Factoring we get s 1 r a 1 rn assuming r 1 we

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Unformatted text preview: 2) = x E 3 (1 − 2)2 + y 2 = 1 Equation E 1 gives us x − 2 = x or −2 = 0, which is a contradiction. This means we have no solution to the system in this case, even though E 3 is solvable and gives y = 0. Hence, our final answer is {(1, 0, −1), (3, 0, 3)}. These points are easy enough to check algebraically in our three original equations, so that is left to the reader. As for verifying these solutions graphically, they require plotting surfaces in three dimensions and looking for intersection points. While this is beyond the scope of this book, we provide a snapshot of the graphs of our three equations near one of the solution points, (1, 0, −1). Example 8.7.2 showcases some of the ingenuity and tenacity mentioned at the beginning of the section. Sometimes you just have to look at a system the right way to find the most efficient method to solve it. Sometimes you just have to try something. 538 Systems of Equations and Matrices Next, we use systems of nonlinear equations to solve some classic...
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