**Unformatted text preview: **always positive, we need to restrict our values of
s and t. We know R3 = s > 0 and when we combine that with R1 = −0.7s + 1.7 > 0,
we get 0 < s < 16 . Similarly, R6 = t > 0 and with R4 = −0.6t + 1.6 > 0, we ﬁnd
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0 < t < 8 . In order visualize the inequality R2 = 3.5s + 1.5t − 4 > 0, we graph the
3
line 3.5s + 1.5t − 4 = 0 on the st-plane and shade accordingly.8 Imposing the additional
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conditions 0 < s < 16 and 0 < t < 3 , we ﬁnd our values of s and t restricted to the region
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depicted on the right. Using the roster method, the values of s and t are pulled from the region
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(s, t) : 0 < s < 16 , 0 < t < 3 , 3.5s + 1.5t − 4 > 0 . The reader is encouraged to check that
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the solution presented in 1(a), namely all resistance values equal to 1, corresponds to a pair
(s, t) in the region.
8 See Section 2.4 for a review of this procedure. 504 Systems of Equations and Matrices s s 3
2 2 1 −2 3 1 −1 1 2 4 −1 t −2 −1 s= 1 2 4 −1...

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