Question 1014408
All you need is [ distance ] = [ rate ] x [ time ]
Each of the cars needs it's own equation.
Let {{{ d }}} = distance traveled by the slower car
until they are {{{ 200 }}} km apart
{{{ 200 - d }}} = distance traveled by faster car
until they are {{{ 200 }}} km apart
Let {{{ t }}} = time in hrs when they are {{{ 200 }}} km  apart
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Equation for slower car:
(1) {{{ d = 35t }}}
Equation for faster car:
(2) {{{ 200 - d = 40t }}}
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Substitute (1) into (2)
(2) {{{ 200 - 35t = 40t }}}
(2) {{{ 75t = 200 }}}
(2) {{{ t = 8/3 }}} hrs
(2) {{{ t = 2 + 2/3 }}} hrs
{{{ (2/3)*60 = 40 }}} min
They will be 200 km apart in 2 hrs and 40 min
--------------------------
check:
(1) {{{ d = 35t }}}
(1) {{{ d = 35*( 8/3 ) }}}
(1) {{{ d = 93.333 }}}
and
(2) {{{ 200 - d = 40t }}}
(2) {{{ 200 - d = 40*(8/3) }}}
(2) {{{ 200 - d = 106.667 }}}
(2) {{{ d = 93.333 }}}
OK
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Be careful in the 2nd problem because the cars get
further apart as time goes on, but not as much as 
when they were going in opposite directions
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Let {{{ d }}} = distance in miles traveled by slower car
until they are {{{ 15 }}} miles apart
{{{ d + 15 }}} =  distance in miles traveled by faster car
until they are {{{ 15 }}} miles apart
Let {{{ t }}} = time in hrs until they are {{{ 15 }}} miles apart
--------------------
Equation for slower car:
(1) {{{ d = 35t }}}
Equation for faster car:
(2) {{{ d + 15 = 40t }}}
--------------------
Substitute (1) into (2)
(2) {{{ 35t + 15 = 40t }}}
(2) {{{ 5t = 15 }}}
(2) {{{ t = 3 }}} hrs
In 3 hrs they will be 15 mi apart
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check:
(1) {{{ d = 35t }}}
(1) {{{ d = 35*3 }}}
(1) {{{ d = 105 }}}
and
(2) {{{ d + 15 = 40t }}}
(2) {{{ d + 15 = 40*3 }}}
(2) {{{ d + 15 = 120 }}}
(2) {{{ d = 105 }}}
OK
Hope this helps