Question 728481
Let {{{ t }}} = Doreen's time in hrs
{{{ t - 3 }}} = Sue's time in hrs ,   
Let {{{ s }}} = Doreen's speed in mi/hr
{{{ s + 15 }}} = Sue's speed in mi/hr
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Doreen's equation:
(1) {{{ 42 = s*t }}}
Sue's equation:
(2) {{{ 84 = ( s + 15 )*( t - 3 ) }}}
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(2) {{{ 84 = s*t + 15t - 3s - 45 }}}
(1) {{{ t = 42/s }}}
Substitute (1) into (2)
(2) {{{ 84 = s*(42/s) + 15*(42/s) - 3s - 45 }}} 
(2) {{{ 84 = 42 + 630/s - 3s -45 }}}
(2) {{{ 87 = 630/s - 3s }}}
Multiply both sides by {{{ s }}}
(2) {{{ 87s = 630 - 3s^2 }}}
(2) {{{ 3s^2 + 87s - 630 = 0 }}}
(2) {{{ s^2 + 29s - 210 = 0 }}}
Use the quadratic equation
{{{ s = (-b +- sqrt( b^2 - 4*a*c ))/(2*a) }}}
{{{ a = 1}}}
{{{ b = 29 }}}
{{{ c = -210 }}}
{{{ s = (-29 +- sqrt( 29^2 - 4*1*(-210) ))/(2*1) }}}
{{{ s = (-29 +- sqrt( 841 + 840))/2 }}}
{{{ s = (-29 +- sqrt( 1681))/2 }}} 
{{{ s = (-29 + 41)/2 }}} ( ignore negative sqrt- can't have negative speed )
{{{ s = 12/2 }}}
{{{ s = 6 }}}
{{{ s + 15 = 21 }}}
and, since
(1) {{{ t = 42/s }}}
(1) {{{ t = 42/6 }}}
(1) {{{ t = 7 }}}
{{{ t - 3 = 4 }}}
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Sue's time: 4 hrs
Doreen's time: 7 hrs
Sue's speed: 21 mi/hr
Doreen's speed: 6 mi/hr
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check:
(2) {{{ 84 = ( s + 15 )*( t - 3 ) }}}
(2) {{{ 84 = ( 6 + 15 )*( 7 - 3 ) }}}
(2) {{{ 84 = 21*4 }}}
(2) {{{ 84 = 84 }}}
and
(1) {{{ 42 = s*t }}}
(1) {{{ 42 = 6*7 }}}
(1) {{{ 42 = 42 }}}
OK