution.(c) 3 2 289 10, 2 is a solution. 2e0(c) log 1 3 3 0, 3 is not a solution.19 1 (d) 2 2e 1 3 289 1, 3 is a solution. 1, 2 is not a solution.(d) log 2 3 4 2, 4 is not a solution.5.7.2x y 6 x y 0 6. x y 4x 2y 5Equation 1Equation 2Equation 1Equation 2Solve for y in Equation 1: y 6 2xSolve for x in Equation 1: x y 4Substitute for y in Equation 2: x (6 2x) 0Substitute for x in Equation 2: y 4 2y 5Solve for x: 3x 6 0 x 2Solve for y: 3y 4 5 y 3Back-substitute x 2: y 6 2(2) 2Back-substitute y 3: x 3 4 1Solution: 2, 2 Solution: 1, 3 xx yy 4 22Equation 1Equation 2Solve for y in Equation 1: y x 4Substitute for y in Equation 2: x2 (x 4) 2Solve for x: x2 x 2 0 x 1 x 2 0 x 1, 2Back-substitute x 1: y 1 4 3Back-substitute x 2: y 2 4 6Solutions: 1, 3 , 2, 6 8. 3x y 2x3 2 y 0Equation 1Equation 2Solve for y in Equation 1: y 2 3xSubstitute for y in Equation 2: x 3 2 2 3x 0x 3 3x 0Solve for x: x 3 3x 0 x x2 3 0 x 0, 3Back-substitute x 0: y 2 3 0 2Back-substitute x 3: y 2 3 3Back-substitute x 3: y 2 3 3 2 3 3Solutions: 0, 2 , 3, 2 3 3 , 3, 2 3 3

Section 7.19. 2x y 5Equation 1 y 25Equation 2 x2210.Linear and Nonlinear Systems of Equationsx y 0 x 5x y 03Equation 1Equation 2Solve for y in Equation 1: y 2x 5Solve for y in Equation 1: y xSubstitute for y in Equation 2: x2 2x 5 2 25Substitute for y in Equation 2: x 3 5x x 0Solve for x:Solve for x:5x2 20x 0 5x x 4 0 x 0, 4613x3 4x 0 x x2 4 0 x 0, 2Back-substitute x 0: y 2 0 5 5Back-substitute x 0: y 0 0Back-substitute x 4: y 2 4 5 3Back-substitute x 2: y 2Solutions: 0, 5 , 4, 3 Back-substitute x 2: y 2 2Solutions: 0, 0 , 2, 2 , 2, 2 11. x2 y 0x2 4x y 0Equation 112.Equation 2Solve for y in Equation 1: y x24 2x 2 1 Equation 22 x 4 2x 2 1 2x 2 2Solve for x: 2x2 4x 0 2x x 2 0 x 0, 2Back-substitute x 0: y y 2 xEquation 1Substitute for y in Equation 1:Substitute for y in Equation 2: x2 4x x2 0 02y 2x2 2Solve for x: x 4 2x 2 1 x 2 1 0 0x4 x2 0Back-substitute x 2: y 2 42x 2 x 2 1 0 x 0, 1Solutions: 0, 0 , 2, 4 Back-substitute x 0: y 2 0 2 2 2Back-substitute x 1: y 2 1 2 2 0Back-substitute x 1: y 2 1 2 2 0Solutions: 0, 2 , 1, 0 , 1, 0 13. y x 3x 1y x3 3x2 12Equation 114.Equation 2Substitute for y in Equation 2: y 2x 4Equation 1Equation 2Substitute for y in Equation 1: 2x 4 x 3 3x 2 4x3 3x2 1 x2 3x 1Solve for x: 0 x 3 3x 2 2xx3 4x2 3x 00 x x 2 3x 2 x x 1 x 3 0 x 0, 1, 3Back-substitute x 0: y 15.y x 3 3x 2 403 3 0 1 120 x x 2 x 1 x 0, 1, 2Back-substitute x 0: y 2 0 4 4Back-substitute x 1: y 13 3 1 2 1 1Back-substitute x 1: y 2 1 4 2Back-substitute x 3: y 33 3 3 2 1 1Back-substitute x 2: y 2 2 4 0Solutions: 0, 1 , 1, 1 , 3, 1 Solutions: 0, 4 , 1, 2 , 2, 0 x y 0 5x 3y 10Equation 1Equation 2Solve for y in Equation 1: y xSubstitute for y in Equation 2: 5x 3x 10Solve for x: 2x 10 x 5Back-substitute in Equation 1: y x 5Solution: 5, 5

