Question 862389
Let {{{ t }}} = the time in hours to row upstream
{{{ 18 - t }}} = the time in hours to row downstream
Let {{{ s }}} = the rate of the boat in still water in mi/hr
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Rowing upstream:
(1) {{{ 32 = ( s - 7 )*t }}}
Rowing downstream:
(2) {{{ 32 = ( s + 7 )*( 18 - t ) }}}
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(1) {{{ t = 32 / ( s-7 ) }}}
and
(2) {{{ 32 = 18s + 126 - st - 7t }}}
(2) {{{ t*( s + 7 ) - 18s = 94 }}}
(2) {{{ t = ( 18s - 94 ) / ( s + 7 ) }}}
(2) {{{ 32 / ( s - 7 ) = ( 18s - 94 ) / ( s + 7 ) }}}
(2) {{{ 32*( s + 7 ) = ( 18s - 94 )*( s - 7 ) }}}
(2) {{{ 32s + 224 = 18s^2 - 94s - 126s + 658 }}}
(2) {{{ 18s^2 - 252s + 434 = 0 }}}
(2) {{{ 9s^2 - 126s + 217 = 0 }}}
Use the quadratic formula to solve, then
find {{{ t }}}
Check my math,I could have made an error
The method should be OK