Question 567816
Let {{{ s }}} = original speed in km/hr
Let {{{ t }}} = time in hrs to cover {{{ 90 }}} km at speed {{{ s }}}
given:
(1) {{{ 90 = s*t }}}
(2) {{{ 90 = ( s + 15 )*( t - .5 ) }}}
------------------------
(1) {{{ t = 90/s }}}
Substitute (1) into (2)
(2) {{{ 90 = s*t + 15t - .5s - 7.5 }}}
(2) {{{ 90 = s*(90/s)  + 15*(90/s) - .5s - 7.5 }}}
(2) {{{ 90 = 90 + 1350/s - .5s - 7.5 }}}
(2) {{{ 0 = 1350/s -.5s  - 7.5 }}}
(2) {{{ 1350/s = .5s + 7.5 }}}
(2) {{{ 1350 = .5s^2 + 7.5s }}}
(2) {{{ 5s^2 + 75s - 13500 = 0 }}}
(2) {{{ s^2 + 15s - 2700 = 0 }}}
Use quadratic formula
{{{ s = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{ a = 1 }}}
{{{ b = 15 }}}
{{{ c = -2700 }}}
{{{ s = ( -15 +- sqrt( 15^2 - 4*1*(-2700) ))/(2*1) }}}
{{{ s = ( -15 +- sqrt( 225 + 10800 )) / 2 }}}
{{{ s = ( -15 +- sqrt( 11025 )) / 2 }}} ( I can't use the negative square root )
{{{ s = ( -15 + 105 ) / 2 }}}
{{{ s = 90/2 }}}
{{{ s = 45 }}}
The original speed is 45 km/hr
check answer:
(1) {{{ 90 = s*t }}}
(1) {{{ 90 = 45t }}}
(1) {{{ t = 2 }}} hrs
and
(2) {{{ 90 = ( s + 15 )*( t - .5 ) }}}
(2) {{{ 90 = ( 45 + 15 )*( 2 - .5 ) }}}
(2) {{{ 90 = 60*1.5 }}}
(2) {{{ 90 = 90 }}}
OK