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4/11/2012 7.4 Systems of Nonlinear Equations in Two Variables. Objective 1: Recognize systems of nonlinear equations in two Objective 1: Recognize systems of nonlinear equations in two variables. A system of two nonlinear equations in two variables, also called a nonlinear system, contains at least one equation that cannot be expressed in the form Ax+By=C. A solution of a nonlinear system in two variables is an ordered pair of real numbers that satisfies both equations in the system. The solution set of the system is the set of all such ordered pairs. 1 4/11/2012 Objective 2: Solve nonlinear systems by substitution. The substitution method involves converting a nonlinear system to one equation in one variable by an appropriate substitution. The steps in the solution process are exactly the same as those used to solve a linear system by substitution. Since at least one equation is nonlinear you may get more than one point of intersection. Solve by substitution y-x=-2 y=x 2 − 4 y – x = ‐2 2 2 x – x – 2 = 0 Since the second equation has y = x2 – 4, we can substitute x2 – 4 for the y in the first equation. 2 – x – 2 = 0 for x. Now, solve x N l 2 0f (x – 2)(x + 1) = 0 x – 2 = 0 or x + 1 = 0 x = 2 or x = ‐1 Use y – x = ‐2 to find the y‐values for x = 2 and x = ‐1. y – 2 y 2 = ‐2 2 y y – ((‐1) 1) = ‐2 2 y = 0 y + 1 = ‐2 y = ‐3 The solutions are: (2, 0) and (‐1, ‐3) Sometimes when solving nonlinear systems, you may get extra solutions that do not satisfy both equations in the system. Because of this, you should check all solution pairs in each of the two equations. 2 4/11/2012 Solve each system by the substitution method ⎧ x − y = −1 ⎨ 2 ⎩ y = x +1 ⎧2 x + y = −5 ⎨ 2 ⎩ y = x + 6x + 7 3 4/11/2012 ⎧⎪ y = x 2 + 4 x + 5 ⎨ 2 ⎪⎩ y = x + 2 x − 1 ⎧ xy = −12 ⎨ ⎩ x − 2 y + 14 = 0 4 4/11/2012 Objective 3: Solve nonlinear systems by addition. For nonlinear systems, the addition method can be used when each equation is in the form Ax 2 + By 2 = C. If necessary, we will multiply either equation or both equations by appropriate numbers so that the coefficients coefficients of x 2 or y 2 will have a sum of 0. 0 Again you may get more than one point of intersection. Solve each system by the addition method. Solve the system: x 2 + y 2 = 16 y = x −4 2 x2 + y2 = 16 ‐x2 + y = ‐4 y + y2 = 12 Arrange like terms in columns. Add the equations. Now, solve this quadratic equation. y2 + y – 12 = 0 Next substitute each of the values (y + 4)(y – 3) = 0 for y into either of the original y = ‐4 or y = 3 equations. y = x2 – 4 y = x y = x 4 y = x2 ‐ 4 2 ‐4 = x ‐4 3 = x2 ‐ 4 x2 = 0 7 = x2 x = 0 x = ‐√7 and x = √7 The solutions are: (0, ‐4), (‐√7, 3), and √7, 3) 5 4/11/2012 ⎧⎪4 x 2 − y 2 = 4 ⎨ 2 2 ⎪⎩4 x + y = 4 ⎧⎪16 x 2 − 4 y 2 − 72 = 0 ⎨ 2 2 ⎪⎩ x − y − 3 = 0 6 4/11/2012 ⎧⎪ x 2 + y 2 = 5 ⎨ 2 2 ⎪⎩ x + ( y − 8) = 41 Solve each system by the method of your choice. ⎧⎪ x 2 − y 2 − 4 x + 6 y − 4 = 0 ⎨ 2 2 ⎪⎩ x + y − 4 x − 6 y + 12 = 0 7

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