Question 1054600
Her driving time to Shelbyville was
10:45 minus 9:00 = {{{ 1.75 }}} hrs
( I converted 45 min to {{{ .75 }}} hrs )
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Her driving time back to Springfield was
4:15 minus 2:15 = {{{ 2 }}} hrs
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Let {{{ s }}} = her speed in mi/hr
driving to Shelbyville
{{{ s - 10 }}} = her speed driving back 
to Springfield in mi/hr
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Let {{{ d }}} = the distance between 
the two cities in miles
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Equation for drive to Shelbyville:
(1) {{{ d = s*1.75 }}}
Equation for drive back to Springfield:
(2) {{{ d = ( s - 10 )*2 }}}
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Substitute (1) into (2)
(2) {{{ 1.75s = 2*( s - 10 ) }}}
(2) {{{ 1.75s = 2s - 20 }}}
(2) {{{ .25s = 20 }}}
(2) {{{ s = 80 }}}
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(1) {{{ d = 1.75s }}}
(1) {{{ d = 1.75*80 }}}
(1) {{{ d = 140 }}}
the distance between the 2 cities is 140 mi
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check answer:
(2) {{{ d = ( s - 10 )*2 }}}
(2) {{{ d = ( 80 - 10 )*2 }}}
(2) {{{ d = 2*70 }}}
(2) {{{ d = 140 }}}
OK