Stitz-Zeager_College_Algebra_e-book

3 and simplify some familiar equations arise which

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Unformatted text preview: oe − river speed = 6 miles − R miles hour hour Proceeding as before, we get 5 miles = rate traveling upstream · time traveling upstream 5 miles = (6 − R) miles · time traveling upstream hour The last piece of information given to us is the total trip lasted 3 hours. If we let tdown denote the time of the downstream trip and tup the time of the upstream trip, we have: tdown + tup = 3 hours. Substituting tdown and tup into the ‘distance-rate-time’ equations, we get (suppressing the units) the following system of three equations and three unknowns: E 1 (6 + R) tdown = 5 E2 (6 − R) tup = 5 E3 tdown + tup = 3 We begin by solving E 1 for tdown by dividing both sides by the quantity (6 + R). While we usually discourage dividing both sides of an equation by a variable expression, we know (6 + R) = 0 since 5 otherwise we couldn’t possibly multiply it by tdown and get 5. Hence, tdown = 6+R . Similarly, we 5 solve E 2 for tup and get tup = 6−R . Substituting these into E 3, we get:...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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