Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 2 - Solving Equations - 2-4 Solving Equations With Variables on Both Sides - Practice and Problem-Solving Exercises - Page 106: 41

Answer

a) $2$ hours b) $d=40t+40$ c) $120$ miles; $48\text{ mi/hr}$

Work Step by Step

Using $d=rt$, with $r_{\text{visit}}=60$, then $$ d_{\text{visit}}=60t .$$ With $r_\text{return}=40$, $$ d_{\text{return}}=40(t+1) .$$ a) Since the distances of the trip visit and the return trip are the same, then $$\begin{aligned} d_\text{visit}&=d_\text{return} \\ 60t&=40(t+1) \\ 60t&=40t+40 \\ 20t&=40 \\ t&=2 .\end{aligned}$$Hence, the travel time, $t$, to visit is $2$ hours. b) The return trip, $d_\text{return}$, is given by $$\begin{aligned} d_\text{return}&=40(t+1) \\&= 40t+40 .\end{aligned}$$Hence, in terms of $d$, the return trip is given by $d=40t+40$. c) Using $d_\text{visit}=60t$, with $t=2$, then $$ d_\text{visit}=60(2)=120 .$$Hence, the distance the family drove to visit their relatives is $120$ miles. The total distance of the entire trip is $2(120)=240\text{ mi}$. The total time of the entire trip is $t+(t+1)=2+3=5\text{ hours}$. Hence, the average rate of the entire trip is $$\begin{aligned} d&=rt \\ 240&=r(5) \\ r&=48 .\end{aligned}$$Hence, the average rate, $r$, of the entire trip is $48\text{ mi/hr}$.
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