To solve for the speed of the current, let's denote the speed of the boat in still water as [tex]\(10\)[/tex] miles per hour and let [tex]\(x\)[/tex] be the speed of the current in miles per hour.
When the boat travels with the current, its effective speed is [tex]\((10 + x)\)[/tex] miles per hour. The boat covers a distance of [tex]\(6\)[/tex] miles with the current. The time taken to travel these 6 miles is:
[tex]\[
\text{time with current} = \frac{6}{10 + x}
\][/tex]
When the boat travels upstream against the current, its effective speed is [tex]\((10 - x)\)[/tex] miles per hour. The boat covers a distance of [tex]\(4\)[/tex] miles upstream. The time taken to travel these 4 miles is:
[tex]\[
\text{time upstream} = \frac{4}{10 - x}
\][/tex]
Since the times taken to travel both distances are the same, we can set up the following rational equation:
[tex]\[
\frac{6}{10 + x} = \frac{4}{10 - x}
\][/tex]
