Question 685815
You need 2 equations, one for regular commute
and one for icy roads
Let {{{ s }}} = her regular speed in mi/hr
Let {{{ t }}} = her regular time in hrs
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given:
For regular commute:
(1) {{{ 6 = s*t }}}
For icy roads:
(2) {{{ 6 = ( s - 20 )*( t + 3/20 ) }}}
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(2) {{{ 6 = s*t - 20t + (3/20)*s - 3 }}}
Make these substitutions:
(1) {{{ s*t = 6 }}}
and
(1) {{{ s = 6/t }}}
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(2) {{{ 6 = 6 - 20t + (3/20)*(6/t) - 3 }}}
(2) {{{ 3 = -20t + ( 9/10t ) }}}
Multiply both sides by {{{ 10t }}}
(2) {{{ 30t = -200t^2 + 9 }}}
(2) {{{ 200t^2 + 30t - 9 = 0 }}}
Use quadratic formula
{{{ t = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}} 
{{{ a = 200 }}}
{{{ b = 30 }}}
{{{ c = -9 }}}
{{{ t = (-30 +- sqrt( 30^2 - 4*200*(-9) )) / (2*200) }}} 
{{{ t = (-30 +- sqrt( 900 + 7200 )) / 400 }}} 
{{{ t = (-30 +- sqrt( 8100 )) / 400 }}} 
{{{ t = ( -30 + 90 ) / 400 }}} ( ignore the negative square root )
{{{ t = 60 / 400 }}}
{{{ t = .15 }}} hrs
{{{ .15*60 = 9 }}} min
When roads are icy,
{{{ t + 9 = 9 + 9 }}}
{{{ t + 9 = 18 }}}
When roads are icy, it takes 18 minutes
check:
(1) {{{ 6 = s*t }}}
(1) {{{ 6 = s*.15 }}}
(1) {{{ s = 40 }}}
and
(2) {{{ 6 = ( s - 20 )*( t + 3/20 ) }}}
(2) {{{ 6 = ( 40 - 20 )*( .15 + .15 ) }}}
(2) {{{ 6 = 20 * .3 }}}
(2) {{{ 6 = 6 }}}
OK