Question 769712
Let {{{ c }}} = the speed of the current
Let {{{ d }}} = the one-way distance of the trip
You must convert the minutes to hours
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Going upstream:
(1) {{{ d = ( 16 - c )*( 20/60 ) }}}
Going downstream:
(2) {{{ d = ( 16 + c )*( 15/60 ) }}}
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Substitute (1) into (2)
(2) {{{ ( 16 - c )*( 20/60 ) = ( 16 + c )*( 15/60 ) }}}
(2) {{{ ( 16 - c )*( 1/3 ) = ( 16 + c )*( 1/4 ) }}}
Multiply both sides by {{{ 12 }}}
(2) {{{ 4*( 16 - c ) = 3*( 16 + c ) }}}
(2) {{{ 64 - 4c = 48 + 3c }}}
(2) {{{ 7c = 16 }}}
(2) {{{ c = 16/7 }}} mi/hr
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check:
(2) {{{ d = ( 16 + 16/7 )*( 15/60 ) }}}
(2) {{{ d = 16*( 1 + 1/7 )*( 1/4 ) }}}
(2) {{{ d = 4*( 8/7 ) }}}
(2) {{{ d = 32/7 }}}
and
(1) {{{ d = ( 16 - 16/7 )*( 20/60 ) }}}
(1) {{{ d = 16*( 1 - 1/7 )*( 1/3 ) }}}
(1) {{{ d = 16*( 6/7 )*(1/3 ) }}}
(1) {{{ d = 16*( 2/7 ) }}}
(1) {{{ d = 32/7 }}}
OK