Question 1024657
The hiker had a head start of 
{{{ d[1] = 3*1 }}}
{{{ d[1] = 3 }}} mi
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Start a stop watch when the cyclist leaves
Let {{{ t }}} = the time in hrs on the stop watch
when the cyclist catches up with the hiker
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Let {{{ d }}} = distance cyclist traveled until
cyclist caught up with hiker
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Hiker's equation:
(1) {{{ d - 3 = 3t }}}
Cyclist's equation:
(2) {{{ d = 6t }}}
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Substitute (2) into (1)
(1) {{{ 6t - 3 = 3t }}}
(1) {{{ 3t = 3 }}}
(1) {{{ t = 1 }}}
The cyclist takes 1 hr to catch up with hiker
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check:
(2) {{{ d = 6t }}}
(2) {{{ d = 6*1 }}}
(2) {{{ d = 6 }}}
and
(1) {{{ d - 3 = 3t }}}
(1) {{{ d - 3 = 3*1 }}}
(1) {{{ d - 3 = 3 }}}
(1) {{{ d = 6 }}}
OK