Question 1054463
Let {{{ t }}} = time in hrs the teacher takes
when she arrives on time
Let {{{ s }}} = her speed in mi/hr when she 
arrives {{{ 1 }}} hr late
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Equation for arriving 1 hr late:
(1) {{{ 280 = s*( t + 1 ) }}}
Equation for arriving on time:
(2) {{{ 280 = ( s + 5 )*t }}}
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(1) {{{ s = 280/( t + 1 ) }}}
and
(2) {{{ 280 = s*t + 5t }}}
(2) {{{ 280 = 280t / ( t+1 ) + 5t }}}
Multiply both sides by {{{ t + 1 }}}
(2) {{{ 280*( t+1 ) = 280t + 5t*( t+1 ) }}}
(2) {{{ 280t + 280 = 280t + 5t^2 + 5t }}}
(2) {{{ 280 = 5t^2 + 5t }}}
(2) {{{ t^2 + t = 56 }}}
complete the square
(2) {{{ t^2 + t + (1/2)^2 = 56 + (1/2)^2 }}}
(2) {{{ t^2 + t + 1/4 = 224/4 + 1/4 }}}
(2) {{{ ( t + 1/2 )^2 = 225/4 }}}
(2) {{{ ( t + 1/2 )^2 = ( 15/2 )^2 }}}
(2) {{{ t + 1/2 = 15/2 }}}
(2) {{{ t = 14/2 }}}
(2) {{{ t = 7 }}} ( can only use positive time here )
( cannot use negative square root of {{{ ( 15/2 )^2 }}} )
(1) {{{ 280 = s*( t + 1 ) }}}
(1) {{{ 280 = s*( 7 + 1 ) }}}
(1) {{{ 280 = 8s }}}
(1) {{{ s = 35 }}}
Her average speed when arriving 1 hr late was 35 mi/hr
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check answer:
Equation for arriving on time:
(2) {{{ 280 = ( s + 5 )*t }}}
(2) {{{ 280 = ( 35 + 5 )*t }}}
(2) {{{ 280 = 40t }}}
(2) {{{ t = 7 }}} hrs
OK