Unformatted text preview: − (E 3)
−z = −6 Finally, we apply the second move from Theorem 8.1 one last time and multiply E 3 by −1
to satisfy the conditions of Deﬁnition 8.3 for the variable z . 5 (E 1) x − y + z =
(E 2)
y − 1 z = −3
2 (E 3)
−z = −6 5 (E 1) x − y + z =
1
(E 2)
y − 2 z = −3
−− − − − − −→
−−−−−−− (E 3)
z=
6
Replace E 3 with −1E 3 Now we proceed to substitute. Plugging in z = 6 into E 2 gives y − 3 = −3 so that y = 0.
With y = 0 and z = 6, E 1 becomes x − 0 + 6 = 5, or x = −1. Our solution is (−1, 0, 6).
We leave it to the reader to check that substituting the respective values for x, y , and z into
the original system results in three identities. Since we have found a solution, the system is
consistent; since there are no free variables, it is independent.
2. Proceeding as we did in 1, our ﬁrst step is to get an equation with x in the E 1 position with
1
1 as its coeﬃcient. Since there is no easy ﬁx, we multiply E 1 by 2 . (E 1) 2x + 3y − z = 1
(E 2)
10x − z...
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 Fall '13
 Wong
 Algebra, Trigonometry, Cartesian Coordinate System, The Land, The Waves, René Descartes, Euclidean geometry

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