Question 179311
we know the distance are equal 3 miles
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lets call kims rate r and her time t
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Bryan's r would be r+10 and his time would be t+1/4(remember this is all in hours)
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Kim:::: 3=rt................eq 1
Bryan:: 3=(r+10)(t-(1/4))...eq 2
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lets rewrite eq 1 to t=3/r and plug it into eq 2
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3=(r+10)((3/r)-(1/4))
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3=(r+10)((12-r)/4r)found common denominator 2nd term on the right
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12r=(r+10)(12-r)....multiplied both sides by 4r
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{{{12r=12r-r^2+120-10r}}}multiplied out right side
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{{{r^2+10r-120=0}}}put all terms on one side
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r=7.04mph-Kim's rate, (as the negative value is thrown out)
:
r+10=17.04mph-Bryans rate
*[invoke quadratic "r", 1, 10, -120]