Question 1026000
Let {{{ s }}} = his speed walking into country in km/hr
Let {{{ t }}} = his time walking into country in hrs
{{{ s - 3 }}} = his speed returning in km/hr
{{{ 7 - t }}} = his time returning in hrs
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His equation for walking into country:
(1) {{{ 10 = s*t }}}
His equation for returning from country
(2) {{{ 10 = ( s - 3 )*( 7 - t ) }}}
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(1) {{{ t = 10/s }}}
(2) {{{ 10 = 7s - 21 - s*t + 3t }}}
(2) {{{ t*( 3 - s ) + 7s = 31 }}}
Substitute (1) into (2)
(2) {{{(10/s)*( 3 - s ) + 7s = 31 }}}
(2) {{{ 10*( 3 - s ) + 7s^2 = 31s }}}
(2) {{{ 30 - 10s + 7s^2 - 31s = 0 }}}
(2) {{{ 7s^2 - 41s + 30 = 0 }}}
use the quadratic formula
{{{ s = ( -b +- sqrt( b^2 - 4*a*c )) / (2*a) }}} 
{{{ a = 7 }}}
{{{ b = -41 }}}
{{{ c = 30 }}}
{{{ s = ( -(-41) +- sqrt((-41)^2 - 4*7*30 )) / (2*7) }}} 
{{{ s = ( 41 +- sqrt( 1681 - 840 )) / 14 }}}
{{{ s = ( 41 +- sqrt( 841)) / 14 }}} 
{{{ s = ( 41 + 29) / 14 }}} 
{{{ s = 70/14 }}}
{{{ s = 5 }}}
and
{{{ s = ( 41 - 29) / 14 }}} 
{{{ s = 12/14 }}}
{{{ s = 6/7 }}}
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I can't subtract {{{ 3 }}} from {{{ 6/7 }}},
so {{{ s = 5 }}} must be the answer
{{{ s - 3 = 2 }}}
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his speed walking into country was 5 km/hr
his speed returning was 2 km/hr
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check:
(1) {{{ 10 = s*t }}}
(1) {{{ 10 = 5*t }}}
(1) {{{ t = 2 }}}
and
(2) {{{ 10 = ( 5 - 3 )*( 7 - t ) }}}
(2) {{{ 10 = 2*( 7 - t ) }}}
(2) {{{ 10 = 14 - 2t }}}
(2) {{{ 2t = 4 }}}
(2) {{{ t = 2 }}}
OK