Question 1148943
Let {{{ t }}} = the amount of time after they both 
leave until they are {{{ 15 }}} miles apart
Let {{{ d }}} = the distance in miles that Danica
( the slower one ) has traveled when Mario has traveled 
{{{ d + 15 }}} miles
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Danica's equation:
(1) {{{ d = 42t }}}
Mario's equation:
(2) {{{ d + 15 = 48t }}}
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(2) {{{ d = 48t - 15 }}}
{{{ 42t = 48t - 15 }}}
{{{ 6t = 15 }}}
{{{ t = 2.5 }}}
5 PM + 2.5 hrs = 7:30 PM 
when they are 15 mi apart
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check answer:
(1) {{{ d = 42t }}}
(1) {{{ d = 42*2.5 }}}
(1) {{{ d = 105 }}} mi
and
(2) {{{ d + 15 = 48t }}}
(2) {{{ d = 48t - 15 }}}
(2) {{{ d = 48*2.5 - 15 }}}
(2) {{{ d = 120 - 15 }}}
(2) {{{ d = 105 }}} mi
OK
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Let {{{ t }}} = time in hrs for Sally's plane to 
catch up with Jim's
Let {{{ d }}} = distance in miles Sally's plane
travels to catch up with Jim's
Let {{{ d[1] = 40*1 }}}. This is Jimmy's head start in miles.
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Sally's equation:
(1) {{{ d = 60t }}}
Jimmy's equation:
(2) {{{ d - d[1] = 40t }}}
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(2) {{{ d - 40 = 40t }}}
(2) {{{ d = 40t + 40 }}}
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{{{ 60t = 40t + 40 }}}
{{{ 20t = 40 }}}
{{{ t = 2 }}}
Sally catches up at 9 AM + 1 + 2 = 12 noon
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check answer:
(1) {{{ d = 60t }}}
(1) {{{ d = 60*2 }}}
(1) {{{ d = 120 }}}  mi
and
(2) {{{ d - d[1] = 40t }}}
(2) {{{ d - 40 = 40*2 }}}
(2) {{{ d = 80 + 40 }}}
(2) {{{ d = 120 }}} mi
OK