Question 1010050
Let {{{ s }}} = the average speed of the bus
(1) {{{ 90 = s*t }}}
(2) {{{ 90 = ( s + 3 )*( t - 1 ) }}}
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(2) {{{ 90 = t*( s + 3 ) - s - 3 }}}
and
(1) {{{ t = 90/s }}}
Substitute (1) into (2)
(2) {{{ 90 = ( 90/s )*( s + 3 ) - s - 3 }}}
(2) {{{ 93 = ( 90/s )*( s + 3 ) - s }}}
(2) {{{ 93 = 90 + 270/s - s }}}
(2) {{{ 3 = 270/s - s
Multiply both sides by {{{ s }}}
(2) {{{ 3s = 270 - s^2 }}}
(2) {{{ s^2 + 3s - 270 = 0 }}}
Use quadratic formula
{{{ s = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}  
{{{ a = 1 }}}
{{{ b = 3 }}}
{{{ c = -270 }}}
{{{ s = (-3 +- sqrt( 3^2 - 4*1*(-270) )) / (2*1) }}}  
{{{ s = (-3 +- sqrt( 9 + 1080 )) /2 }}}  
{{{ s = (-3 + 33 ) /2 }}}  
{{{ s = 30/2 }}}
{{{ s = 15 }}}
The usual speed is 15 mi/hr
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check:
(1) {{{ t = 90/s }}}
(1) {{{ t = 90/15 }}}
(1) {{{ t = 6 }}} hrs
and
(2) {{{ 90 = ( s + 3 )*( t - 1 ) }}}
(2) {{{ 90 = ( 15 + 3 )*( t - 1 ) }}}
(2) {{{ 90 = 18t - 18 }}}
(2) {{{ 18t = 108 }}}
(2) {{{ t = 6 }}}
OK