Ex 3: Solve a System of Nonlinear Equations (Elimination) — Transcript

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1. 0:00- WELCOME TO ANOTHER EXAMPLE OF SOLVING A SYSTEM
2. 0:03OF NONLINEAR EQUATIONS.
3. 0:05WE WANT TO FIND ALL THE SOLUTIONS TO THE SYSTEM
4. 0:072X SQUARED - Y SQUARED = 23 AND X SQUARED + 2Y SQUARED = 34.
5. 0:15WE'LL FIRST SOLVE THIS SYSTEM ALGEBRAICALLY
6. 0:17USING SUBSTITUTION OR ELIMINATION
7. 0:20AND THEN WE'LL VERIFY THE SOLUTION GRAPHICALLY.
8. 0:22IN THE FIRST TWO EXAMPLES,
9. 0:24WE SOLVE THE SYSTEMS USING SUBSTITUTION
10. 0:27SO FOR THIS PROBLEM WE'LL USE ELIMINATION
11. 0:28OR THE ADDITION METHOD WHICH MEANS WE WANT TO ADD
12. 0:31THE TWO EQUATIONS TOGETHER TO ELIMINATE
13. 0:33EITHER THE X SQUARED TERMS OR THE Y SQUARED TERMS.
14. 0:36FOR THIS EXAMPLE, LET'S FOCUS ON ELIMINATING THE Y SQUARED TERMS.
15. 0:41SO WE WANT THE COEFFICIENTS OF THE Y SQUARED TERMS
16. 0:44TO BE OPPOSITES.
17. 0:45WELL RIGHT NOW IN THE FIRST EQUATION, WE HAVE -1Y SQUARED.
18. 0:49IN THE SECOND EQUATION WE HAVE 2Y SQUARED.
19. 0:52THEY'RE ALREADY OPPOSITE SIGNS BUT WE WANT THE Y SQUARED TERM
20. 0:55IN THE FIRST EQUATION TO BE -2Y SQUARED.
21. 0:59SO WE'RE GOING TO MULTIPLY THIS FIRST EQUATION BY +2
22. 1:04AND WE'LL LEAVE THE SECOND EQUATION THE SAME.
23. 1:07SO IF WE MULTIPLY THE FIRST EQUATION BY +2,
24. 1:09WE WOULD HAVE 4X SQUARED - 2Y SQUARED = 2 x 23 IS 46.
25. 1:19WE'LL LEAVE THE SECOND EQUATION THE SAME.
26. 1:25NOW I DO WANT TO MENTION TO USE THE ELIMINATION METHOD
27. 1:28FOR A NONLINEAR SYSTEM,
28. 1:30WE DO WANT TO MAKE SURE THE DEGREE OF THE Y TERMS
29. 1:32ARE THE SAME, MEANING BOTH OF THEM ARE Y SQUARED IN THIS CASE.
30. 1:35IF ONE WAS Y, ONE WAS Y SQUARED. WE WOULD NOT BE ABLE
31. 1:39TO ELIMINATE THE Y TERMS.
32. 1:41SO NOW WE'LL ADD THESE TWO EQUATIONS TOGETHER.
33. 1:44NOTICE WHEN WE DO THIS,
34. 1:45THE Y SQUARED TERMS ARE OPPOSITES
35. 1:47SO THEY'LL ADD TO ZERO.
36. 1:49SO WE'LL HAVE 5X SQUARED = 46 + 34 = 80.
37. 1:57TO SOLVE THIS FOR X, WE'LL FIRST DIVIDE BY 5 ON BOTH SIDES.
38. 2:04SO WE HAVE X SQUARED = 80 DIVIDED BY 5 = 16.
39. 2:10AT THIS POINT WE SHOULD RECOGNIZE
40. 2:12WE'LL HAVE TWO VALUES OF X THAT SATISFY THIS EQUATION.
41. 2:15SINCE WE HAVE X SQUARED AND WE WANT X
42. 2:17WE'LL TAKE THE SQUARE ROOT OF BOTH SIDES OF THE EQUATION
43. 2:20AND WE'LL INCLUDE A + OR - SIGN HERE TO MAKE SURE
44. 2:23WE OBTAIN BOTH SOLUTIONS.
45. 2:25SO THE SQUARE ROOT OF X SQUARED = X,
46. 2:29IT'S ACTUALLY THE ABSOLUTE VALUE OF X,
47. 2:30ANOTHER REASON WHY WE HAVE TWO SOLUTIONS.
48. 2:33AND THEN THE SQUARE ROOT OF 16 = 4.
49. 2:36SO X = EITHER + OR -4 OR + OR - 4.
50. 2:40AND NOW TO FIND THE CORRESPONDING VALUES OF Y
51. 2:43THAT GIVE US THE SOLUTIONS FOR THE SYSTEM,
52. 2:45WE'LL PERFORM BACK SUBSTITUTION.
53. 2:48LET'S GO AHEAD AND USE THE FIRST EQUATION
54. 2:50HERE TO FIND THE CORRESPONDING VALUES OF Y.
