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Two buses leave a station at the same time and travel in opposite directions. One bus travels 16(km)/(h) slower than the other. If the two buses are 1040 kilometers apart after 4 hours, what is the rate of each busSolve using a system of linear equations

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Let x be the velocity (rate of change ) of one of the buses. We know that the other one travels 16 km/h slower; this means that the second velocity is:


x-16

Now the combined velocity would be:


2x-16

We know that the distance is equal to time by velocity, then we have that:


4(2x-16)=1040

Solving for x we have:


\begin{gathered} 4(2x-16)=1040 \\ 2x-16=(1040)/(4) \\ 2x-16=260 \\ 2x=260+16 \\ 2x=276 \\ x=(276)/(2) \\ x=138 \end{gathered}

Therefore the rate of the faster bus is 138 km/h and the rate for the slower bus is 122 km/h.

User Ola Sundell
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