61416.18.Chapter 7x 2y Systems of Equations and Inequalities1 5x 4y 2317. 2x y 2 0 4x y 5 0Equation 1Equation 2Equation 1Equation 2Solve for x in Equation 1: x 1 2ySolve for y in Equation 1: y 2x 2Substitute for x in Equation 2: 5 1 2y 4y 23Substitute for y in Equation 2: 4x 2x 2 5 0Solve for y: 14y 28 y 21Solve for x: 6x 3 0 x 2Back-substitute y 2: x 1 2y 1 2 2 311Back-substitute x 2: y 2x 2 2 2 2 3Solution: 3, 2 Solution: 6x 3y 4 0Equation 1x 2y 4 0Equation 2Solve for x in Equation 2: x 4 2ySubstitute for x in Equation 1: 6 4 2y 3y 4 04Solve for y: 24 12y 3y 4 0 15y 20 y 3444Back-substitute y 3: x 4 2y 4 2 3 3Solution: 43, 43 19. 1.5x 0.8y 2.3 0.3x 0.2y 0.1Equation 1Equation 2Multiply the equations by 10.15x 8y 23Revised Equation 13x 2y 1Revised Equation 231Solve for y in revised Equation 2: y 2 x 231Substitute for y in revised Equation 1: 15x 8 2 x 2 23Solve for x: 15x 12x 4 23 27x 27 x 131Back-substitute x 1: y 2 1 2 1Solution: 1, 1 20. 0.5x 3.2y 9.00.2x 1.6y 3.6Equation 1Equation 2Multiply the equations by 10.5x 32y 902x 16y 36Revised Equation 1Revised Equation 2Solve for x in revised Equation 2: x 8y 18Substitute for x in revised Equation 1: 5 8y 18 32y 90Solve for y: 40y 90 32y 90 72y 180 y 5255Back-substitute y 2: x 8 2 18 25Solution: 2, 2 12, 3

Section 7.121. xx yy 2081512Equation 1Equation 222.Solve for x in Equation 2: x 20 ySubstitute for x in Equation 1:Solve for y: 4 310 yBack-substitute y 15 20 8 y 403: y 8 34 y 10Equation 1 y 4Equation 23 3Substitute for y in Equation 1: 12x 4 4x 4 1040319Solve for x: 2x 16x 3 10 x 20 y 20 403 203Back-substitute x Solution: x6x 5yy 3 7Equation 1Equation 256Solve for x in Equation 2: x 7 24.56ySubstitute for x in Equation 1: 6 7 25.12x34x56y6153Solve for y in Equation 2: y 4x 412y40Solution: 203, 3 23.Linear and Nonlinear Systems of Equations20817 : 13 x 2081788y 34 20817 4 1788 20817 , 17 y 2 2 xx 3y 6231716 xEquation 1Equation 2Solve for y in Equation 1: y 23x 2 5y 3Substitute for y in Equation 2: 2x 3 23x 2 6Solve for y: 42 5y 5y 3 42 3 (False)Solve for x: 2x 2x 6 6 0 12 InconsistentNo solutionNo solutionx2 y 0 2x y 0Equation 1Equation 2Solve for y in Equation 2: y 2xSubstitute for y in Equation 1: x2 2x 0Solve for x: x2 2x 0 x x 2 0 x 0, 2Back-substitute x 0: y 2 0 0Back-substitute x 2: y 2 2 4Solutions: 0, 0 , 2, 4 26.x 2y 0 3x y2 0Equation 127.Equation 2x y 1Equation 1 y 4Equation 2 x2Solve for x in Equation 1: x 2ySolve for y in Equation 1: y x 1Substitute for x in Equation 2: 3 2y y 2 0Substitute for y in Equation 2: x 2 x 1 4Solve for y: 6y y 2 0 y 6 y 0 y 0, 6Solve for x: x 2 x 1 4 x 2 x 3 0Back-substitute y 0: x 2 0 0The Quadratic Formula yields no real solutions.Back-substitute y 6: x 2 6 12Solutions: 0, 0 , 12, 6 28. y x y x3Equation 1 3x2 2x29.Equation 2Substitute for y in Equation 2: x x 3 3x2 2x x 2y 2 y 3x y 15 y 3x 15ySolve for x: x 3 3x2 3x 0 x x2 3x 3 0 x 0, x 2y 265 3 i 32Point of intersection: 4, 3 4(4, 3)32Back-substitute x 0: y 0The only real solution is 0, 0 .x 22 2 1 2x12343x y 156

