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**Unformatted text preview: **solutions. We
4 As
are 536 Systems of Equations and Matrices
√ √ left with ex = −1+ 5 , so that x = ln −1+ 5 . We now return to y = 1 − 4e2x to ﬁnd the
4
4
accompanying y values for each of our solutions for x. For x = − ln(2), we get
y=
=
=
=
=
For x = ln √
−1+ 5
4 1 − 4e2x
1 − 4e−2 ln(2)
1 − 4eln(1/4)
1−4 1
4
0 , we have
y = 1 − 4e2x
2 ln = 1 − 4e ln = 1 − 4e
= 1−4
= 1−4
= √
−1+ 5
2
√ √
−1+ 5
4
√
−1+ 5
4 2 √2
−1+ 5
4
√
3− 5
8 √ We get two solutions, (0, − ln(2)), ln −1+ 5 , −1+ 5 . It is a good review of the prop4
2
erties of logarithms to verify both solutions, so we leave that to the reader. We are able to
sketch y = 1 − 4e2x using transformations, but the second equation is more diﬃcult and we
resort to the calculator. We note that to graph y 2 + 2ex = 1, we need to graph both the
√
positive and negative roots, y = ± 1 − 2ex . After some careful zooming,2 we conﬁrm our
solutions. √
The graphs of y = 1 − 4e2x and y = ± 1 − 2ex .
3. Our last system involves three variables and gives some insight on how to keep such systems
organized...

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