Question 679635
Let {{{ s }}} = the speed of the boat in still water
{{{ s + 4 }}} = speed of canoe going downstream
{{{ s - 4 }}} = speed of canoe going upstream
Let {{{ t }}} = time to go upstream
{{{ 7 - t }}} = time to go downstream
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Equation for going upstream:
(1) {{{ 10 = ( s - 4 )*t }}}
Equation for going downstream:
(2) {{{ 10 = ( s + 4 )*( 7 - t ) }}}
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(1) {{{ t = 10 / ( s - 4 ) }}}
and
(2) {{{ 10 = 7s + 28 - s*t - 4t }}}
(2) {{{ 10 = 7s + 28 - t*( s + 4 ) }}}
Substitute (1) into (2)
(2) {{{ 10 = 7s + 28 - ( 10 / ( s - 4 ) )*( s + 4 ) }}}
(2) {{{ 7s + 18 =   ( 10 / ( s - 4 ) )*( s + 4 ) }}}
Multiply both sides by {{{ s - 4 }}}
(2) {{{ ( s - 4 )*( 7s + 18 ) = 10s + 40 }}}
(2) {{{ 7s^2 - 28s + 18s - 72 = 10s + 40 }}}
(2) {{{ 7s^2 - 20s -112 = 0 }}}
Use quadratic formula
{{{ s = ( -b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 7 }}}
{{{ b = -20 }}}
{{{ c = -112 }}}
{{{ s = ( -(-20) +- sqrt( (-20)^2 - 4*7*(-112) )) / (2*7) }}}
{{{ s = ( 20 +- sqrt( 400 + 3136 )) / 14 }}}
{{{ s = ( 20 + sqrt( 3536 )) / 14 }}}
{{{ s = ( 20 + 59.464 ) / 14 }}}
{{{ s = 79.464 / 14 }}}
{{{ s = 5.676 }}}
The speed in still water is 5.676 mi/hr
check answer:
(1) {{{ 10 = ( s - 4 )*t }}}
(1) {{{ 10 = ( 5.676 - 4 )*t }}}
(1) {{{ 10 = 1.676*t }}}
(1) {{{ t = 5.967 }}} hrs
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(2) {{{ 10 = ( s + 4 )*( 7 - 5.967 ) }}}
(2) {{{ 10 = ( s + 4 )*1.033 }}}
(2) {{{ 9.681 = s + 4 }}}
(2) {{{ s = 5.681 }}}
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This is pretty close. Error could be
due to rounding off. I think the method
is correct anyway.