the following exercises, translate to a system of equations and solve

Marcus can drive his boat 36 miles down the river in three hours but takes four hours to return upstream. Find the rate of the boat in still water and the rate of the current.

Based on the question we have to find out rate of the boat in still water and the rate of the current.

Take x= the rate of the boat in still water.

y= the rate of the current

The boat goes downstream and then upstream. Going downstream,

then boat’s actual rate is x + y. Going upstream, the actual rate is x-y

The distance is 36 miles and downstream takes 3 hours, but while upstream it takes 4 hours.

We know that

here system of equations are $$\begin{gathered} 3\mathrm{x}+3\mathrm{y}=36 \\ 4\mathrm{x}-4\mathrm{y}=36 \end{gathered}$$

To solve this equation multiply first equation by 4 and second equation by 3 and then add , thus we get

$$\begin{gathered} 3\mathrm{x}+3\mathrm{y}=36\left(1\right) \\ 4\mathrm{x}-4\mathrm{y}=36\left(2\right) \\ \left(1\right)\times 4+\left(2\right)\times 3,\mathrm{then}\mathrm{we}\mathrm{get}\mathrm{value}\mathrm{for}\mathrm{x} \\ \Rightarrow 12\mathrm{x}+12\mathrm{y}=144 \\ 12\mathrm{x}-12\mathrm{y}=10 \\ \Rightarrow 24\mathrm{x}=252 \\ \mathrm{x}=10.5 \\ \mathrm{substitute}\mathrm{x}=10.5\mathrm{in}\left(1\right)\mathrm{we}\mathrm{get}\mathrm{value}\mathrm{for}\mathrm{y} \\ 3\times 10.5+3\mathrm{y}=36 \\ 3\mathrm{y}=36-31.5 \\ \mathrm{y}=1.5 \end{gathered}$$

The downstream rate would be 10.5 + 1.5 = 12 mph. In 3 hours the boat would travel $12\times 3=36$miles.

The upstream rate would be 10.5 − 1.5 = 9mph. In 4 hours the boat would travel $9\times 4=36$miles

Hence the rate of boats in still water is 10.5mph and the rate of current is 1.5mph