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**Unformatted text preview: **E1 + E2 (E 1) x2 + y 2 = 4
(E 2) y 2 + y = 4
√ From y 2 + y = 4 we get y 2 + y − 4 = 0 which gives y = −1± 17 . Due to the complicated
2
nature of these answers, it is worth our time to make a quick sketch of both equations to head
oﬀ any extraneous solutions we may encounter. We see that the circle x2 + y 2 = 4 intersects
the parabola y = x2 exactly twice,√and both of these points have a positive y value. Of the
two solutions for y , only y = −1+ 17 is positive, so to get our solution, we substitute this
2
1 We encourage the reader to solve the system using substitution to see that you get the same solution. 534 Systems of Equations and Matrices
√ √
−1+ 17
2 √ = 0 and solve for x. We get x = ±
= ± −2+2 17 . Our solution is
2
√
√
√
√
√
−2+2 17 −1+ 17
−2+2 17 −1+ 17
,−
, which we leave to the reader to verify.
,
,
2
2
2
2 into y −
√ x2 y y 1 −3 −2 −1 1 1 2 3 −3 −2 x −1 1 2 3 x −1 Graphs for x2 + y 2 = 4
y − 2x = 0 Graphs for x2 + y 2 = 4
y − x2 = 36 A couple of remarks about Example 8.7.1 are in order. First note that, unlike systems of linear
equations, it is possible for a system of non-linear equations to have more than one solution without
having inﬁnitely many solutio...

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