Applications Involving Systems of Equations — Transcript

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1. 0:01- LET'S TAKE A LOOK AT SOME APPLICATIONS INVOLVING
2. 0:03SYSTEMS OF EQUATIONS.
3. 0:08OUR FIRST EXAMPLE, FIND TWO NUMBERS FOR WHICH THE SUM IS 93
4. 0:14AND THE DIFFERENCE IS 9.
5. 0:18SO WE HAVE TWO UNKNOWNS BECAUSE WE HAVE TWO NUMBERS.
6. 0:21SO THE SUM, THAT WOULD MEAN ADDITION.
7. 0:24X + Y = 93 AND THEIR DIFFERENCE IS 9,
8. 0:32SO WE WOULD HAVE X - Y = 9.
9. 0:37NOW HERE'S OUR SYSTEM OF EQUATIONS.
10. 0:39REMEMBER WE HAVE A VARIETY OF METHODS FOR SOLVING THIS,
11. 0:43BUT WHAT STANDS OUT FOR ME IS THE ELIMINATION METHOD.
12. 0:47IF WE WERE TO ADD THESE TWO EQUATIONS TOGETHER
13. 0:49NOTICE THE Y TERMS ARE OPPOSITES.
14. 0:52THIS WOULD SIMPLIFY TO ZERO, THEREFORE IF WE ADD X + X = 2X.
15. 0:5893 + 9 = 102.
16. 1:02THIS WILL ALLOW US TO EASILY SOLVE FOR X BY DIVIDING BY 2.
17. 1:07X = 51.
18. 1:10NOW IN ORDER TO FIND THE OTHER NUMBER
19. 1:11WE NEED TO SUBSTITUTE X = 51 INTO EITHER OF THESE EQUATIONS.
20. 1:15I WILL GO AHEAD AND JUST USE THE FIRST EQUATION.
21. 1:19SO SINCE X = 51,
22. 1:23WE CAN SOLVE THIS EQUATION FOR Y BY SUBTRACTING 51 ON BOTH SIDES.
23. 1:28AND WE CAN SEE HERE THAT OUR SECOND NUMBER
24. 1:30WILL BE EQUAL TO 42.
25. 1:37SO WE CAN SAY THE TWO NUMBERS WHOSE SUM IS 93
26. 1:40AND WHOSE DIFFERENCE IS 9 WOULD BE 42 AND 51.
27. 1:49LET'S TAKE A LOOK AT ANOTHER.
28. 1:51THE PERIMETER OF A RECTANGLE IS 160 YARDS.
29. 1:55THE WIDTH IS 4 MORE THAN HALF THE LENGTH.
30. 1:58FIND THE LENGTH AND THE WIDTH.
31. 2:00SO LET'S LET X = WIDTH AND Y = LENGTH.
32. 2:05LET'S LABEL OUR RECTANGLE.
33. 2:08I'M GOING TO LABEL THIS Y FOR THE LENGTH
34. 2:12AND THIS X AS THE WIDTH.
35. 2:16SO THE FIRST SENTENCE STATES THE PERIMETER IS 160 YARDS.
36. 2:22SO REMEMBER THE PERIMETER IS THE DISTANCE AROUND THE OUTSIDE.
37. 2:27SO THE PERIMETER OF THIS WOULD BE 2X + 2Y MUST EQUAL 160.
38. 2:37THE SECOND SENTENCE STATES THE WIDTH
39. 2:38IS 4 MORE THAN HALF THE LENGTH.
40. 2:42SO THE WIDTH IS 4 MORE THAN HALF THE LENGTH.
41. 2:47WHEN I READ THE WIDTH IS,
42. 2:49THAT'S TELLING ME THAT I'M GOING TO HAVE AN EQUATION
43. 2:51IN THE FORM OF X =.
44. 2:55SINCE Y IS THE LENGTH, TO EXPRESS 4 MORE THAN HALF
45. 3:01I'LL START BY FINDING HALF OF THE LENGTH.
