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Section 10-8 Systems of second-degree equations and inequalities If you had three nonconcentric (not having the same center)circles, how many points of intersections can you possibly have? 2 3 1 0 4 5 6 The number of solutions of a system of second degree equations equals the number of times the graphs of the equations intersect. If the system is composed of three circles you can have 0, 1,2,3,4,5,or 6 possible solutions. If the system is composed of a line and a circle there may be 0, 1 or 2 solutions. Line and a conic: Possible solutions 0, 1, 2 Two conics: 0, 1, 2, 3, or 4 (example: parabola and ellipse) While you can determine the possible number of solutions, to find the exact solution you must use algebra. 2 2 2 Graph the system of equations 5x - 4y =20 and 4x + 16y 2= 64. Use the graph to find approximate solutions. Then solve it algebraically. We have an ellipse and a hyperbola. Putting them in standard form, we have (x-0) 2 /16 + (y-0) 2 /4 = 1 center (0,0) vertices (±4,0) and (0,±2) Also (x-0) 2 /4 - (y-0)2 /5 = 1 Hyperbola with center (0,0) vertices (±2,0) 2 -4 -2 2 4 Approximate intersection points (±2.5, 1.5) and (±2.5, -1.5 To get the exact points of intersection, we need to solve the system algebraically. 5x - 4y = 20 2 2 4x + 16y = 64 2 2 2 2 Multiple thru by 4 20x -16y = 80 2 2 4x +16y =64 __________________ 2 2 24x = 144 x =6 so x = ± 6 2 2 2 Now find y. 5(6)-4y = 20 30 -4y =20 -4y = -10 y 2 = 10/4 y=± 10 /2 So the points of intersection are (± 6 , 10 /2 ) and (± 6 , - 10 /2) There are 4 solutions! 2 Graph the solutions for the system of inequalities. y-1 > 2x and x2 /9 + y /4 < 1. 2 The first one is a parabola and the second one is an ellipse. Both boundary lines are dashed. 2 -3 3 -2 Section 10-8 Pp. 682-684 #13-23 odds, 34,36,45