Question 517775
Let {{{t}}} = the usual driving time
in hours to to arrive on schedule
Let {{{s}}} = the bus's usual speed in mi/hr
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given:
For the usual trip:
(1) {{{ 170 = s*t }}}
For the late trip:
(2) {{{ 170 = ( s + 9 )*( t - .5 ) }}}
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(2) {{{ 170 = s*t + 9t - .5s - 4.5 }}}
Substitute (1) into (2) 
(2) {{{ 170 = 170 + 9t - .5s - 4.5 }}}
(2) {{{ .5s = 9t - 4.5 }}}
also
(1) {{{ s = 170/t }}}
(2) {{{ .5*(170/t) = 9t - 4.5 }}}
Multiply both sides by {{{t}}}
(2) {{{ 85 = 9t^2 - 4.5t }}}
(2) {{{9t^2 - 4.5t - 85 = 0 }}}
Use quadratic equation
{{{t = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{ a = 9 }}}
{{{ b = -4.5 }}}
{{{ c = -85 }}}
{{{t = (-(-4.5) +- sqrt( (-4.5)^2-4*9*(-85) ))/(2*9) }}} 
{{{t = ( 4.5 +- sqrt( 20.25 + 3060 )) / 18 }}} 
{{{t = ( 4.5 +- sqrt( 3080.25 )) / 18 }}} 
{{{ t = (  4.5 + 55.5 ) / 18 }}}
{{{ t = 60/18 }}}
{{{ t = 10/3 }}} 
and, from (1)
(1) {{{ s = 170/t }}}
(1) {{{ s = 170*(3/10) }}}
(1) {{{ s = 51 }}}
The usual speed is 51 mi/hr
check answer:
(2) {{{ 170 = ( s + 9 )*( t - .5 ) }}}
(2) {{{ 170 = ( 51 + 9 )*( 20/6 - 3/6 ) }}}
(2) {{{ 170 = 60*( 17/6 ) }}}
(2) {{{ 170 = 60*(17/6) }}}
(2) {{{ 170 = 170 }}}
OK