Question 820274
When the boat goes upstream, the rate 
of the current gets subtracted
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When the boat goes downstream, the
rate of the current gets added
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Let {{{ s }}} = the speed of the boat in still water
Let {{{ t[1] }}} = the time to go downstream
Let {{{ t[2] }}} = the time to go upstream
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Equation for going downstream:
(1) {{{ 20 = ( s + 5 )*t[1] }}}
Equation for going upstream:
(2) {{{ 20 = ( s - 5 )*t[2] }}}
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What you're looking for is {{{ T = t[1] + t[2] }}},
the time for the whole trip
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(1) {{{ t[1] = 20/( s+5 ) }}}
and
(2) {{{ t[2] = 20/( s-5 ) }}}
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{{{ T = 20/( s+5 ) + 20/( s-5 ) }}}
{{{ T = 20*( 1/(s+5) + 1/(s-5)) }}}
This is {{{ T = f(s) }}}