Question 414524
Let {{{s}}} = the speed of the boat in still water in mi/hr
Let {{{d}}} = the one-way trip in miles 
Going downstream, speed = {{{s + 5}}} mi/hr
Going upstream, speed = {{{s - 5}}} mi/hr
given:
downstream:
{{{d = (s + 5)*2}}}
upstream:
{{{d = (s - 5)*4}}}
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{{{d}}} is the same in both equations
{{{(s + 5)*2 = (s -5)*4}}}
{{{2s + 10 = 4s - 20}}}
{{{2s = 30}}}
{{{s = 15}}}
The speed of the boat in still water is 15 mi/hr
check:
{{{d = (15 + 5)*2}}}
{{{d = 40}}}
upstream:
{{{d = (15 - 5)*4}}}
{{{d = 40}}}
OK