61630.Chapter 7Systems of Equations and Inequalities x y 03x 2y 1031.x 3y 2 y 13 x 2 5x 3y 17 y 13 5x 17 yy5x 3y 1714x 11234(2, 2) 21x y 0 3 4( 52 , 32 )x 3y 2 2 1 2 1x1 1233x 2y 10 2Point of intersection: 2, 2 Point of intersection: 32. x 2y 1x y 233. x y2y4 52, 32 x y 4 y x 42 4x 0 x 2 2 y 2 4yx y 46644(2, 2) x 2y 12(5, 3)2(4, 0) 246 4x y 2 x 6x x27 y y 32 02x 2 y 2 4x 035.x y 3 0 y x 3 y x 4x 7 y x 2 322yy12810(1, 4)(4, 7)68( 3, 0)x8Points of intersection: 2, 2 , 4, 0 Point of intersection: 5, 3 34.6 2x22(3, 6)6x y 3 0 4y x 2 4x 7x2 46 68 10 12x 2246 2 4 6Points of intersection: 3, 0 , 3, 6 36. y2 4x 11 0 12x y 12yPoints of intersection: 1, 4 , 4, 7 37.7x 8y 24 y 78x 3 x 8y 1x y 122918x8 y y(15, 7)7x 8y 24632(3, 1)(4, 21 )x69121518 3 2 6 9 2y2 4x 11 0Points of intersection: 3, 1 , 15, 7 xx 8y 8 41Point of intersection: 4, 2 1

Section 7.138. x y 05x 2y 639.yLinear and Nonlinear Systems of Equations 61733x 2y 0 y x2x2 y2 4 x2 y2 1444y3x 2y 032(2, 2)1 2 121x234x 1 4 3 1 2134 2 3Point of intersection: 2, 2 4x 2 y2 4No points of intersection No solution40. x 2x y y 4x3 002y254No points of intersectionso, no solution31 3x 11325 2 341. 3xx 16yy 250 y 222Algebraically we have:3 216 xx2 25 y2y163y6 416y 0(4, 3)3y2 16y 75 02 2x24 3y 25 y 3 06 24002y 253 x 9 , 4 6x 2 y 2 25 x 2 y 2 25 x 8 2 y 2 41Solutions: 4, 3 y43.121086Points of intersection: 3, 4 , 3, 4 6 8 10 12 y 3x 8y 0 4e x16 7608Point of intersection: 0.49, 6.53 106 2(3, 4) 6 8 10 12x 6x 2 x y 1y 0e y x 1Point of intersection: 0, 1 (3, 4)2 644.No real solutiony 3 x2 16Points of intersection: 4, 3 and 4, 3 42. 25 y216y 75 3y24( 4, 3) 63x 2

61845.Chapter 7 Systems of Equations and Inequalities1 y x 42ln xy log2 x y ln 2x 2y 846.y 2 ln x 1 3y 2x 945 1 21014 6Point of intersection: 5.31, 0.54 3Point of intersection: 4, 2 47. x2 y2 169 y1 169 x2 and y2 169 x2x2 8y 104 y3 18x2 1316Points of intersection: 0, 13 , 12, 5 2424 1648. x2 y2 4 y1 4 x2, y2 4 x22x2 y 2 y3 2x2 2Points of intersection:49. 662Equation 1 1Equation 2Solve for x: x2 2x 1 x 1 2 0 x 1Back-substitute x 1 in Equation 1: y 2x 2Solution: 1, 2 4 xx yy 242Substitute for y in Equation 2: 2x x2 14 0, 2 , 1.32, 1.5 , 1.32, 1.5 50.y 2x y xEquation 1Equation 2Solve for y in Equation 1: y 4 xSubstitute for y in Equation 2: x2 4 x 2Solve for x: x2 x 2 0No real solutions because the discriminant in the Quadratic Formula is negative.Inconsistent; no solution51. 3x 7y 6 0Equation 1x2 y2 4Equation 2 Solve for y in Equation 1: y 3x 67Substitute for y in Equation 2: x2 Solve for x:x2 9x2 3x 67 2 4 36x 36 449 Back-substitute x 49x2 9x2 36x 36 196Back-substitute x 2: y 40x2 36x 232 04 10x 29 x 2 0 x 3x 6 3(29 10) 6 2129:y 10771029, 210Solutions: 10, 10 , 2, 0 29 213x 6 3 2 6 077

