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有理式的應用 3 (英)

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有理式的應用 3 (英) : 有理式的應用 3

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Two different hoses are being used to fill a fish pond.
Used together, the two hoses take 12
minutes to fill the pond.
If used alone, one hose is able to fill the pond 10
minutes faster than the other.
So 10 minutes faster than the other hose.
How long does each hose take to fill the pond by itself?
So let's think about each of the hoses.
We have a faster hose, and then we have a slower hose.
And let's say that the faster hose fills the pond.
Let's say it takes him f minutes per pond.
Now how long is it going to take the slower hose?
Well, the faster hose does it in 10 minutes less.
It's 10 minutes faster.
So the slower hose is going to take 10 minutes more.
So the slower hose is going to take f plus 10 minutes per
fish pond or per pond.
Now this is in minutes per pond, but if we want to be
able to add rates together, we should really think about in
terms of ponds per minute.
So let's rewrite each of these statements
as ponds per minute.
You could write this as f minutes per 1 pond or f plus
10 minutes per 1 pond.
And if you just take the inverse of each of these
statements, these ratios are equivalent to saying 1 pond
per f minutes.
So it's really not saying anything else.
I'm just inverting the ratio.
Or you could think of it as 1/f ponds per minute.
Same logic right here.
We could essentially rewrite this ratio as 1 over f plus 10
ponds per minute.
So now we have the rate of the faster hose.
We have the rate of the slower hose.
How many pounds per minute for the faster hose?
How many pounds per minute for the slower hose?
If we add these two rates, we'll know the ponds per
minute when they're acting together.
So if we have 1/f ponds per minute plus 1 over f plus 10
ponds per minute.
This is the faster hose, this is the slower hose.
This'll tell us how many ponds for
minute they can do together.
Now, we know that information.
They say the two hoses take 12 minutes.
So let me write that over here.
So combined.
The combined take 12 minutes per pond.
So what is their combined rate in terms of ponds per minute?
So you could view this as 12 minutes per one pond.
You could take the inverse of this or take the ratio in
terms of ponds per minute instead and
you get 1/12 minutes.
Sorry, 1/12 ponds per minute.
In one minute combined, they'll fill 1/12 of a pond.
Which makes complete sense.
Because it takes them 12 minutes to
fill the whole thing.
So in one minute they'll only do 1/12 of it.
So this is their combined rate in ponds per minute.
This is also their combined rate in ponds per minute.
So this is going to be equal to 1/12.
And now we just have to solve for f and then f plus 10 is
going to be how long it takes the slower hose.
So let's multiply.
Let's see what we could do.
We could multiply both sides of this equation times f and
times f plus 10.
So let's do that.
So I'm going to multiply both sides of this equation times f
and f plus 10 times both sides of this equation.
So f and f plus 10.
Scroll down a little bit.
Scroll to the left, so we have some real estate.
So let's distribute this f times f plus 10.
So if we multiply f times f plus 10 times 1/f, that f and
that f will cancel out and we're just going to be left
with an f plus 10.
That's when you multiply that term times the f
times f plus 10.
Now, when you multiply this term, when you multiply 1 over
f plus 10 times f times f plus 10, this and this will cancel
out and you're just left with an f.
So you have plus f is equal to-- and you have over 12.
Actually, well in a second, let me multiply all sides of
this equation by 12.
I'll do that next in a second.
So let's just say this is going to be equal to 1/12
times f squared.
f times f is f squared.
Plus 10f.
And now let's just multiply both sides of
this equation by 12.
I could've done it in the last step so that we don't have any
fractions here.
And so the left-hand side we get 12 times f.
We get 12f plus 120 plus 12f.
The right-hand side, that and that cancels out and you are
left with f squared plus 10f.
And now we have a quadratic.
We just have to get it into a form that we know how to
manipulate or deal with.
Before that, we can simplify it.
We have a 12f and a 12f.
So this becomes 24f plus 120 is equal to f
squared plus 10f.
And then let's get all of this stuff out of
the left-hand side.
They'll get all on the right-hand side.
So from both sides of this equation, let's subtract 24f
and a negative 120 or minus 120.
So you have minus 24f minus 120.
The left-hand side just becomes 0.
That was the whole point.
Right-hand side is f squared.
10 minus 24f is negative 14f.
Minus 120.
Now we could factor this.
Let's see, if you do 20 timea-- yeah, that looks like
it would work.
So negative 20 and 6, when you take their product, give you
negative 120.
And negative 20 plus 6 is negative 14.
So we could factor this right-hand side as 0 is equal
to f minus 20 times f plus 6.
When you multiply negative 20 times 6, you get negative 120.
Negative 20 plus 6 is negative 14.
And the only way that that's going to be equal to 0 is if f
minus 20 is equal to 0 or f plus 6 is equal to 0.
Add 20 to both sides of this equation.
You get f is equal to 20.
Remember, f is how many minutes does it take for the
fast hose to fill the pond.
And then, if you take this one, you
subtract 6 from both sides.
You get f is equal to negative 6.
Now, when we're talking about how many minutes does it take
for the fast hose to fill the pond, it doesn't make any
sense to say that it takes it negative 6
minutes to fill the pond.
So we can't use this answer.
We need a positive answer.
So this is how many minutes it takes the fast hose to fill
the pond. f is equal to 20.
So this right here, the faster hose takes
20 minutes per pond.
I'll write it here.
20 minutes per pond is the fast hose.
And then the slower hose, it takes 10 minutes more.
It's f plus 10.
So it takes 30 minutes per pond.
And we're done.
I don't want to confuse you with this stuff.
The faster hose takes 20 minutes.
Slower hose takes 30 minutes per pond.
If they were to do it together, it would take 12
minutes, which is a little bit more than half.
If you had two faster hoses, it would take 10 minutes.
But this guy's a little bit slower, so it's taking you a
little bit more than 10 minutes.
So it makes sense.
It takes you 12 minutes when they're working together.

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有理式的應用 3 (英)

有理式的應用 3 (英) : 有理式的應用 3