Question 1110171
Let {{{ d }}} = distance traveled by the slower car 
in {{{ t }}} hrs
{{{ d + 42 }}} = distance traveled by the faster car 
in {{{ t }}} hrs
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Equation for slower car:
(1) {{{ d = 48t }}}
Equation for faster car:
(2) {{{ d + 42 = 62t }}}
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Plug (1) into (2)
{{{ 48t + 42 = 62t }}}
{{{ 14t = 42 }}}
{{{ t = 3 }}}
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In 3 hrs the faster car is 42 mi ahead of the slower car
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check:
(1) {{{ d = 48*3 }}}
(1) {{{ d = 144 }}}
and
(2) {{{ d + 42 = 62*3 }}}
(2) {{{ d = 186 - 42 }}}
(2) {{{ d = 144 }}}
OK