Question 1046112
Let {{{ t }}} = time in hrs going TO the cafe
{{{ 6 - t }}} = time in hrs returning FROM cafe
Let {{{ d }}} = the one-way distance to cafe
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Going to cafe:
(1) {{{ d = 20t }}}
Returning from cafe:
(2) {{{ d = 10*( 6-t ) }}}
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Plug (1) into (2)
(2) {{{ 20t = 10*( 6-t ) }}}
(2) {{{ 20t = 60 - 10t }}}
(2) {{{ 30t = 60 }}}
(2) {{{ t = 2 }}} hrs
and
{{{ 6 - t = 6 - 2 }}}
{{{ 6 - t = 4 }}} hrs
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Leah took 2 hrs going 20 mi/hr
Leah took 4 hrs going 10 mi/hr
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check:
(1) {{{ d = 20t }}}
(1) {{{ d = 20*2 }}}
(1) {{{ d = 40 }}} mi
and
(2) {{{ d = 10*( 6-t ) }}}
(2) {{{ d = 10*( 6-2 ) }}}
(2) {{{ d = 10*4 }}}
(2) {{{ d = 40 }}} mi
OK