Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 2 - Systems of Linear Equations and Inequalities - 2.5 Absolute Value Equations and Inequalities - 2.5 Exercises - Page 186: 20

Answer

$6.09$ hours or $7.18$ hours

Work Step by Step

William wants to know the time it will take him to be $50$ miles away from Boise. Substitute $30$ in place of $D(t)$ and solve for $t$. $$\begin{aligned} D(t) &= \lvert 365-55t\rvert\\ 30 &= \lvert 365-55t\rvert. \end{aligned}$$ Rewrite into two equations and solve. $$\begin{aligned} 365-55t&= 30\\ -55t&=30-365\\ -55t& = -335\\ 55t& = 335\\ t&= \frac{335}{55}\\ &= \frac{67}{11}\\ &\approx 6.09 \end{aligned}$$ or $$\begin{aligned} 365-55t&= -30\\ -55t&=-30-365\\ -55t& = -395\\ 55t& = 395\\ t&= \frac{395}{55}\\ &= \frac{79}{11}\\ &\approx 7.18. \end{aligned}$$ Check the solution. $$\begin{aligned} D(67/11) &= \lvert 365-55\cdot \frac{67}{11}\rvert\\\\ & = \lvert 365-5\cdot 67\rvert\\\\ & = \lvert 365-335\rvert\\ & = \lvert 30\rvert\\ & = 30 \end{aligned}$$ and $$\begin{aligned} D(79/11) &= \lvert 365-55\cdot \frac{79}{11}\rvert\\\\ & = \lvert 365-5\cdot 79\rvert\\\\ & = \lvert 365-395\rvert\\ & = \lvert -30\rvert\\ & = 30. \end{aligned}$$ Hence, William will be $30$ miles away from Boise at either $6.09$ hours later of $7.18$ hours later.
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