Question 772975
Let {{{ t }}} = the time in hours for airplane A
{{{ t + 1 }}} = the time in hours  for airplane B
Let {{{ s }}} = the speed of A in mi/hr
{{{ s + 50 }}} = the speed of B in mi/hr
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Equation for A:
(1) {{{ 600 = s*t }}}
(2) {{{ 1000 = ( s + 50 )*( t + 1 ) }}}
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(1) {{{ t = 600/s }}}
Substitute (1) into (2)
(2) {{{ 1000 = ( s + 50 )*( 600/s + 1 ) }}}
(2) {{{ 1000 = 600 + 30000/s + s + 50 }}}
(2) {{{ 350 = 30000/s + s }}}
(2) {{{ 350s = 30000 + s^2 }}}
(2) {{{ s^2 - 350s + 30000 = 0 }}}
Use the quadratic formula
{{{ s = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 1 }}}
{{{ b = -350 }}}
{{{ c = 30000 }}}
{{{ s = (-(-350) +- sqrt( (-350)^2 - 4*1*(30000) )) / (2*1) }}}
{{{ s = ( 350 +- sqrt( 122500 - 120000 )) / 2 }}}
{{{ s = ( 350 +- sqrt( 2500 )) / 2 }}}
{{{ s = ( 350 + 50 ) / 2 }}}
{{{ s = 400 / 2 }}}
{{{ s = 200 }}}
and
{{{ s = ( 350 - 50 ) / 2 }}}
{{{ s = 150 }}}
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I'll choose 150 mi/hr for A
and 200 mi/hr for B
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check:
(1) {{{ 600 = s*t }}}
(1) {{{ 600 = 150*t }}}
(1) {{{ t = 4 }}} hrs
and
(2) {{{ 1000 = ( s + 50 )*( t + 1 ) }}}
(2) {{{ 1000 = ( 150 + 50 )*( 4 + 1 ) }}}
(2) {{{ 1000 = 200*5 }}}
(2) {{{ 100 = 1000 }}}
OK