Question 800317
{{{ 500 *1 = 500 }}} km is the distance the Austin to LAX
plane travels in the hour before the LAX to Austin takes off
Let {{{ d }}} = the distance LAX to Austin travels until they meet
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Start a stopwatch when the LAX to Austin plane takes off
Let {{{ t }}} = the time in hours for them to pass each other
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Equation for Austin to LAX:
(1) {{{ 2100 - d - 500 = 500t }}}
Equation for LAX to Austin:
(2) {{{ d = 700t  }}}
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Substitute (2) into (1)
(1) {{{ 2100 - 700t - 500 = 500t }}}
(1) {{{ 1200t = 1600 }}}
(1) {{{ t = 1.333 }}}
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It was {{{ t + 1 = 2.333 }}} earlier at noon 
when Austin top LAX took off
{{{ .333*60 = 20 }}} minutes
12:00 + 2.333 = 2:20 PM when they passed each other
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check:
(2) {{{ d = 700t  }}}
(2) {{{ d = 700*1.333 }}}
(2) {{{ d = 933.333 }}}
and
(1) {{{ 2100 - d - 500 = 500*1.333 }}}
(1) {{{ 1600 - d = 666.667 }}}
(1) {{{ d = 933.333 }}}
OK