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PP Section 10.6

# PP Section 10.6 - -= = Multiply(1 by-1(1(2 = = 2 6 12 2 3 4...

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Nonlinear Systems of Equations

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We want to solve systems of equations that involve non-linear equations. Geometrically, what we want to do is to find the points of intersection of two curves. = - = + 0 2 4 : Solve 2 2 y x y x For example, consider the system We want to find the points of intersection of the line and the circle.
= - = + ) 2 ( 0 2 ) 1 ( 4 : Solve 2 2 y x y x Solve for y in (2). (3) 2 0 2 x y y x = = - Substitute y = 2x in (1). 4 5 4 4 ) 2 ( 2 2 2 2 2 = = + = + x x x x x Solve for x . 5 2 5 4 4 5 2 ± = ± = = x x Since from (2) we can solve for y , the substitution method is the most appropriate.

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Use (3) to find y . 5 4 5 2 2 2 ± = ± = = y x y - - 5 4 , 5 2 , 5 4 , 5 2 : Solution Check your answer. 4 5 20 5 16 5 4 5 4 5 2 2 2 = = + = ± + ±
Solve: (1) (2) 2 6 12 0 2 3 4 4

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Unformatted text preview: -=-+ + = Multiply (1) by -1 (1) (2)--+ =-+ + = 2 6 12 2 3 4 4 2 2 2 2 x y y x y y Add (1) and (2)- + + = 9 16 4 2 y y Use the elimination method when both x and y are squared in each equation. ( 29 ( 29- + + = - - - = 9 16 4 9 2 2 2 y y y y y y = -= 2 9 2 or For y = -2/9: 12 6 2 2 2 =-+ y y x 2 6 2 9 12 2 9 2 2 x +- -- = 2 80 27 2 x + = No Solution For y = 2: 12 6 2 2 2 =-+ y y x 2 6 2 12 2 2 2 x + - = ( ) ( ) 2 24 24 2 x +-= 2 0 2 x = x = 0 Solution: (0, 2)...
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