Question 1049953
Let {{{ s }}} = her speed walking
{{{ s + 4 }}} = her speed biking
Let {{{ t }}} = her time walking
{{{ 2 - t }}} = her time biking
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Melissa walking:
(1) {{{ 3 = s*t }}}
Melissa biking:
(2) {{{ 3 = ( s+4 )*( 2-t ) }}}
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(2) {{{ 3 = 2s + 8 - s*t - 4t }}}
(2) {{{ 3 = 2s + 8 - t*( s + 4 ) }}}
and
(1) {{{ t = 3/s }}}
By substitution:
(2) {{{ 3 = 2s + 8 - (3/s)*( s + 4 ) }}}
(2) {{{ 2s = (3/s)*( s+4 ) - 5 }}}
Multiply both sides by {{{ s }}}
(2) {{{ 2s^2 = 3*( s+4 ) - 5s }}}
(2) {{{ 2s^2 = 3s + 12 - 5s }}}
(2) {{{ 2s^2 + 2s - 12 = 0 }}}
(2) {{{ s^2 + s - 6 = 0 }}}
(2) {{{ ( s - 2 )*( s + 3 ) }}}
(2) {{{ s = 2 }}} ( speed has to be positive )
Her walking speed is 2 mi/hr
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check:
(1) {{{ t = 3/2 }}}
and
(2) {{{ 3 = ( s+4 )*( 2-t ) }}}
(2) {{{ 3 = ( 2+4 )*( 2-t ) }}}
(2) {{{ 3 = 6*( 2-t ) }}}
(2) {{{ 1 = 2*( 2-t ) }}}
(2) {{{ 1 = 4 - 2t }}}
(2) {{{ 2t = 3 }}}
(2) {{{ t = 3/2 }}}
OK