Question 854011
Let {{{ s }}} = his speed riding to the city
{{{ s-30 }}} = his speed returning
Let {{{ t }}} = his time in hours to get to the city
{{{ t + 3 }}} = his time returning
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Equation for 1st trip:
(1) {{{ 100 = s*t }}}
Equation for return trip
(2) {{{ 100 = ( s - 30 )*( t + 3 ) }}}
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(2) {{{ 100 = s*t - 30t + 3s - 90 }}}
(2) {{{ 100 = ( t + 3 )*s - 30t - 90 }}}
(2) {{{ ( t + 3 )*s = 30t + 190 }}}
(2) {{{ s = ( 30t + 190 ) / ( t + 3 ) }}}
Substitute this result into (1)
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(1) {{{ 100 = ( t*( 30t + 190 )) / ( t + 3 ) }}}
(1) {{{ 100t + 300 = 30t^2 + 190t }}}
(1) {{{ 30t^2 + 90t - 300 = 0 }}}
(1) {{{ t^2 + 3t - 10 = 0 }}}
I can see by looking at it, this is:
(1) {{{ ( t + 5 )*( t - 2 ) }}}
{{{ t = 2 }}} ( because time is positive, not negative )
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{{{ t = 2 }}} hrs is his time to get to the city
{{{ t + 3 = 5 }}} hrs is his time to return
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check:
(1) {{{ 100 = s*t }}}
(1) {{{ 100 = s*2 }}}
(1) {{{ s = 50 }}}
and
(2) {{{ 100 = ( s - 30 )*( t + 3 ) }}}
(2) {{{ 100 = ( s - 30 )*5 }}}
(2) {{{ 20 = s - 30 }}}
(2) {{{ s = 20 + 30 }}}
(2) {{{ s = 50 }}}
OK