Question 833439
Uniform rates for travel, using {{{R*T=D}}}.
This example is in kilometers and hours.
8 minutes is {{{8/60=2/15}}} hour
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Let t = time for John to finish the race



______________speed_____________time_______________distance(km)
Danny_________(___)_____________t-8/60_____________20
John_________(____)______________t_________________20



Danny is at a rate r+5, if John went at a rate of r.  The rates are unknown.
This and also reducing the time expression for Danny,...


______________speed_____________time_______________distance(km)
Danny_________(r+5)_____________t-2/15_____________20
John_________(r)_________________t_________________20


A rate*time=distance equation can be made for each of Danny and John.


John: {{{highlight_green(rt=20)}}}, which can also be used as {{{highlight_green(t=20/r)}}}


Danny:  {{{(r+5)(t-2/15)=20}}}
{{{rt+5t-2r/15-2/3=20}}}
Substituting with John's equation,
{{{20+5t-2r/15-2/3=20}}}
{{{5t-2r/15-2/3=0}}}
Multiply members by 15 to clear the denominators
{{{75t-2r-10=0}}}
Make use of John's equation:
{{{75(20/r)-2r-10=0}}}
{{{1500/r-2r-10=0}}}
{{{1500-2r^2-10r=0}}}
{{{-2r^2-10r+1500=0}}}
{{{highlight_green(r^2+5r-750=0)}}}----Quadratic equation in r, the rate for John.
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Use the * general solution to a quadratic equation to solve for r, even through the question really just asks for Danny's rate; both rates are given a relationship in the description.
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* The quadratic expression on the left hand side is factorable, so you might be able to explore the factorizations of 750, but you will find that the equation is {{{highlight((r-30)(r+25)=0)}}}.
The meaningful answer is, {{{highlight(highlight(r=30))}}}.  This means, Danny's rate was 35 kilometers per hour.