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45 = rt usual
45 = (r+5)t if she does increase her speed
Solve both for t
45/r = t
45 /(r+5) = t
45/(r + 5) + 1/10 = 45/r multiply each term by 10r(r+5) which is your common denominator
450r + r^2 + 5r = 450r + 2250
r^2 + 5r - 2250 = 0
(r + 50)(r - 45) = 0
r = -50 or r = 45
You throw out the negative number be it cannot be her speed.
Original speed was 45 mph
thanks but if her time is decreased by six, woudnt the one formula be 45/(r+5)=t-6?
No because it says 6 minutes and everything else is in hours... so you need to keep everything in hours.
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In the formula d = rt time is always in hours.
ok but still then 45/(r+5)=t-1/60 ? or am I just being weird
6 minutes is 6 minutes out of 60 minutes which is 1 hour
One minute would be 1/60 of an hour
Six minutes would be 6/60 of an hour which reduces to 1/10
....45/(r+5)=t-1/10? sorry I'm really tired :p
oh haha wait nvm I see you added that in in the next step
The way you have to look at it is two compare the two ways of driving.
normal 45 = rt
Going faster 45 = (r+5)t
because we are comparing times between driving to work the two different ways we solve both sides for t.
45/r = t
45/(r + 5) = t
Usually these problems say it took the same amount of time so you would put the left side of those equations equal to each other, but on this one it says that when she "sped up", it took her 6 minutes less to get to work. (which makes sense). The problem is "HOW DO WE get these equal to each other?" If "speeding" makes your time 6 minutes less than the other... then to make them equal you would add 6 to the "normal" driving.
Ex. I am doing this in minutes because it is easier
So to make these "equal", you would add 6 minutes to the speeding side..