Section 7.152. x2x yy 251022Linear and Nonlinear Systems of Equations619Equation 1Equation 2Solve for y in Equation 2: y 10 2xSubstitute for y in Equation 1: x2 10 2x 2 25Solve for x: x2 100 40x 4x2 25 x2 8x 15 0 x 5 x 3 0 x 3, 5Back-substitute x 3: y 10 2 3 4Back-substitute x 5: y 10 2 5 0Solutions: 3, 4 , 5, 0 53. x 2y 4 x y 0Equation 12Equation 2Solve for y in Equation 2: y x2Substitute for y in Equation 1: x 2x2 4Solve for x: 0 2x2 x 4 x 1 1 4 2 4 1 31 x 2 2 4The discriminant in the Quadratic Formula is negative.No real solution54. y x 1 3y x 1 55.y e x 1 y e x 1 y ln x 3y y ln x 3y362514x 3 11 12321 2x–2 –1 3 2345Point of intersection: approximately 0.287, 1.751 No points of intersection, so no solution56. x2 y 4 y 4 x2ex y 0 y ex157.y x4 2x2 1 y 1 x25Solve for x: x4 x2 0 x2 x2 1 0 x 0, 13Back-substitute x 0: 1 x2 1 02 12x 1Equation 2Substitute for y in Equation 1: 1 x2 x4 2x2 1y 3Equation 113 1Points of intersection (solutions):approximately 1.96, 0.14 , 1.06, 2.88 Back-substitute x 1: 1 x2 1 12 0Back-substitute x 1: 1 x2 1 1 2 0Solutions: 0, 1 , 1, 0

620Chapter 7Systems of Equations and Inequalities 58. y x3 2x2 x 1y x2 3x 1Equation 1Equation 259.Substitute for y in Equation 1: x2 3x 1 x3 2x2xy 1 0 2x 4y 7 0 x 11 x , 42y 02 3 0 1 1Back-substitute x 2 in Equation 2:11Back-substitute x : y 221 2y 22 3 2 1 1Back-substitute x 4: y Back-substitute x 1 in Equation 2:y 1 2 3 1 1 5Solutions:Solutions: 0, 1 , 2, 1 , 1, 5 x 1 2 x 1 x 1 2 4 x 1 x2 2x 1 4x 4x2 6x 5 0 x 1 x 5 0 x 1, 5Back-substitute x 1: y 1 1 0Back-substitute x 5: y 5 1 2Solutions: 1, 0 , 5, 2 62. C 5.5 x 10,000, R 3.29xR C3.29x 5.5 x 10,0003.29x 5.5 x 10,000 0Let u x.3.29u2 5.5u 10,000 0u 5.5 5.5 2 4 3.29 10,000 2 3.29 u 5.5 131,630.256.58Choosing the positive value for u, we havex u2 x 55.974 2 3133 units.11 44 2, 2 , 4, 4 1161. C 8650x 250,000, R 9950xSubstitute for y in Equation 1: x 2 x 1 1u 55.974, 54.30212x2 4 7x 0 2x 1 x 4 0Back-substitute x 0 in Equation 2:Solve for x: x 7 0Solve for x:0 x x 2 x 1 x 0, 2, 1Equation 1Equation 21xSubstitute for y in Equation 2: 2x 40 x x2 x 2 Equation 2Solve for y in Equation 1: y Solve for x: 0 x3 x2 2x60. x 2y 1y x 1Equation 1R C9950x 8650x 250,0001300x 250,000x 192 units

