anonymous
  • anonymous
Larry's time to travel 357 miles is 3 hours more than Terrell's time to travel 220 miles. Terrell drove 4 miles per hour faster than Larry. How fast did each one travel?
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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jdoe0001
  • jdoe0001
so, if terrell traveled "h" hours, larry traveled "h+3" terrel drove 4 miles faster so if larry drove "r" m/h fast, terrell drove "r+4" m/h faster
jdoe0001
  • jdoe0001
so $$ \begin{matrix} & d & t & r \\ \hline larry & 357 & h+3 & r \\ terrell & 220 & h & r+4\\ \end{matrix} $$
jdoe0001
  • jdoe0001
so, using the d = rt distance formula larry traveled "t" distance, or t = d/r or \(h+3 = \cfrac{357}{r} \implies h = \cfrac{357}{r}-3\)

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jdoe0001
  • jdoe0001
terrell's rate is r+4 which \(r+4 =\cfrac{220}{h}\) but \( h = \cfrac{357}{r}-3\) from the 1st equation so $$ r+4 =\cfrac{220}{\cfrac{357}{r}-3} $$
jdoe0001
  • jdoe0001
$$ r+4 =\cfrac{220}{\cfrac{357}{r}-3} \implies r+4 = \cfrac{220r}{357-3r}\\ \implies (r+4)(357-3r) = 220r\\ \implies -3r^2-12r+357r+1428-220r = 0\\ \implies \color{blue}{-3r^2+125r+1428=0}\\ \text{from there you'd need to use the quadratic formula}\\ \text{to get "r" or "larry's rate"} $$
jdoe0001
  • jdoe0001
@inprogress you'd get 2 values, discard the negative one, since "rate" is positive
jdoe0001
  • jdoe0001
@inprogress $$ \text{quadratic formula for it}\\ \cfrac{-125\sqrt{15625+17136}}{-6}\\ \cfrac{-125\pm 181}{-6} $$
anonymous
  • anonymous
A solution using Mathematica to perform the calculations is attached.
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