Question 675983
Let {{{ s }}} = Frank's speed in mi/hr
{{{ s + 1 }}} = Joe's speed in mi/hr
{{{ s*1 = s }}} = Frank's head start in miles
{{{ t }}} = Frank's time after his head start
{{{ t - 1 }}} = Joe's time after Frank's head start
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Frank's equation:
(1) {{{ 30 - s  = s*t }}}
Joe's equation:
(2) {{{ 30 = ( s + 1 )*( t - 1 ) }}}
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(2) {{{ 30 = s*t + t - s - 1 }}}
(2) {{{ 30 = t*( s + 1 ) - ( s + 1 ) }}}
and
(1) {{{ t = ( 30 - s ) / s }}}
Substitute this into (2)
(2) {{{ 30 = (1/s)*( 30 - s )*( s + 1 ) - ( s + 1 ) }}}
(2) {{{ 30 = (1/s)*( 30s - s^2 + 30 - s ) - ( s + 1 ) }}}
(2) {{{ 30 = 30 - s + 30/s - 1 - s - 1 }}}
(2) {{{ 30/s = 2s + 2 }}}
(2) {{{ 30 = 2s^2 + 2s }}}
(2) {{{ s^2 + s - 15 = 0 }}}
Use quadratic formula
{{{ s = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 1 }}}
{{{ b = 1 }}}
{{{ c = -15 }}}
{{{ s = (-1 +- sqrt( 1^2 - 4*1*(-15) )) / (2*1) }}}
{{{ s = (-1 +- sqrt( 1 + 60 )) / 2 }}}
{{{ s = (-1 + sqrt( 61 )) / 2 }}}
{{{ s = ( -1 + 7.8103 ) / 2 }}}
{{{ s = 6.8103/2 }}}
{{{ s = 3.4051 }}} mi/hr = Frank's speed
{{{ s + 1 = 4.4051 }}} mi/hr = Joe's speed
My math could have got messed up- check it