Section 7.163. C 35.45x 16,000, R 55.95x(a)Linear and Nonlinear Systems of Equations64. C 2.16x 5000, R 3.49xR CR C(a)55.95x 35.45x 16,0002.16x 5000 3.49x20.50x 16,0005000 1.33xx 781 units(b)621x 3760P R C3760 items must be sold to break even.60,000 55.95x 35.45x 16,000 (b) P R C60,000 20.50x 16,0008500 3.49x 2.16x 5000 76,000 20.50x8500 1.33x 500013,500 1.33xx 3708 units10,151 x10,151 items must be sold to make a profit of 8500.65. R 360 24x R 24 18xEquation 1Equation 2(a) Substitute for R in Equation 2: 360 24x 24 18xSolve for x: 336 42x x 8 weeks(b)66. (a)Weeks12345678910R 360 24x336312288264240216192168144120R 24 18x42607896114132150168186204S 25x 100 S 50x 475The rentals are equal whenx 8 weeks.Rock CDRap CD25x 100 50x 47575x 100 47575x 375x 5Conclusion: It takes 5 weeks for the sales of the two CDs to become equal.(b)Number of weeks, x0123456Sales, S (rock)100125150175200225250Sales, S (rap)47542537532527522517567. 0.06x 0.03x 3500.03x 350x 11,666.67To make the straight commission offer the better offer,you would have to sell more than 11,666.67 per week.By inspecting the table, we can see that thetwo sales figures are equal when x 5.68. p 1.45 0.00014x 210p 2.388 0.007x 2The market equilibrium(point of intersection) isapproximately 99.99, 2.85 .01500

622Chapter 7Systems of Equations and Inequalities69. (a) 0.06x 0.085y 2000(c) The point of intersection occurs when x 5000, sothe most that can be invested at 6% and still earn 2000 per year in interest is 5000.x y 25,000(b) y1 25,000 x27,0002000 0.06xy2 0.085As the amount at 6% increases, the amount at 8.5%decreases. The amount of interest is fixed at 2000.70.D 4 , VV 0.79D 2D 4,22(a)5 D 405 D 40012,00010,000Doyle Log RuleScribner Log Rule(b) The graphs intersect when D 24.7 inches.1500V1(c) For large logs, the Doyle Log Rule gives a greater volume for a given 31465768105108(a) Solar: C 0.1429t 2 4.46t 96.8(d) 0.1429t 2 4.46t 96.8 16.371t 102.7Wind: C 16.371t 102.7(b)0.1429t 2 20.831t 199.5 0By the Quadratic Formula we obtain t 10.3 andt 135.47.1508130(c) Point of intersection: 10.3, 66.01 (e) The results are the same for t 10.3. The other “solution”, t 135.47, is too large to consideras a reasonable answer.(f) Answers will vary.During the year 2000, the consumption of solarenergy will equal the consumption of wind energy.72. (a) For Alabama, P 17.4t 4273.2.4800For Colorado, P 84.9t 3467.9.(b) The lines appear to intersect at (11.93, 4480.79).Colorado’s population exceeded Alabama’s just after this point.(c) Using the equations from part (a),17.4t 4273.2 84.9t 3467.94273.2 67.5t 3467.9805.3 67.5t11.93 t.9400013

Section 7.173. 2l 2w 30 Linear and Nonlinear Systems of Equations74. 2l 2w 280l w 15 l w 140l w 3 w 3 w 15w l 20 l l 20 1402w 122l 160l 80w 6w l 20 80 20 60l w 3 9Dimensions: 60 80 centimetersDimensions: 6 9 meters76. 2l 2w 210 75. 2l 2w 42 l w 21w 34l l 34l 2174l 21l 32w l w 10532w w 10552wl 12w 34l 105w 42 9l Dimensions: 9 12 inches77. 2l 2w 40 62332 42 63Dimensions: 42 63 feetl w 20 w 20 l78. A 12bhlw 96 l 20 l 961 12a2a2 220l l 2 96a 20 l 2 20l 962aThe dimensions are 2 2 2 inches.0 l 8 l 12 l 8 or l 12aIf l 8, then w 12.If l 12, then w 8.Since the length is supposed to be greater than the width,we have l 12 kilometers and w 8 kilometers.Dimensions: 8 12 kilometers79. False. To solve a system of equations by substitution, youcan solve for either variable in one of the two equationsand then back-substitute.80. False. The system can have at most four solutionsbecause a parabola and a circle can intersect atmost four times.81. To solve a system of equations by substitution, use the following steps.1. Solve one of the equations for one variable in terms of the other.2. Substitute this expression into the other equation to obtain an equation in one variable.3. Solve this equation.4. Back-substitute the value(s) found in Step 3 into the expression found in Step 1 to find the value(s) of the other variable.5. Check your solution(s) in each of the original equations.82. For a linear system the result will be a contradictory equation such as 0 N, where N is a nonzero real number.For a nonlinear system there may be an equation with imaginary solutions.83. y x2(a) Line with two points of intersection(b) Line with one point of intersectiony 2xy 0 0, 0 and 2, 4 0, 0 (c) Line with no points of intersectiony x 2

