Question 696737
Let {{{ t }}} = the boat's time in hrs to travel 63 km downstream
Let {{{ c }}} = the speed of the current in km/hr
The boat's speed going downstream is {{{ 15 + c }}}
The boat's speed going upstream is {{{ 15 - c }}}
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given:
Equation for going upstream is
(1) {{{ 63 = ( 15 - c )*( t + 4 ) }}}
Equation for going downstream is
(2) {{{ 63 = ( 15 + c )*t }}}
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(1) {{{ 63 = 15t - c*t + 60 - 4c }}}
(1) {{{ 15t - c*t = 3 + 4c }}}
(1) {{{ t*( 15 - c ) = 3 + 4c }}}
(1) {{{ t = ( 3 + 4c ) / ( 15 - c ) }}}
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Substitute (1) into (2)
(2) {{{ 63 = ( 15 + c )*( 3 + 4c ) / ( 15 - c )  }}}
(2) {{{ 63*( 15 - c ) =  ( 15 + c )*( 3 + 4c ) }}}
(2) {{{ 945 - 63c = 45 + 3c + 60c - 4c^2 }}}
(2) {{{ 4c^2 - 126c + 900 = 0 }}}
(2) {{{ 2c^2 - 63c + 450 = 0 }}}
Use quadratic equation:
{{{ c = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{ a = 2 }}}
{{{ b = -63 }}}
{{{ c = 450 }}}
{{{ c = (-(-63) +- sqrt( (-63)^2-4*2*450 ))/(2*2) }}}
{{{ c = ( 63 +- sqrt( 3969 - 3600 ))/4 }}}
{{{ c = ( 63 +- sqrt( 369 ))/4 }}}
{{{ c = ( 63 +- sqrt( 369 ))/4 }}}
You'll have to finish, but this is the basic idea
of how to solve. Then check the work.