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Unformatted text preview: ide’ of the circle - with the circle itself as the
boundary between the two. Just like we used test values to determine whether or not an interval
belongs to the solution of the inequality, we use test points in the each of the regions to see which
of these belong to our solution set.7 We choose (0, 0) to represent the region inside the circle and
(0, 3) to represent the points outside of the circle. When we substitute (0, 0) into x2 + y 2 − 4 < 0,
we get −4 < 4 which is true. This means (0, 0) and all the other points inside the circle are part of
the solution. On the other hand, when we substitute (0, 3) into the same inequality, we get 5 < 0
which is false. This means (0, 3) along with all other points outside the circle are not part of the
solution. What about points on the circle itself? Choosing a point on the circle, say (0, 2), we get
0 < 0, which means the circle itself does not satisfy the inequality.8 As a result, we leave the circle
dashed in the ﬁnal diagram.
2 −2 2 x −2 The solution to x2 &...
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