624Chapter 7Systems of Equations and Inequalities84. (a) b 1b 26b 4b 366 66 6 266 6 666 2 2 2(b) Three85. 2, 7 , 5, 5 m 86. 3.5, 4 , 10, 6 5 72 5 2 7m 2y 7 x 2 7y 6 7y 49 2x 42x 7y 45 02 x 10 6.56.5y 39 2x 202x 6.5y 19 087. 6, 3 , 10, 3 m 6 42 10 3.5 6.588. 4, 2 , 4, 5 3 3 0 The line is horizontal.10 6x 4x 4 0y 3y 3 089. 5, 0 , 4, 6 m y 6 6 0630 4 3 5 17 5 17m 30 x 4 17y 17y 102 30x 12029y 0 30x 17y 1830x 17y 18 091. f x 5x 6Domain: All real numbers except x 6Horizontal asymptote: y 0Vertical asymptote: x 693. f x 75 190. , 8 , ,32 23x2 2x2 168 1 2 15 245 7 3 5 2 29 629 1455 x 2292 29225 45x 2245x 29y 127 092. f x 2x 73x 2Domain: All real numbers except x 232Vertical asymptote: x 3Horizontal asymptote: y 94. f x 3 2x2Domain: All real numbers except x 4.Domain: All real numbers except x 0Horizontal asymptote: y 1Horizontal asymptote: y 3Vertical asymptotes: x 4Vertical asymptote: x 023

Section 7.2Two-Variable Linear Systems625Two-Variable Linear SystemsSection 7.2 You should be able to solve a linear system by the method of elimination.1. Obtain coefficients for either x or y that differ only in sign. This is done by multiplying all the terms of one or bothequations by appropriate constants.2. Add the equations to eliminate one of the variables and then solve for the remaining variable.3. Use back-substitution into either original equation and solve for the other variable.4. Check your answer. You should know that for a system of two linear equations, one of the following is true.1. There are infinitely many solutions; the lines are identical. The system is consistent. The slopes are equal.2. There is no solution; the lines are parallel. The system is inconsistent. The slopes are equal.3. There is one solution; the lines intersect at one point. The system is consistent. The slopes are not equal.Vocabulary Check1. elimination2. equivalent3. consistent; inconsistent4. equilibrium price1. 2x y 5 x y 1Equation 1Equation 2Add to eliminate y: 3x 6 x 22. x 2y 4x 3y 1Equation 1Equation 2x 3y 1Add to eliminate x: x 2y 4Substitute x 2 in Equation 2: 2 y 1 y 1Solution: 2, 1 y5y 5 y 1Substitute y 1 in Equation 1: x 3 1 1 x 2x y 1Solution: 2, 1 43y21 2 1x12454x 3y 16 x 2y 42x y 5 3 4 6 4x 2 23.x y 0 3x 2y 1Equation 1y4Equation 23Multiply Equation 1 by 2: 2x 2y 0Add this to Equation 2 to eliminate y: x 1Substitute x 1 in Equation 1: 1 y 0 y 1Solution: 1, 1 23x 2y 1x y 0 4 3 2 1 2 3 4x234

