Question 1051270
Let {{{ t }}} = time in hrs for the 1st trip
{{{ t - 1 }}} = time in hrs for the 2nd trip
Let {{{ s }}} = speed in km/hr for the 1st trip
{{{ s + 10 }}} = speed in km/hr on the 2nd trip
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For 1st trip:
(1) {{{ 120 = s*t }}}
For 2nd trip:
(2) {{{ 120 = ( s+10 )*( t - 1 ) }}}
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(2) {{{ 120 = s*t + 10t - s - 10 }}}
(2) {{{ 120 + 10 = t*( s + 10 ) - s }}}
and
(1) {{{ t = 120/s }}}
Substitute (1) into (2)
(2) {{{ 130 = ( 120/s )*( s + 10 ) - s }}}
Multiply both sides by {{{ s }}}
(2) {{{ 130s = 120*( s + 10 ) - s^2 }}}
(2) {{{ 130s = 120s + 1200 - s^2 }}}
(2) {{{ s^2 + 10s = 1200 }}}
Complete the square
(2) {{{ s^2 + 10s + (10/2)^2 = 1200 + (10/2)^2 }}}
(2) {{{ s^2 + 10s + 25 = 1225 }}}
(2) {{{ ( s + 5 )^2 = 35^2 }}}
(2) {{{ s + 5 = 35 }}}
{{{ s = 30 }}}
and
{{{ s + 10 = 40 }}}
He drove 30 km/hr for the 1st trip and
40 km/hr for the 2nd trip
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check:
(1) {{{ 120 = 30t }}}
(1) {{{ t = 4 }}} hrs
and
(2) {{{ 120 = ( 30+10 )*( t - 1 ) }}}
(2) {{{ 120 = 40*( t - 1 ) }}}
(2) {{{ 3 = t - 1 }}}
(2) {{{ t = 4 }}} hrs
OK