Question 760016
Let {{{ t }}} = her time in hrs for the 2nd day
{{{ t + 2 }}} = her time in hrs for the 1st day
{{{ r }}} = her speed on both days
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1st day:
(1) {{{ 600 = r*( t + 2 ) }}}
2nd day:
(2) {{{ 500 = r*t }}}
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This is 2 equations with 2 unknowns, so it's solvable
(2) {{{ r = 500/t }}}
and, by substitution:
(1) {{{ 600 = ( 500/t )*( t + 2 ) }}}
(1) {{{ 600 = 500 + 1000/t }}}
(1) {{{ 100 = 1000/t }}}
(1) {{{ 100t = 1000 }}}
(1) {{{ t = 10 }}}
{{{ t + 2 = 12 }}}
Her time for the 1st day was 12 hrs
Her time for the 2nd day was 10 hrs
check:
(1) {{{ 600 = r*( t + 2 ) }}}
(1) {{{ 600 = r*( 10 + 2 ) }}}
(1) {{{ 12r = 600 }}}
(1) {{{ r = 50 }}}
and
(2) {{{ 500 = r*t }}}
(2) {{{ 500 = r*10 }}}
(2) {{{ r = 50 }}}
OK