Systems of equations

Question

Going agasint the current in my motorboat,  I can go 20 miles in 10 hours. Going with the same current, I can do the 32 miles in 8 hours.  How fast does my boat going still water ? Please work out the problem,  thanks

Answer 1

I am not going to work out the entire problem, but I will tell you that to attack a problem of this type, you should set up a chart as follows:

 

                       Rate            Time           Distance

With Current

Against Current

 

Fill in the table with the information that you have been given.  Let one unknown variable (X) be equal to the rate of the boat in still water, and the other (Y) be equal to the rate of the current.  Then your rate going with the current will be X + Y, and your rate going against the current will be X - Y.  The formula that ties this together is Distance = Rate x Time.