46. 3:03THAT WOULD BE 1/2Y,
47. 3:06BUT SINCE WE WANT 4 MORE THAN THAT I'D ADD 4 TO THAT.
48. 3:12HERE'S OUR SYSTEM OF EQUATIONS.
49. 3:14THE METHOD THAT STICKS OUT HERE FOR ME WOULD BE SUBSTITUTION
50. 3:17SINCE THIS IS ALREADY SOLVED FOR X.
51. 3:19SO WHEREVER WE SEE AN X IN THE FIRST EQUATION
52. 3:22WE CAN REPLACE IT WITH 1/2Y + 4.
53. 3:26SO LET'S GO AHEAD AND DO THIS SUBSTITUTION
54. 3:29AND THEN SOLVE THE RESULTING EQUATION.
55. 3:32SO WE WOULD HAVE 2 TIMES, INSTEAD OF X,
56. 3:361/2Y + 4 + 2Y = 160.
57. 3:45LET'S DISTRIBUTE AND SOLVE FOR Y.
58. 3:50WE'D HAVE Y + 8 + 2Y = 160.
59. 3:58SO WE'LL COMBINE OUR LIKE TERMS AND THEN SUBTRACT 8.
60. 4:04SO THAT WOULD GIVE US 3Y.
61. 4:06IF WE SUBTRACT 8 FROM BOTH SIDES THAT WOULD LEAVE US WITH 152,
62. 4:12DIVIDING BY 3, Y = 50 2/3 YARDS. NOW WE'RE NOT DONE.
63. 4:24WE STILL HAVE TO FIND X.
64. 4:28SO OVER WE HAVE AN EQUATION THAT RELATES X TO Y
65. 4:31SO LET'S GO AHEAD AND USE THIS.
66. 4:33X = 1/2 x Y + 4.
67. 4:41SINCE THIS IS IN FRACTION FORM I'M GOING TO GO AHEAD
68. 4:43AND USE THE FRACTION FORM OF Y OR 152/3.
69. 4:49SIMPLIFYING HERE.
70. 4:50THIS CHANGES TO 1 AND THIS CHANGES TO 76.
71. 4:55SO WE'D HAVE 76/3 + 4.
72. 5:00WELL 4 IS THE SAME AS 12/3 SO WE'D HAVE 88/3,
73. 5:07WHICH IS EQUAL TO 29 1/3 YARDS.
74. 5:13LET'S GO AHEAD AND WRITE THIS ONE MORE TIME TO CLEAN IT UP.
75. 5:16WE KNOW THE WIDTH NOW, WHICH IS X,
76. 5:19EQUAL TO 29 1/3 YARDS.
77. 5:24AND THE LENGTH, OR Y, IS EQUAL TO 50.
78. 5:29THIS ONE'S A LITTLE MORE INVOLVED
79. 5:30BECAUSE OF THE FRACTIONS, BUT I DON'T THINK IT WAS THAT BAD.
80. 5:34WHAT WE COULD DO IS REPLACE X AND Y IN THIS FIRST EQUATION
81. 5:37TO MAKE SURE THAT THEIR PERIMETER STILL WOULD BE 160,
82. 5:40SO I'M GOING TO GO AHEAD AND DO THAT. WE'D HAVE 2 x--
83. 5:44REMEMBER THAT 29 1/3 WAS THE SAME AS 88/3.
84. 5:52AND 50 2/3 WAS THE SAME AS 152/3.
85. 5:58THIS SHOULD GIVE US THE PERIMETER,
86. 6:00WHICH THEY STATED AS 160,
87. 6:03WHICH VERIFIES OUR SOLUTION IS CORRECT.
88. 6:06LET'S GO AHEAD AND TAKE A LOOK AT ONE MORE.
89. 6:08SUNSET RENTS AN 18 FOOT TRUCK
90. 6:12FOR $49.95 PLUS 75 CENTS PER MILE.
91. 6:15CACTUS RENTS AN 18 FOOT VAN FOR $59.95 PLUS 50 CENTS PER MILE.
92. 6:22FOR WHAT MILEAGE IS THE COST THE SAME?