55. 2:54SO IF WE USE THE EQUATION 2X SQUARED - Y SQUARED = 23,
56. 3:00LET'S FIRST SOLVE THIS FOR Y SQUARED
57. 3:03SO WE'RE ACTUALLY GOING TO ADD Y SQUARED TO BOTH SIDES
58. 3:06AND SUBTRACT 23 ON BOTH SIDES
59. 3:08THAT WOULD GIVE US 2X SQUARED - 23 = Y SQUARED.
60. 3:14LET'S GO AHEAD AND SWITCH THIS AROUND.
61. 3:16WE HAVE Y SQUARED = 2X SQUARED - 23.
62. 3:22NOW WE KNOW THAT X CAN BE EITHER A +4 OR A -4.
63. 3:26LET'S SUBSTITUTE X = +4 FIRST
64. 3:29TO FIND THE CORRESPONDING VALUES OF Y.
65. 3:32SO WE WOULD HAVE Y SQUARED = 2 x 4 SQUARED - 23.
66. 3:40SO WE HAVE Y SQUARED = 4 SQUARED IS 16, 16 x 2 = 32.
67. 3:50WELL 32 - 23 = 9.
68. 3:52SO WE HAVE Y SQUARED = 9.
69. 3:56SO AGAIN TO SOLVE FOR Y
70. 3:57WE'LL TAKE THE SQUARE ROOT OF BOTH SIDES OF THE EQUATION.
71. 4:00WE'LL HAVE TWO SOLUTIONS, ONE POSITIVE AND ONE NEGATIVE.
72. 4:04SINCE THE SQUARE ROOT OF 9 = 3, WE'LL HAVE Y = + OR - 3.
73. 4:10SO FROM THIS OUTCOME, WE KNOW WE HAVE AT LEAST TWO SOLUTIONS.
74. 4:13WHEN X IS +4, Y CAN BE EITHER +3 OR -3.
75. 4:20SO THIS WOULD BE ONE SOLUTION AND THE SECOND SOLUTION
76. 4:23WOULD BE FOR -3.
77. 4:26REMEMBER WHEN GIVING THE SOLUTION AS ORDERED PAIRS,
78. 4:29THE FIRST COORDINATE IS ALWAYS THE X COORDINATE OR X VALUE
79. 4:32AND THE SECOND COORDINATE IS THE Y COORDINATE OR Y VALUE.
80. 4:37REMEMBER WE ALSO SAID THAT X CAN BE -4
81. 4:40SO USING THIS EQUATION AGAIN, WE'LL SUBSTITUTE -4 FOR X.
82. 4:47SO WE WOULD HAVE Y SQUARED = 2 x - 4 SQUARED - 23.
83. 4:55WELL -4 SQUARED = +4 SQUARED.
84. 4:59SO IF WE SIMPLIFY THIS,
85. 5:01WE'RE GOING TO GET Y SQUARED = 9 AGAIN.
86. 5:03SO IF WE TAKE THE SQUARE ROOT OF BOTH SIDES OF THE EQUATION,
87. 5:11AGAIN WE'LL HAVE Y = + OR -3.
88. 5:15SO THIS OUTCOME GIVES US TWO MORE SOLUTIONS.
89. 5:17WHEN X IS -4, Y CAN BE +3 OR -3.
90. 5:23SO THE THIRD SOLUTION WOULD BE (-4,3)
91. 5:27AND THE FOURTH SOLUTION WOULD BE (-4,-3).
92. 5:32SO THIS SYSTEM ACTUALLY HAS FOUR SOLUTIONS
93. 5:35GIVEN BY THESE ORDERED PAIRS.
94. 5:40AND TO VERIFY THIS WE'RE GOING TO GRAPH BOTH EQUATIONS
95. 5:42ON THE SAME COORDINATE PLANE AND OUR GRAPHS SHOULD INTERSECT
96. 5:46AT THESE FOUR POINTS.
97. 5:47LET'S TAKE A LOOK.
98. 5:50THE FIRST EQUATION IS ACTUALLY A HYPERBOLA IN RED
99. 5:53AND THE SECOND EQUATION IS ACTUALLY AN ELLIPSE IN BLUE.
100. 5:57NOTICE HOW WE DO HAVE FOUR POINTS OF INTERSECTION,
101. 6:00ONE, TWO, THREE, FOUR,
102. 6:05WHEREIN THE FIRST QUADRANT THE POINT OF INTERSECTION IS (4,3).
103. 6:09IN THE SECOND QUADRANT THE POINT OF INTERSECTION IS (-4,3).
104. 6:14IN THE THIRD QUADRANT THE INTERSECTION IS (-4,-3).
105. 6:19AND FINALLY IN THE FOURTH QUADRANT THE INTERSECTION
106. 6:21IS (4,-3).
107. 6:24SO THESE FOUR POINTS OF INTERSECTION VERIFY
108. 6:27THAT OUR SOLUTION IS CORRECT.
109. 6:29OKAY, I HOPE YOU FOUND THIS EXPLANATION HELPFUL.