Question 889930
The time actually spent biking was:
{{{ 6 - 2 = 4 }}}
Let {{{ d }}} = his one-way distance
Let {{{ t }}} = time in hrs  spent biking at {{{ 5 }}} km/hr
{{{ 4 - t }}} = time in hrs  spent biking at {{{ 3 }}} km/hr
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given:
(1) {{{ d = 5t }}}
(2) {{{ d = 3*( 4 - t ) }}}
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By substitution:
(2) {{{ 5t = 3*( 4 - t ) }}}
(2) {{{ 5t = 12 - 3t }}}
(2) {{{ 8t = 12 }}}
(2) {{{ t = 3/2 }}}
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(1) {{{ d = 5t }}}
(1) {{{ d = 5*(3/2) }}}
(1) {{{ d = 15/2 }}}
The total distance was:
{{{ 15/2 + 15/2 = 15 }}}
The boy went 15 km
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check:
(2) {{{ d = 3*( 4 - t ) }}}
(2) {{{ 15/2 = 3*( 4 - 3/2 ) }}}
(2) {{{ 15/2 = 3*(5/2) }}}
(2) {{{ 15/2 = 15/2 }}}
OK