Question 794073
Let {{{ c }}} = the rate of the current
{{{ 15 - c }}} = rate going upriver
{{{ 15 + c }}} = rate going downriver
Let {{{ t }}} = time in hours to go downriver
{{{ t + 3 }}} = time in hours to go upriver
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Equation for going upriver:
(1) {{{ 60 = ( 15 - c )*( t + 3 ) }}}
Equation for going downriver:
(2) {{{ 60 = ( 15 + c )*t }}}
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(1) {{{ 60 = 15t - c*t + 45 - 3c }}}
(1) {{{ 15 = 15t - c*( t + 3 ) }}}
(1) {{{ c*( t + 3 ) = 15*( 1 - t ) }}}
(1) {{{ c = 15*( 1 - t ) / ( t + 3 ) }}}
and
(2) {{{ 60 = 15t + c*t }}}
(2) {{{ c = ( 60t - 15 ) / t }}}
(2) {{{ c = 15*( 4t - 1 ) / t }}}
by substitution:
(1) {{{ 15*( 4t - 1 ) / t = 15*( 1 - t ) / ( t + 3 ) }}}
(1) {{{ ( 4t - 1 ) / t = ( 1 - t ) / ( t + 3 ) }}}
multiply both sides by {{{ t*( t+3 ) }}}
(1) {{{ ( 4t - 1 )*( t + 3 ) = t*( -t + 1 ) }}}
(1) {{{ 4t^2 - t + 12t - 3  = -t^2 + t }}}
(1) {{{ 5t^2 + 10t - 3 = 0 }}}
Use the quadratic equation:
{{{ t = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{ a = 5 }}}
{{{ b = 10 }}}
{{{ c = -3 }}}
{{{ t = (-10 +- sqrt( 10^2-4*5*(-3) ))/(2*5) }}}
{{{ t = (-10 +- sqrt( 100 + 60 )) / 10 }}}
{{{ t = (-10 +- sqrt( 160 )) / 10 }}}
{{{ t = ( -10 + 4*sqrt( 10 ) )/10 }}}
{{{ t = -1 + (2/5)*sqrt(10) }}}
You can finish this. Find {{{t }}}, and then plug 
value into (2) to find {{{ c }}}
Hopefully, I didn't make a mistake.