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Tammy drove to the mount last weekend. There was heavy traffic on the way there, the trip took 6 hours. When she drove home there was no traffic and it only took 4 hours to drive home. If her average rate was 22 miles per hour faster on the trip home, how far does she live from the mountains?

User Adam Smaka
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1 Answer

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Let's call the average rate on trip to the mount as x, the average rate on the trip back home as y, and the distance to the mount as d.

The distance to the mount, is equal to the average rate times the amount of time it took to do the trip, then, the relation between the distance and the rates can be written as


\begin{gathered} 6x=d \\ 4y=d \end{gathered}

We also know that her average rate was 22 miles per hour faster on the trip home, which means that


y=x+22

If we substitute this y value in our previous set of equations, we're going to have


4(x+22)=d

Expanding this result and writing x as a function of d:


\begin{gathered} 4(x+22)=d \\ x+22=(d)/(4) \\ x=(d)/(4)-22 \end{gathered}

Using this x value in our first equation, we're going to get an equation only for d.


\begin{gathered} 6x=d \\ 6((d)/(4)-22)=d \\ (6)/(4)d-d=132 \\ ((3)/(2)-1)d=132 \\ (d)/(2)=132 \\ d=264 \end{gathered}

She lives 264 miles away from the mountain.

User Paul Nearney
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