Question 888635
Let {{{ s }}} = the speed of the slower car
{{{ ( 8/3 )*s }}} = the speed of the faster car
Let {{{ d }}} = the distance the faster car travels in {{{ 1 }}} hr
{{{ 50 - d }}} = the distance the slower car travels in {{{ 1 }}} hr
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Equation for slower car:
(1) {{{ 50 - d = s*1 }}}
Equation for faster car:
(2) {{{ d = ( 8/3 )*s*1 }}}
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Substitute (2) into (1)
(1) {{{ 50 - ( 8/3 )*s = s }}}
(1) {{{ 150 - 8s = 3s }}}
(1) {{{ 11s = 150 }}}
(1) {{{ s = 13 636 }}} mi/hr
{{{ ( 8/3)*s = ( 8/3 )*( 150/11 ) }}}
{{{ ( 8/3 )*s = 36.364 }}} mi/hr
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The faster car's speed is 36.364 mi/hr
The slower car's speed is 13.636 mi/hr
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check:
(1) {{{ 50 - d = s*1 }}}
(1) {{{ 50 - d = 150/11 }}}
(1) {{{ d = 550/11 - 150/11 }}}
(1) {{{ d = 400/11 }}}
and
(2) {{{ d = ( 8/3 )*s*1 }}}
(2) {{{ d = ( 8/3 )*( 150/11 ) }}}
(2) {{{ d = 400/11 }}}
OK