algebra 2: driver drives 72 miles before running out of gas

[lesliebrown892]

a motorist drives 72 miles before running out of gas then walking 4 miles to a gas station. the driving rate of the motorist was 12 times the walking rate. the time spent driving and walking was 2.5 hours. find the rate at which the motorist walks

 

[soroban]

Re: algebra 2 word problem

Hello, lesliebrown892!

A motorist drives 72 miles before running out of gas, then walking 4 miles to a gas station.
The driving rate of the motorist was 12 times the walking rate.
The time spent driving and walking was 2.5 hours.
Find the rate at which the motorist walks

\(\displaystyle \text{We will use: }\text{Distance} \:=\:\text{Speed} \times \text{Time} \quad\Rightarrow\quad T \:=\:\frac{D}{S}\)

\(\displaystyle \begin{array}{c}\text{Let }\,w = \text{walking speed}\\ \\[-4mm] \text{Then: }\,12w = \text{driving speed} \end{array}\)


\(\displaystyle \begin{array}{c}\text{He drove 72 miles at }12w\text{ mph.} \\ \\[-3mm] \text{This took: }\:\frac{72}{12w} \:=\:\frac{6}{w}\text{ hours.} \end{array}\)

\(\displaystyle \begin{array}{c}\text{He walked 4 miles at }w\text{ mph.} \\ \\[-3mm] \text{This took}\:\frac{4}{w}\text{ hours.} \end{array}\)


\(\displaystyle \begin{array}{cccc}\text{The total time is }\frac{5}{2}\text{ hours.} \\ \text{There is our equation!}& \hdots & \dfrac{6}{w} + \dfrac{4}{w} \:=\:\dfrac{5}{2} \end{array}\)


Can you finish it now?