Question 1011029
Let {{{ c }}} = the speed of the current in mi/hr
{{{ 15 + c }}} = the speed of the boat going with the current
{{{ 15 - c }}} = the speed of the boat going against the current
Let {{{ d }}} = the distance in miles traveled against the current
{{{ d + 20 }}} = the distance in miles traveled with the current
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Equation for going with the current:
(1) {{{ d + 20 = ( 15 + c )*6 }}}
Equation for going against the current:
(2) {{{ d = ( 15 - c )*10 }}}
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(1) {{{ d + 20 = 90 + 6c }}}
(1) {{{ d - 6c = 70 }}}
and
(2) {{{ d = 150 - 10c }}}
(2) {{{ d + 10c = 150 }}}
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Subtract (1) from (2)
(2) {{{ d + 10c = 150 }}}
(1) {{{ -d + 6c = -70 }}}
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{{{ 16c = 80 }}}
{{{ c = 5 }}}
The speed of the current is 5 mi/hr
check:
(1) {{{ d + 20 = ( 15 + c )*6 }}}
(1) {{{ d + 20 = ( 15 +5 )*6 }}}
(1) {{{ d + 20 = 120 }}}
(1) {{{ d = 100 }}}
and
(2) {{{ d = ( 15 - c )*10 }}}
(2) {{{ d = ( 15 - 5 )*10 }}}
(2) {{{ d = 10*10 }}}
(2) {{{ d = 100 }}}
OK