Question 521581
Let {{{t}}} = time for both cars
Let {{{d}}} = distance traveled by slower car
For faster car:
(1) {{{ d + 15 = 65t }}}
For slower car:
(2) {{{ d = 55t }}}
Substitute (2) into (1)
(1) {{{ 55t + 15 = 65t }}}
(1) {{{ 10t = 15 }}}
(1) {{{ t = 1.5 }}}
and, since
(2) {{{ d = 55t }}}
(2) {{{ d = 55*1.5 }}}
(2) {{{ d = 82.5 }}}
{{{ d + 15 = 97.5 }}}
The faster car traveled 97.5 mi
The slower car traveled 82.5 mi
check:
(1) {{{ d + 15 = 65t }}}
(1) {{{ 82.5 + 15 = 65*1.5 }}}
(1) {{{ 97.5 = 97.5 }}}
OK