Question 1026577
Let {{{ s }}} = the speed of the slower car in mi/hr
{{{ 2s }}} = the speed of the faster car in mi/hr
Let {{{ d }}} = distance in miles the slower car travels in 6 hrs
{{{ 204 + d }}} = distance in miles the faster car travels in 6 hrs
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Equation for slower car:
(1) {{{ d = s*6 }}}
Equation for faster car:
(2) {{{ 204 + d = 2s*6 }}}
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Substitute (1) into (2)
(2) {{{ 204 + 6s = 2s*6 }}}
(2) {{{ 204 + 6s = 12s }}}
(2) {{{ 6s = 204 }}}
(2) {{{ s = 34 }}}
and
{{{ 2s = 68 }}}
The slower car moves at 34 mi/hr
The faster car moves at 68 mi/hr
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check:
(1) {{{ d = s*6 }}}
(1) {{{ d = 34*6 }}}
(1) {{{ d = 204 }}}
and
(2) {{{ 204 + d = 2s*6 }}}
(2) {{{ 204 + d = 12*34 }}}
(2) {{{ 204 + d = 408 }}}
(2) {{{ d = 408 - 204 }}}
(2) {{{ d = 204 }}}
OK