6264.Chapter 7Systems of Equations and Inequalities 2x4x 3yy 213Equation 1Equation 2y2x y 36Multiply Equation 1 by 3: 6x 3y 946x 3y 9Add this to Equation 2 to eliminate y:4x 3y 214x 3y 212 3010xx2 x 34 2Substitute x 3 in Equation 1: 2 3 y 3 y 3Solution: 3, 3 5.x y 2 2x 2y 5 6. 3x 2y 36x 4y 14Equation 1Equation 2Equation 1Equation 2Multiply Equation 1 by 2: 2x 2y 4Multiply Equation 1 by 2: 6x 4y 6Add this to Equation 2: 0 9Add this to Equation 2: 6x 4y 66x 4y 14There are no solutions.0 yThere are no solutions.48y 2x 2y 56x 4y 141 4x 2 12 234x 2x y 242 2 43x 2y 3 47.3x 2y 6x 4y 10Equation 158.Equation 2 3x9x 3yy 155Equation 1Equation 2Multiply Equation 1 by 2 and add to Equation 2: 0 0Multiply Equation 2 by 3: 9x 3y The equations are dependent. There are infinitelymany solutions.Add this to Equation 1:Let x a, then y Solution: 3a 5 35 a .222 2y1x234 3x y 5856 2150 0Solution: a, 3a 5 , where a is any real number.3 3 2 1 9x 3y 3a y 5 y 3a 53x 2y 54a is any real number.9x 3y 15There are infinitely many solutions. Let x a.y53a, a where2215 6x 4y 109x 3y 15x 8 6 42 4 6 8468

Section 7.29. 9x 3y 1 3x 6y 5Equation 149x 3y 3Add to eliminate x: 21y 14 y 32in Equation 1: 9x 3 23Substitute y Solution:10. 5x 3y 18Equation 11Equation 2 2x 6y 10x 6y 362x 6y 2 1349x 3y 1 2 3 4y5x 3y 18 4Multiply Equation 1 by 2: 10x 6y 36Add this to Equation 2 to eliminate y:x 4 3 2 1x 13 13, 23 3x 6y 51 9x 18y 15 23627yEquation 2Multiply Equation 2 by 3 :Two-Variable Linear Systems2 62x 6y 1x 42 21 35 x 351212x35Substitute x 12 in Equation 2:412 3512 6y 1 y 363541Solution: 12 , 36 11. x 2y 4 x 2y 1Equation 112. 3x 5y 2 2x 5y 13Equation 2Add to eliminate y:2x 5y 1352Substitute x 3 in Equation 1: 3 3 5y 2 y 57Solution: 3, 75 2y 4 y 43Solution: 52, 34 13. 2x 3y 18Equation 1y 11Equation 2 5x 15 x 35x5Substitute x 2 in Equation 1:52Equation 2Add to eliminate y: 3x 5y 22x 5x Equation 114.x 7y 12 3x 5y 10Equation 1Equation 2Multiply Equation 2 by 3: 15x 3y 33Multiply Equation 1 by 3: 3x 21y 36Add this to Equation 1 to eliminate y:Add this to Equation 2 to eliminate x:17x 51 x 3Substitute x 3 in Equation 1:6 3y 18 y 4Solution: 3, 4 3x 21y 363x 5y 10 26y 26 y 1Substitute y 1 in Equation 1: x 7 12 x 5Solution: 5, 1

628Chapter 7Systems of Equations and Inequalities15. 3x 2y 10 2x 5y 1316.Equation 1Equation 2Multiply Equation 1 by 2 andEquation 2 by 3 :2r 4s 5Equation 1 16r 50s 55Equation 2Multiply Equation 1 by 8 : 16r 32s 40Add this to Equation 2 to eliminate r:6x 4y 20 6x 15y 9 16r 32s 4016r 50s 5515Substitute y 1 in Equation 1:18s s 563x 2 10 x 456Add to eliminate x: 11y 11 y 1Substitute s in Equation 1:2r 4 56 5 r 65Solution: 4, 1 Solution:17. 5u 6v 24 3u 5v 18Equation 118.Equation 2 2x3x 11y5y 49Equation 1Equation 2Multiply Equation 1 by 2 and Equation 2 by 3:Multiply Equation 1 by 5 and Equation 2 by 6:25u 30v 56, 56 8 6x6x 22y15y 27 18u 30v 108120Add to eliminate x:Add to eliminate v: 7u 12 u 127Substitute u 127 in Equation 1:7y 35 y 5108185 127 6v 24 6v 7 v 7Solution:6x 22y 8 6x 15y 27Substitute y 5 in Equation 1: 3x 11 5 4 127, 187 x 17Solution: 17, 5 19.95x 65 y 4 9x 6y 3Equation 120.Equation 2 x 3yy 3x4941838Equation 1Equation 2Multiply Equation 1 by 3:Multiply Equation 1 by 10 and Equation 2 by 2:18x 12y 40 18x 12y 6 94 x 3y 3894x 3y 38Add to eliminate x and y: 0 34Add these two together to obtain 0 0.InconsistentThe original equations are dependent. They have infinitelymany solutions.No solution31Set x a in 4x y 8 and solve for y.13The points on the line have the form a, 8 4a .21. xy 14 6x y 3Equation 1Equation 2Multiply Equation 1 by 6:3x y 62Add this to Equation 2 to eliminate y:518x 9 x 25Substitute x 18in Equation 2:518 y 353y 518 3Solution:,5 5

