Question 1105126
Let {{{ s }}} = her normal speed in km/hr
Let {{{ t }}} = her normal time in hrs
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(1) {{{ 5 = s*t }}}
(2) {{{ 5 = ( s+2 )*( t - 3 ) }}}
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(2) {{{ 5 = s*t + 2t - 3s - 6 }}}
(2) {{{ s*t + 2t - 3s = 11 }}}
(2) {{{ s*( t - 3 ) + 2t = 11 }}}
and
(1) {{{ s = 5/t }}}
(2) {{{ (5/t)*( t - 3 ) + 2t = 11 }}}
(2) {{{ 5*( t - 3 ) + 2t^2 = 11t }}}
(2) {{{ 5t - 15 + 2t^2 - 11t = 0 }}}
(2) {{{ 2t^2 - 6t - 15 = 0 }}}
use quadratic formula
{{{ t = ( -b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{ a = 2 }}}
{{{ b = -6 }}}
{{{ c = -15 }}}
{{{ t = ( -(-6) +- sqrt( (-6)^2-4*2*(-15) ))/(2*2) }}}
{{{ t = ( 6 +-sqrt( 36 + 120 )) / 4 }}}
{{{ t = ( 6 + sqrt( 156 ) )/4 }}}
{{{ t = ( 6 + 12.49 )/4 }}}
{{{ t = 18.49/4 }}}
{{{ t = 4.6225 }}} 
and
(1) {{{ s = 5/t }}}
(1) {{{ s = 5/4.6225 }}}
(1) {{{ s = 1.0817 }}}
She normally traveled 1.0817 km/hr
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check:
(2) {{{ 5 = ( s+2 )*( t - 3 ) }}}
(2) {{{ 5 = ( 1.0817+2 )*( 4.6225 - 3 ) }}}
(2) {{{ 5 = 3.0817*1.6225 }}}
(2) {{{ 5 = 5.00006 }}}
close enough!
check the math and get another
opinion if needed