Question 1056739
Let {{{ s }}} = Rachel's speed in mi/hr
{{{ s + 2 }}} = Floyd's speed in mi/hr
Let {{{ t }}} = Floyd's time in hrs
{{{ t + 3 }}} = Rachel's time in mi/hr
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Floyd's equation:
(1) {{{ 12 = ( s + 2 )*t }}}
Rachel's equation:
(2) {{{ 12 = s*( t + 3 ) }}}
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(1) {{{ t = 12/( s + 2 ) }}}
Substitute (1) into (2)
(2) {{{ 12 = s*t + 3s }}}
(2) {{{ 12 = s*( 12/( s + 2 ) ) + 3s }}}
Multiply both sides by {{{ s + 2 }}}
(2) {{{ 12*( s + 2 ) = 12s + 3s*( s + 2 ) }}}
(2) {{{ 12s + 24 = 12s + 3s^2 + 6s }}}
(2) {{{ 3s^2 + 6s - 24 = 0 }}}
(2) {{{ s^2 + 2s - 8 = 0 }}}
(2) {{{ ( s + 4 )*( s - 2 ) = 0 }}}
{{{ s = 2 }}} ( I can't use {{{ s = -4 }}}, since it's negative )
and
{{{ s + 2 = 4 }}}
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Rachel's speed is 2 mi/hr
Floyd's speed is 4 mi/hr
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check:
(1) {{{ 12 = ( s + 2 )*t }}}
(1) {{{ 12 = 4t }}} 
(1) {{{ t = 3 }}}
and
(2) {{{ 12 = s*( t + 3 ) }}}
(2) {{{ 12 = 2*( t + 3 ) }}}
(2) {{{ 6 = t + 3 }}}
(2) {{{ t = 3 }}}
OK