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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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