Question 873342
_________________speed______________time__________distance
CAR______________r+25_______________{{{t-1&1/4}}}_______{{{(r+25)(t-1&1/4)=550}}}
TRAIN____________r__________________t_____________{{{rt=550}}}


Two equations occur based on knowing the one-way distance.


CAR equation:  {{{(r+25)(t-5/4)=550}}}
{{{rt+25t-5r/4-125/4=550}}}
{{{rt+25t-5r/4-125/4=rt}}}, the substitution using the TRAIN equation.
{{{25t-5r/4-125/4=0}}}
{{{100t-5r-125=0}}}
{{{4t-r-5=0}}}
{{{r=4t-5}}} allowing for substitution now into the TRAIN equation.
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{{{rt=550}}}
{{{(4t-5)t=550}}}
{{{4t^2-5t=550}}}
{{{highlight_green(4t^2-5t-550=0)}}} ----You want to solve this for time t, and use it to find the speeds of the train and CAR.
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discrim, {{{25^2+4*4*550=9425}}}
discrim, {{{9425=25*377=5^2*13*29}}}
{{{highlight(t=(5+5sqrt(377))/8)}}} HOURS
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{{{r=550/t}}}
{{{r=550/((5+5sqrt(377))/8)}}}
{{{r=(550*8)/(5+5sqrt(377))}}}
{{{r=(110/8)/(1+sqrt(377))}}}
{{{r=(55/4)/(1+sqrt(77))}}}
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TRAIN speed, not quite yet in proper simplified radical form,
{{{r+25=25+(55/4)/(1+sqrt(377))}}}
{{{highlight(r+25=25+55/(4(1+sqrt(377))))}}}