Question 614456
Let {{{ s }}} = Austin's speed
Let {{{ t }}} = Austin's time
{{{ s - 2 }}} = Doug's speed
{{{ t + 1/6 }}} = Doug's time ( 1/6 hr = 10 min )
{{{ d = 4 }}} It's a 4 mile race
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Austin's equation:
(1) {{{ 4 = s*t }}}
Doug's equation:
(2) {{{ 4 = ( s - 2 )*( t + 1/6 ) }}}
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(2) {{{ 4 = s*t - 2t + s/6 - 1/3 }}}
(2) {{{ s/6 - 2t = 4 + 1/3 - s*t }}}
Substitute (1) into (2)
(2) {{{ s/6 - 2t = 4 + 1/3 - 4 }}}
(2) {{{ s/6 - 2t = 1/3 }}}
(2) {{{ s - 12t = 2 }}}
And also from (1):
(1) {{{ t = 4/s }}}
Substituting:
(2) {{{ s - 12*(4/s) = 2 }}}
(2) {{{ s^2 - 48 = 2s }}}
(2) {{{ s^2 - 2s - 48 = 0 }}}
Use quadratic formula
{{{s = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{ a = 1 }}}
{{{ b = -2 }}}
{{{ c = -48 }}}
{{{s = (-(-2) +- sqrt( (-2)^2-4*1*(-48) ))/(2*1) }}} 
{{{s = ( 2 +- sqrt( 4 + 192 )) / 2 }}} 
{{{s = ( 2 +- sqrt( 196 )) / 2 }}} 
{{{s = ( 2 + 14) / 2 }}} 
{{{ s = 16/2 }}}
{{{ s = 8 }}}
Austin's speed was 8 mi/hr
From (1):
(1) {{{ t = 4/s }}}
(1) {{{ t = 4/8 }}}
(1) {{{ t = 1/2 }}}
{{{ t + 1/6 }}} = Doug's time
{{{ t + 1/6 = 1/2 + 1/6 }}}
{{{ t + 1/6 = 2/3 }}}
and
{{{ (2/3)*60 = 40 }}} 
Doug's time is 2/3 hr, or 40 min
check:
(2) {{{ 4 = ( s - 2 )*( t + 1/6 ) }}}
(2) {{{ 4 = ( 8 - 2 )*( 1/2 + 1/6 ) }}}
(2) {{{ 4 = 6*( 4/6 ) }}}
(2) {{{ 4 = 4 }}}
OK