Question 1108948
Let {{{ c }}} = the speed of the current in mi/hr
Let {{{ t }}} = time in hrs to go upstream
----------------------------------------------
Convert {{{ 36 }}} min to hrs:
{{{ 36/60 = .6 }}}
{{{ 3.6 }}} hrs = total times for both trips
------------------------------------------
Equation for going upstream:
(1) {{{ 3 = ( 3 - c )*t }}}
Equation for going downstream:
(2) {{{ 3 = ( 3 + c )*( 3.6 - t ) }}}
----------------------------------
(2) {{{ 3 = 10.8 + 3.6c - t*( 3 + c ) }}}
(2) {{{ t*( 3 + c ) = 7.8 + 3.6c }}}
and
(1) {{{ t = 3/( 3 - c ) }}}
and
(2) {{{ ( 3/( 3 - c ) )*( 3 + c ) = 7.8 + 3.6c }}}
(2) {{{ 3*( 3 + c ) = ( 7.8 + 3.6c )*( 3 - c ) }}}
(2) {{{ 9 + 3c = 23.4 + 10.8c - 7.8c - 3.6c^2 }}}
(2) {{{ 3.6c^2 + 3c - 10.8c + 7.8c - 23.4 + 9 = 0 }}}
(2) {{{ 3.6c^2 - 23.4 + 9 = 0 }}}
(2) {{{ 3.6c^2 = 14.4 }}}
(2) {{{ c^2 = 4 }}}
(2) {{{ c = 2 }}} mi/hr
The speed of the current is 2 mi/hr
--------------------------------------
check the answer:
(1) {{{ 3 = ( 3 - c )*t }}}
(1) {{{ 3 = ( 3 - 2 )*t }}}
(1) {{{ t = 3 }}} hrs
and
(2) {{{ 3 = ( 3 + c )*( 3.6 - t ) }}}
(2) {{{ 3 = ( 3 + 2 )*( 3.6 - t ) }}}
(2) {{{ 3 = 5*( 3.6 - t ) }}}
(2) {{{ 3 = 18 - 5t }}}
(2) {{{ 5t = 15 }}}
(2) {{{ t = 3 }}} 
and
notice that:
{{{ t + (3.6 - t) = 3 + 3.6 - 3 }}}
and that = {{{ 3.6 }}} hrs which is the total time
Get another opinion if needed