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Unformatted text preview: the Intermediate Value Theorem - but in this case, applied to functions of more
than one variable.
Another way to see this is that points on the circle satisfy x2 + y 2 − 4 = 0, so they do not satisfy x2 + y 2 − 4 < 0. 542 Systems of Equations and Matrices
parabola. The points on the parabola itself are also part of the solution, since the vertex
(−4, 0) satisﬁes the inequality. We now turn our attention to x < y + 2. Proceeding as before,
we write x − y − 2 < 0 and focus our attention on x − y − 2 = 0, which is the line y = x − 2.
Using the test points (0, 0) and (0, −4), we ﬁnd points in the region above the line y = x − 2
satisfy the inequality. The points on the line y = x − 2 do not satisfy the inequality, since
the y -intercept (0, −2) does not. We see that these two regions do overlap, and to make the
graph more precise, we seek the intersection of these two curves. That is, we need to solve
the system of nonlinear equations
(E 1) y 2 = x + 4
(E 2) y = x − 2
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