93. 6:26SO WE'RE GOING TO HAVE TWO COST EQUATIONS,
94. 6:28ONE FOR SUNSET AND ONE FOR CACTUS.
95. 6:32FOR OUR FIRST EQUATION, THE COST,
96. 6:34C = 49.95 + 75 CENTS PER MILE.
97. 6:41SO WE HAVE A FIXED COST OF 49.95
98. 6:44AND A VARIABLE COST OF 75 CENTS PER MILE.
99. 6:48SO IF WE LET X = THE NUMBER OF MILES,
100. 6:52THE TOTAL COST FOR THE SUNSET TRUCK WOULD BE--
101. 6:56NOW WE NEED TO BE A LITTLE BIT CAREFUL HERE.
102. 6:57THIS IS IN CENTS AND THIS IS IN DOLLARS,
103. 6:59SO WE'RE GOING TO CONVERT 75 CENTS INTO DOLLARS.
104. 7:02SO 0.75 TIMES THE NUMBER OF MILES
105. 7:06PLUS THE FIXED COST OF 49.95.
106. 7:11NOW IF WE GO WITH THE 18 FOOT VAN FROM CACTUS,
107. 7:15THE COST IS GOING TO EQUAL 50 CENTS PER MILE.
108. 7:19SO 0.50 x X + THE FIXED COST OF 59.95.
109. 7:26SO THERE'S OUR SYSTEM OF EQUATIONS,
110. 7:29AND THE QUESTION ASKED, FOR WHAT MILEAGE IS THE COST THE SAME?
111. 7:34SO WE'RE TRYING TO FIND THE VALUES OF X AND C
112. 7:36THAT SATISFY BOTH EQUATIONS AT THE SAME TIME.
113. 7:40AGAIN, WE CAN SOLVE USING A VARIETY OF METHODS,
114. 7:43BUT WHAT STICKS OUT FOR ME IS SUBSTITUTION
115. 7:46BECAUSE BOTH EQUATIONS ARE ALREADY SOLVED FOR C.
116. 7:50SO SINCE C IS EQUAL TO THIS IN THE FIRST EQUATION,
117. 7:53WE'LL PLACE THE C IN THE SECOND EQUATION WITH THIS EXPRESSION
118. 7:57AND THEN SOLVE BY PERFORMING SUBSTITUTION.
119. 8:01SO THE RESULT WOULD BE THIS EQUATION.
120. 8:09NOW, AS BEFORE, IF WE WANTED TO ELIMINATE THE DECIMALS
121. 8:12WE COULD MULTIPLY THROUGH BY 100.
122. 8:14I'M NOT GOING TO DO THAT HERE.
123. 8:16I'M JUST GOING TO SOLVE BY ISOLATING X.
124. 8:18SO I'LL SUBTRACT 0.50X ON BOTH SIDES,
125. 8:27AND I'LL BE LAZY HERE AND ALSO SUBTRACT 49.95 ON BOTH SIDES.
126. 8:38NOTICE I HAVE PERFORMED THE SAME OPERATIONS
127. 8:41ON BOTH SIDES OF THE EQUAL SIGN.
128. 8:44SO WHEN WE COMBINE THE X TERMS WE'RE GOING TO HAVE 0.25X.
129. 8:50THIS SIMPLIFIES TO ZERO THAT'S WHY WE DID THAT.
130. 8:53THIS SIMPLIFIES TO ZERO THAT'S WHY WE DID THAT,
131. 8:56AND THEN LASTLY, IT LOOKS LIKE WE HAVE 10.
132. 8:59SO NOW WE CAN JUST DIVIDE BY 0.25.
133. 9:04X = 40, OR IN THIS CASE, 40 MILES.
134. 9:11THE RENTAL COST WILL BE THE SAME
135. 9:13WHEN THE MILEAGE DRIVEN IS EQUAL TO 40 MILES.
136. 9:18I HOPE THESE EXAMPLES WERE HELPFUL.
137. 9:21THANK YOU AND HAVE A GOOD DAY.