Question 1021345
Let {{{ t }}} = his time in hrs to go downstream
{{{ (3/2)*t }}} = his time in hrs to go upstream
Let {{{ c }}} = the speed of the current
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His equation for going downstream:
(1) {{{ 48 = ( 20 + c )*t }}}
His equation for going upstream:
(2) {{{ 48 = ( 20 - c )*( (3/2)*t ) }}}
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(1) {{{ t = 48/( 20 + c ) }}}
Plug (1) into (2)
(2) {{{ 48 = ( 20 - c )* (3/2)*( 48 / ( 20 + c ) ) }}}
Multiply both sides by {{{ 2*( 20 + c ) }}}
(2) {{{ 2*48*( 20 + c ) = ( 20 - c )*3*48 }}}
Divide both sides by {{{ 48 }}}
{{{ 2*( 20 + c ) = 3*( 20 - c ) }}}
{{{ 40 + 2c = 60 - 3c }}}
{{{ 5c = 20 }}}
{{{ c = 4 }}} 
the speed of the current is 4 mi/hr
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check:
(1) {{{ 48 = ( 20 + c )*t }}}
(1) {{{ 48 = ( 20 + 4 )*t }}}
(1) {{{ 48 = 24t }}}
(1) {{{ t = 2 }}}
and
(2) {{{ 48 = ( 20 - c )*( (3/2)*t ) }}}
(2) {{{ 48 = ( 20 - 4 )*( (3/2)*t ) }}}
(2) {{{ 48 = 16*(3/2)*t }}}
(2) {{{ 48 = 24t }}}
(2) {{{ t = 2 }}}
OK