Question 893076
Call the runners A and B
Let their speeds be {{{ a }}} and {{{ b }}} in feet / minute
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Start a stopwatch when A leaves
In the 1st race:
Let {{{ d }}} = the distance A runs when he overtakes B
A's equation:
(1) {{{ d = a*4 }}}
B's equation:
(2) {{{ d - 100 = b*4 }}}
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In the 2nd race:
I think they both leave at the same time, so
Let {{{ t }}} = time in minutes until A is {{{ 750 }}} ft
ahead of B
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A's equation:
(3) {{{ 9000 + 750 = a*t }}}
B's equation:
(4) {{{ 9000= b*t }}}
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This is actually 4 equations with 4 unknowns,
so it is solvable
Substitute (1) into (2)
(2) {{{ a*4 - 100 = b*4 }}}
(2) {{{ 4a - 4b = 100 }}}
(2) {{{ a - b = 25 }}}
(2) {{{ a = b + 25 }}}
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Substitute (4) into (3)
(4) {{{ t = 9000/b }}}
(3) {{{ 9000 + 750 = a*( 9000/b ) }}} 
(3) {{{ 9750 =  a*( 9000/b ) }}} 
(3) {{{ 9750b = 9000a }}}
Divide both sides by {{{ 750 }}}
(3) {{{ 13b = 12a }}}
Substitute (2) int (3)
(3) {{{ 13b = 12*( b + 25 ) }}}
(3) {{{ 13b = 12b + 300 }}}
(3) {{{ b = 300 }}}
and, since
(2) {{{ a = b + 25 }}}
(2) {{{ a = 325 }}}
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A's speed is 325 ft/min
B's speed is 300 ft/min
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check:
(1) {{{ d = a*4 }}}
(1) {{{ d = 325*4 }}}
(1) {{{ d = 1300 }}} ft
and
(2) {{{ d - 100 = b*4 }}}
(2) {{{ d - 100 = 300*4 }}}
(2) {{{ d = 1200 + 100 }}}
(2) {{{ d = 1300 }}} ft
OK
You can check (3) and (4)