Integrated math 3
- Analyzing structure word problem: pet store (1 of 2)
- Analyzing structure word problem: pet store (2 of 2)
- Rational equations word problem: combined rates
- Rational equations word problem: combined rates (example 2)
- Rational equations word problem: eliminating solutions
- Reasoning about unknown variables
- Reasoning about unknown variables: divisibility
- Structure in rational expression
Rational equations word problem: eliminating solutions
Sal solves a word problem about the combined pool-filling rates of two water hoses, by creating a rational equation that models the situation. The equation has a solution that is eliminated due to the context. Created by Sal Khan and Monterey Institute for Technology and Education.
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- Why is it incorrect to say f+(f+10)=12? That f minutes per one pond plus f+10 minutes per one pond equals 12 minutes per one pond?(30 votes)
- When the two hoses are filling the pond together, all we know is that it takes a total of 12 minutes. We know that one hose can fill a pond 10 minutes faster than the other, but that's only when it's working on it's own.
Your statement basically says that one hose puts in 11 times the work of the other hose when they're both filling a pond together, which is not a given in this problem.
Hope this helps, you had me stumped for a bit!
- 3:37I don't understand why Sal must multiply everything by f(f+10), and how he got the answer.(8 votes)
- The reason you want f(f+10) and not just f or (f+10) is because f(f+10) is a common factor of both f and f+10. That way, when you do f(f+10) [1/(f) + 1/(f+10)] = 1/12, you can simplify the fraction into f+10 + f = 1/12, which will give you 2f + 10 = 1/12. Then you can solve for f.(11 votes)
- At3:35, why did you multiply both sides of the equation? it just sounded random to me.(8 votes)
- Because, whatever you do one side of an equation, you have to do to the other side. It is just basic algebra principles.(5 votes)
- In order to define the variables, Sal defines the variable for the faster hose first "1 / f" and based on that defines the variable for the slower hose "1 / f + 10" (which means it takes the slower hose 10 minutes longer to fill the pond.
I addressed the problem differently. I defined the variable for the slower hose first "1 / s" and based on that I defined the variable for the faster one "1 / s - 10" (which means it takes the faster hose 10 minuets less to fill the pond).
I believe both of these mean the same but only see the problem from different angels. (according to Sal's way if "f = 20" then "f + 10 =30"; according to my way if "s = 30" then "s - 10 = 20") But much to my surprise I came up with a totally different answer: ( x - 4 ) ( x - 30 ) = 0.
Could somebody explain why? I'm absolutely clueless!(7 votes)
- I'm answering to the following question:
"How can you choose between eliminating 4 or 20?"
You can't choose which one, only given ( x - 4 ) ( x - 30 ) = 0.
But if you choose x = 4 and find another hose speed, you will get:
another hose speed = x - 10 = -6
Since negative value is not suitable for this problem, you can deduce to the following solution:
x isn't 4, hence, x = 30
another hose speed = x - 10 = 20(2 votes)
- why does it have to be 1/(x units) + 1/(x units) = 1/(y units). I tried it with the inverse as Ponds per Minute and the answer doesnt work, but I am not sure why? I want to make sure that I correctly apply this type of proportional question in future.(5 votes)
- If I'm not mistaken all 3 denominators are time/pond units:
1/(f minutes/pond) + 1/(f + 10 min/pond) = 1/(12 min/pond).
I also tried to do it by rates (ponds/min) and was at a dead end (and didn't figure out why). I also was trying to find some unifying ideas and was not successful. ;-|(1 vote)
- What does f = -6 mean then since it does not have a real-world significance like f = 20?(2 votes)
- In this example, the f = -6 solution would represent a hose that sucks water OUT of the pond!
You can interpret "filling an empty pond in -6 minutes" as "emptying a full pond in 6 minutes". The other hose would have to fill the pond in -6 + 10 = 4 minutes, when used alone.
Both hoses together would then fill the pond in 12 minutes. This makes sense, because the inlet hose goes faster than the outlet hose, so the pond would eventually get filled, albeit in a longer time than the 4 minutes that the inlet hose alone would take.(2 votes)
- A dog kennel owner plans to build five adjacent rectangular running pens out of 150 meters of fencing. If each pen measures x meters by y meters, with a total area of 468 square meters for the five pens, which of the following quadratic equations can be used to determine the value of x?(1 vote)
- I am assuming that "adjacent" means that the pens share a common side with their neighbors.
This means that the big rectangle the five smaller rectangles form would have a long side of 5x and a small side of y.
Thus 5xy= 468 m²
The total amount of fencing used will be the perimeter of the larger rectangle plus 4y (those are the lengths of fence that subdivide the larger rectangle into 5 smaller rectangles.
The comes up to be 2(5x+y)+4y = 150 meters
Simplifying gives us: 10x + 6y = 150
Solving for y:
6y = 150 - 10x
y = 25 - (5/3)x
5x[25 - (5/3)x] = 468
125x- (25/3)x² = 468
NOTE: The problem didn't say which side was x and which was y, so I picked one at random. So, you really need to solve for both x and y.
If x is taken as the side that is the short side of the combined rectangle, then your quadratic would be this:
5xy = 468 (same as before)
2(5y+x)+4x = 150 meters
10y + 6x = 150
y = 15 - (3/5)x
5x[15-(3/5)x] = 468
75x - 3x² = 468
What you get for x will depend on what side you decide to call x. But, with both versions, you get the same dimensions.(3 votes)
- why do we get a negative answer that's wrong? Is there no way to calculate this problem without getting the wrong (-6) answer?(2 votes)
- Algebra problems often have false solutions. It is like the use of a word -- there might be 4 definitions in the dictionary, so you have to understand the context to know whether the definition applies. So, you sometimes have solutions to a problem that make no sense and have to be tossed out like tossing out a definition to a word that doesn't apply to the situation you're dealing with.
So, to answer your question: while there may sometimes be ways of avoiding false solutions (usually called "extraneous solutions") it is not always possible to void them. In this problem's case, there is no practical way of solving this with the techniques you learned so far to avoid getting the extraneous answer.(2 votes)
- Can someone explain to me why this is wrong?
12*1/f + 12*1/f+10=1
12+12=1 * f * (f+10)
24=f^2+10f <====something's wrong here(0 votes)
- I am assuming you meant"
(1/f) + 1/(f+10) = 12
You did not have a common denominator, so you cannot move the
f+10the way you did. Also, this would be easier if you did not move the
12: Here is how I'd suggest doing this problem:
(1/f) + 1/(f+10) = 12
(f+10)/[f(f+10)] + f/[f(f+10)] = 12
(f+10 + f) / [f(f+10)] = 12
(f+10 + f) = 12[f(f+10)]
2f + 10 = 12[f²+10f]
f + 5 = 6[f² + 10f] ← divide out the common factor of 2
f + 5 = 6f² + 60f
0 = 6f² + 60f - f - 5
0 = 6f² + 59f - 5 where x≠0 and x≠ −10
This cannot be rationally factored, so if you want to solve it you need to use the quadratic formula.(5 votes)
- How do you do variables that are squared and are multiplied? Like 7x^2 -13x = 2
I am really confused.(1 vote)
- You factor it by doing the reverse of FOIL. Bring the 2 to the other side of the equation.(3 votes)
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 times-- 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.