Ryan drives 10mi/h slower than Maya, And it takes ryan 1 hour longer to travel 300 miles. How long does it take Maya to make a trip

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Ryan drives 10mi/h slower than Maya, And it takes ryan 1 hour longer to travel 300 miles. How long does it take Maya to make a trip

Mathematics
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I'll let amistre get this :p
well, you need two equations and two variables to be able to solve this
one variable is the time it takes Maya to take the trip, and one variable is the time it takes ryan to take the trip

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I got 300(x)= 300(x-1) + 10x(x-1)
well, actually, let's make the two variables the speed of each person - then we can easily solve for how long it takes maya to drive 300 miles. So: 1. R = M - 10 2. 300/R - 300/M = 1
i think i have an answer.
what did you get?
i got 60 but it was kind of a pain the way i did it. maybe an easier way.
i used maya's rate as r, ryan's as r - 10 and then \[\frac{300}{r}=\frac{300}{r-10}+1\]
that is ryan's time is one hour more than maya's.
right. yeah that makes sense
sorry but i do not understand
i made a mistake somewhere i am trying again.
first of all the answer: ryan travels at 50 mpr and maya at 60. it is easy to check: \[d=rt\] \[t=\frac{d}{r}\] if maya travels 300 miles at 60 miles an hour she makes the trip in \[\frac{300}{60}=5\] hours while ryan takes \[\frac{300}{50}=6\] hours or one hour longer. so this is the correct answer.
O i understand thank you!
well i got the correct answer, but i am trying to find the right algebra to solve it for you.
whew finally. ok here goes. ready?
yes!
ok we use \[t=\frac{d}{r}\] and we know that ryans rate is r-10 and his maya's time is one less than ryan's. Maya's time is \[\frac{300}{r}\] and ryan's is \[\frac{300}{r-10}\] and \[\frac{300}{r}-1=\frac{300}{r-10}\]
that says maya's time is one less than ryans. to solve for r, easiest to subtract first on the left hand side: \[\frac{300-r}{r}=\frac{300}{r-10}\] now cross multiply: \[(r-10)(300-r)=300r\] mutiply out \[300r-3000-r^2+10r=300r\] subtract 300r \[-3000 -r^2+10r=0\]
damn
thought i had \[(r-60)(r+50)=0\] so r = 60 but i seem to have made another mistake somewhere. i will try to find it.
oh lord. i put the one on the wrong side! should have been \[\frac{300}{r}+1=\frac{300}{r-10}\]
maya's time plus one gives ryans time. ok now. \[\frac{300+r}{r}=\frac{300}{r-10}\] \[(300+r)(r-10)=300r\] \[300r -3000+r^2-10r=300r\] \[r^2-10r-300=0\] \[(r-60)(r+50)=0\] \[r=60\] or \[r=-50\] which makes no sense in this problem so \[r=60\] sorry it took a while.
its okay and thanks again
sorry i made a mistake originally. hope it was not too confusing.
it wasnt thanks

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