Question 751549
Let {{{ s }}} = his speed in mi/hr walking
{{{ s + 10 }}} = his speed in mi/hr riding the bike
Let {{{ t }}} = his time in hrs spent walking
{{{ 5 - t }}} = his time in hrs spent riding bike
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Equation when riding bike:
(1) {{{ 12 = ( s + 10 )*( 5 - t ) }}}
Equation when walking:
(2) {{{ 8 = s*t }}}
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(1) {{{ 12 = 5s + 50 - s*t - 10t }}}
(1) {{{ 5s - s*t = 10t + 12 - 50 }}}
(1) {{{ s*( 5 - t ) = 10t - 38 }}}
and
(2) {{{ 8 = s*t }}}
(2) {{{ t = 8/s }}}
Substitute this into (1)
(1) {{{ s*( 5 - 8/s ) = 10*(8/s) - 38 }}}
(1) {{{ 5s - 8 = 80/s - 38 }}}
Multiply both sides by {{{ s }}}
(1) {{{ 5s^2 - 8s = 80 - 38s }}}
(1) {{{ 5s^2 + 30s = 80 }}}
(1) {{{ s^2 + 6s = 16 }}}
complete the square
(1) {{{ s^2 + 6s + (6/2)^2 = 16 + (6/2)^2 }}}
(1) {{{ s^2 + 6s + 9 = 16 + 9 }}}
(1) {{{ ( s + 3 )^2 = 5^2 }}}
(1) {{{ s + 3 = 5 }}}
(1) {{{ s = 2 }}}
and
{{{ s + 10 = 12 }}}
His speed in mi/hr walking is 2 mi/hr
His speed in mi/hr riding the bike is 12 mi/hr
check:
(2) {{{ 8 = s*t }}}
(2) {{{ t = 8/2 }}}
(2) {{{ t = 4 }}} hrs
and
(1) {{{ 12 = ( s + 10 )*( 5 - t ) }}}
(1) {{{ 12 = ( 2 + 10 )*( 5 - t ) }}}
(1) {{{ 12 = 12*( 5 - t ) }}}
(1) {{{ 12/12 = 5 - t }}}
(1) {{{ 1 - 5 = -t }}}
(1) {{{ -t = -4 }}}
(1) {{{ t = 4 }}} hrs
OK