Question 631628
Let {{{ s }}} = the speed limit
{{{ s - 5 }}} = 5 mi/hr below speed limit
{{{ s + 5 }}} = 5 mi/hr above the speed limit
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Let {{{ t }}} = time spent driving at {{{ s + 5 }}} mi/hr
{{{ t + 24/60 }}} = time spent driving at {{{ s - 5 }}}
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(1) {{{ 390/2 = ( s + 5 )*t }}}
(2) {{{ 390/2 = ( s - 5 )*( t + 24/60 ) }}}
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(1) {{{ 195 = s*t + 5t }}}
(2) {{{ 195 = ( s - 5 )*( t + 2/5 ) }}}
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(2) {{{ 195 = s*t - 5t + (2/5)*s - 2 }}}
(1) {{{ s*t = 195 - 5t }}}
(1) {{{ s = 195/t - 5 }}}
Substitute (1) into (2)
(2) {{{ 195 = ( 195/t - 5 )*t - 5t + (2/5)*( 195/t - 5 ) - 2 }}}
(2) {{{ 195 = 195 - 5t - 5t + 78/t - 2 - 2 }}}
(2) {{{ 0 = -10t + 78/t - 4 }}}
(2) {{{ 10t = 78/t - 4 }}}
(2) {{{ 10t^2 = 78 - 4t }}}
(2) {{{ 10t^2 + 4t - 78 = 0 }}}
(2) {{{ 5t^2 + 2t - 39 = 0 }}}
Use the quadratic formula
{{{ t = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }} 
{{{ a = 5 }}}
{{{ b = 2 }}}
{{{ c = -39 }}}
{{{ t = (-2 +- sqrt( 2^2 - 4*5*(-39) )) / (2*5) }} 
{{{ t = (-2 +- sqrt( 4 + 780 )) / 10 }} 
{{{ t = (-2 +- sqrt( 784 )) / 10 }} 
{{{ t = (-2 + 28 ) / 10 }}  ( ignore the negative root )
{{{ t = 26/10 }}}
{{{ t = 13/5 }}} hrs
and
(1) {{{ 195 = ( s + 5 )*t }}}
(1) {{{ 195 = (  s + 5 )*(13/5) }}}
(1) {{{ 195 = (13/5)*s + 13 }}}
(1) {{{ (13/5)*s = 195 - 13 }}}
(1) {{{ s = (5/13)*182 }}}
(1) {{{ s = 70 }}}
The speed limit is 70 mi/hr