Section 7.222. 23x 16 y 234x y 4Two-Variable Linear Systems 23. 5x 6y 320x 24y 12Equation 1Equation 2Multiply Equation 1 by 6 : 4x y 4Add this to Equation 2: 4x y 44x y 40 0Equation 1Equation 2Multiply Equation 1 by 4: 24y 12 20x20x 24y 12Add to eliminate x and y: 0 0There are infinitely many solutions. Let x a.4a y 4 y 4 4aSolution: a, 4 4a where a is any real numberThe equations are dependent. There are infinitely manysolutions.Let x a, then5a 3 51 a .662 5a 6y 3 y Solution:24.7x 8y 14x 16y 126 a, 65 a 21 where a is any real number25. 0.05x 0.03y 0.21 0.07x 0.02y 0.16Equation 1Equation 2Multiply Equation 1 by 2: 6y 42 10x21x 6y 4812Add to eliminate y: 31x 90Add these two together to obtain 0 0.The original equations are dependent. They have infinitelymany solutions.x 903190Substitute x 31 in Equation 2:0.07 9031 0.02y 0.16Set x a in 7x 8y 6 and solve for y.37The points on the line have the form a, 4 8a .y 6731Solution:26. 0.3x 0.4y 0.2x 0.5y 27.868.7 3b 11m 130.8x 2y 111.2Equation 1Equation 2Multiply Equation 1 by 3 and Equation 2 by 4 :12b 9m 12b 44m 52343.5Add these to eliminate y: 0.8x 2y 111.29Add to eliminate b: 35m 431.5x 2y 343.5m 4335 232.3Substitute m 4335 in Equation 1: x 1012.3xSubstitute x 101 in Equation 1:0.2 101 0.5y 27.8 y 96Solution: 101, 96 9031, 6731 27. 4b 3m 3Equation 1Equation 2Multiply Equation 1 by 4 and Equation 2 by 5: 1.5x 2y Equation 1Equation 2Multiply Equation 1 by 200 and Equation 2 by 300:14x 14x 16y 12 16y 62964b 3 4335 3 b 356 43Solution: 35, 35

630Chapter 7Systems of Equations and Inequalities 28. 2x 5y 85x 8y 10Equation 1Equation 229.Multiply Equation 1 by 5 and Equation 2 by 2 :403x 4y 7Add to eliminate y: 11x 55 x 540Substitute x 5 into Equation 2: 10x 16y 209y Substitute y 20 y 2 5 y 12 y 2209Solution: 5, 2 2020 8in Equation 1: 2x 599 x Solution:30. Equation 2 8x 4y 48Add to eliminate x:10x 25y Equation 1Multiply Equation 1 by 12 and Equation 2 by 4: 10x 16y 2010x 25y x 3 y 1 1432x y 12149 149, 209 x 1 y 2 423Equation 1x 2y 5Equation 231. 2x 5y 0 x Multiply Equation 2 by 5:Multiply Equation 1 by 6:2x 5y 0 5x 5y 153 x 1 2 y 2 24 3x 2y 23Add this to Equation 2 to eliminate y:Add to eliminate y: 3x 15 x 53x 2y 23Matches graph (b).x 2y 5Number of solutions: One 284xy 3Consistent x 7Substitute x 7 in Equation 2: 7 2y 5 y 1Solution: 7, 1 33. 2x 5y 0 2x 3y 4 32. 7x 6y 414x 12y 8 7x 6y 4 6y 7x 4 y 76x 23;The graph contains 0

612 Chapter 7 Systems of Equations and Inequalities 3. (a) is nota solution. (b) is a solution. (c) is nota solution. (d) 1, 2 is nota solution. 2 2e 1 0, 13 3 2e0 0, 2 3 0 28 2 2 2 2e0 2, 0 0 2e 2 3x y 2 y 2ex 4. (a) is nota solution. (b) is a solution. (c) is a solution. (d) 2

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