Question 691629
Let {{{ t }}} = time in hrs for 1st trip
{{{ t + .5 }}} = time in hrs for 2nd trip
Let {{{ s }}} = speed in km/hr for 1st trip
{{{ s - 8 }}} = speed in km/hr for 2nd trip
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1st trip:
(1) {{{ 224 = s*t }}}
2nd trip:
(2) {{{ 224 = ( s - 8 )*( t + .5 ) }}}
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(2) {{{ 224 = s*t - 8t + .5s - 4 }}}
and
(1) {{{ t = 224/s }}}
Substitute (1) into (2)
(2) {{{ 224 = s*(224/s) - 8*(224/s) + .5s - 4 }}}
(2) {{{ 224 = 224 - 8*(224/s) + .5s - 4 }}}
(2) {{{ 4 =  - 8*(224/s) + .5s }}}
Multiply both sides by {{{ 2s }}}
(2) {{{ 8s = -16*224 + s^2 }}}
(2) {{{ s^2 - 8s - 3584 = 0 }}}
I'll complete the square
(2) {{{ s^2 - 8s = 3584  }}}
(2) {{{ s^2 - 8s + (-8/2)^2 = 3584 + (-8/2)^2  }}}
(2) {{{ s^2 - 8s + 16 = 3584 + 16  }}}
(2) {{{ ( s - 4 )^2 = 3600 }}}
(2) {{{ ( s - 4 )^2 = 60^2 }}}
(2) {{{ s - 4 = 60 }}}
(2) {{{ s = 64 }}}  ( ignore the negative square root of {{{60}}} )
and
{{{ s - 8 = 56 }}}
The train's speed on the way back is 56 km/hr
check answer:
(2) {{{ 224 = 56*( t + .5 ) }}} 
(2) {{{ t + .5 = 224/56 }}}
(2) {{{ t + .5 = 4 }}}
(2) {{{ t = 3.5 }}}
and
(1) {{{ 224 = s*t }}}
(1) {{{ 224 = 64t }}}
(1) {{{ t = 224/64 }}}
(1) {{{ t = 3.5 }}}
OK