Stitz-Zeager_College_Algebra_e-book

# As usual some valuable lessons are to be learned

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Unformatted text preview: t of 1 on the leading variable in E 2 (E 1) x + y + w = 500 3 (E 2) 3 y − 10 w = 50 5 Replace E 2 with 5 E2 − − − − − −3− −−−−−−→ (E 1) x + y + w = 500 1 (E 2) y − 2 w = 250 3 Notice that we have no equation to determine w, and as such, w is free. We set w = t and from E 2 1 1 3 get y = 2 t + 250 . Substituting into E 1 gives x + 2 t + 250 + t = 500 so that x = − 2 t + 1250 . This 3 3 3 3 1250 1 250 system is consistent, dependent and its solution set is { − 2 t + 3 , 2 t + 3 , t : −∞ < t < ∞}. While this answer checks algebraically, we have neglected to take into account that x, y and w, being amounts of acid and water, need to be nonnegative. That is, x ≥ 0, y ≥ 0 and w ≥ 0. The 3 1 constraint x ≥ 0 gives us − 2 t + 1250 ≥ 0, or t ≤ 2500 . From y ≥ 0, we get 2 t + 250 ≥ 0 or t ≥ − 500 . 3 9 3 3 The condition z ≥ 0 yields t ≥ 0, and we see that when we take the set theoretic intersection of these intervals, we get 0 ≤ t ≤ 2500 . Our ﬁnal answer is { − 3 t...
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