Question 813685
Let {{{ c }}} = the rate of the current in mi/hr
{{{ 10 + c }}} = speed of the boat going downstream
{{{ 10 - c }}} = speed of the boat going upstream
Let {{{ t }}} = time in hrs for going downstream
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Equation for going upstream:
(1) {{{ 68 = ( 10 - c )*( t + 3 ) }}}
Equation for going downstream:
(2) {{{ 68 = ( 10 + c )*t }}}
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(1) {{{ 68 = 10t - c*t + 30 - 3c }}}
(1) {{{ 68 = ( 10 - c )*t + 30 - 3c }}}
(1) {{{ ( 10 - c )*t = 3c + 38 }}}
(1) {{{ t = ( 3c + 38 ) / ( 10 - c ) }}}
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By substitution:
(2) {{{ 68 = ( 10 + c )*( 3c + 38 ) / ( 10 - c ) }}}
(2) {{{ 68*( 10 - c ) = ( 10 + c )*( 3c + 38 ) }}}
(2) {{{ 680 - 68c = 30c + 3c^2 + 380 + 38c }}}
(2) {{{ 3C^2 + 136C - 300 = 0 }}}
Use the quadratic formula
{{{ C = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} ( note: C and not c is the solution )
{{{ a = 3 }}}
{{{ b = 136 }}}
{{{ c = -300 }}}
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{{{ C = (-136 +- sqrt( 136^2-4*3*(-300) ))/(2*3) }}}
{{{ C = (-136 +- sqrt( 18496 + 3600 ))/6 }}}
{{{ C = ( -136 + sqrt( 22096 ) ) / 6 }}}
{{{ C = ( -136 + 148.65 ) / 6 }}}
{{{ C = 2.108 }}}
The rate of the current is 2.108 mi/hr
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check:
(2) {{{ 68 = ( 10 + c )*t }}}
(2) {{{ 68 = ( 10 + 2.108 )*t }}}
(2) {{{ t = 68 / 12.108 }}}
(2) {{{ t = 5.616 }}} hrs
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(1) {{{ 68 = ( 10 - c )*( t + 3 ) }}}
(1) {{{ 68 = ( 10 - 2.108 )*( 5.616 + 3 ) }}}
(2) {{{ 68 = 7.892*8.616 }}}
(2) {{{ 68 = 67.997 }}}
pretty close
Hope I got it