Question 989299
{{{ 3 }}} mi/hr = speed of the current
Let {{{ s }}} = the speed of the canoe in still water
{{{ s + 3 }}} = the speed of th canoe going downstream
{{{ s - 3 }}} = the speed of the canoe going upstream
Convert minutes to hrs:
{{{ 144/60 = 2.4 }}}
---------------------------
Going downstream:
(1) {{{ 16 = ( s+3 )*t }}}
Going upstream:
(2) {{{ 16 = ( s-3 )*( t + 2.4 ) }}}
---------------------------
(1) {{{ t = 16 / ( s+3 ) }}}
(2) {{{ 16/ ( s-3 ) =  16/( s+3 ) + 2.4 }}}
Multiply both sides by {{{ ( s-3 )*( s+3 ) }}}
(2) {{{ 16*( s+3 ) = 16*( s-3 ) + 2.4*( s-3 )*( s+3 ) }}}
(2) {{{ 16s + 48 = 16s - 48 + 2.4*( s^2 - 9 ) }}}
(2) {{{ 16s + 48 = 16s - 48 + 2.4s^2 - 21.6 ) }}} 
(2) {{{ 48 = 2.4s^2 - 48 - 21.6 }}}
(2) {{{ 2.4s^2 = 96 + 21.6 }}}
(2) {{{ 2.4s^2 = 117.6 }}}
(2) {{{ s^2 = 49 }}}
(2) {{{ s = 7 }}}
The canoeist's rate in still water is 7 mi/hr
check:
(1) {{{ 16 = ( s+3 )*t }}}
(1) {{{ 16 = ( 7+3 )*t }}} 
(1) {{{ 16 = 10t }}}
(1) {{{ t = 1.6 }}} hrs
and
(2) {{{ 16 = ( s-3 )*( t + 2.4 ) }}}
(2) {{{ 16 = ( 7-3 )*( 1.6 + 2.4 ) }}}
(2) {{{ 16 = 4*4 }}}
(2) {{{ 16 = 16 }}}
OK