Question 924458
Let {{{ s }}} = the speed in mi/hr fpr
the 1st part of the trip
Let {{{ t }}} = time in hrs for 
the 1st part of the trip
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(1) {{{ 22 = s*t }}}
and
(2) {{{ 6 = ( s-5 )*( 3-t ) }}}
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(2) {{{ 6 = 3s - 15 - s*t + 5t }}}
(2) {{{ 5t - s*t + 3s = 21 }}}
and
(1) {{{ s = 22/t }}}
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(2) {{{ 5t - (22/t)*t + 3*(22/t) = 21 }}}
(2) {{{ 5t - 22 + 66/t = 21 }}}
(2) {{{ 5t^2 - 22t + 66 = 21t }}}
(2) {{{ 5t^2 - 43t + 66 = 0 }}}
Use quadratic formula
{{{ t = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 5 }}}
{{{ b = -43 }}}
{{{ c = 66 }}}
{{{ t = (-(-43) +- sqrt( (-43)^2 - 4*5*66 )) / (2*5) }}}
{{{ t = ( 43 +- sqrt( 1849 - 1320 )) / 10 }}}
{{{ t = ( 43 +- sqrt( 529 )) / 10 }}}
{{{ t = ( 43 -23 ) / 10 }}}
{{{ t = 2 }}}
and
{{{ 3 - 2 = 1 }}}
The speed for the 1st part is:
(1) {{{ s = 22/t }}}
(1) {{{ s = 22/2 }}}
(1) {{{ s = 11 }}} mi/hr
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The speed for the 2nd part is:
{{{ s - 5 = 6 }}} mi/hr
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check:
(2) {{{ 6 = ( 11-5 )*( 3-2 ) }}}
(2) {{{ 6 = 6*1 }}}
and
(1) {{{ 22 = s*t }}}
(1) {{{ 22 = 11*2 }}}
(1) {{{ 22 = 22 }}}
OK