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## College Algebra

### Unit 8: Lesson 2

Modeling with rational expressions

# 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.

## Want to join the conversation?

• 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?
• 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!
Cheers
• I don't understand why Sal must multiply everything by f(f+10), and how he got the answer.
• 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.
• At , why did you multiply both sides of the equation? it just sounded random to me.
• Because, whatever you do one side of an equation, you have to do to the other side. It is just basic algebra principles.
• 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!
• 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
(1 vote)
• 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.
• 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?
• 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.
• 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
Thus:
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.
• why do we get a negative answer that's wrong? Is there no way to calculate this problem without getting the wrong (-6) answer?
• 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.
• Can someone explain to me why this is wrong?
1/f +1/f+10=1/12
12*1/f + 12*1/f+10=1
12/f+12/f+10=1
12+12=1 * f * (f+10)
24=f^2+10f <====something's wrong here
• I am assuming you meant"
`(1/f) + 1/(f+10) = 12`
You did not have a common denominator, so you cannot move the `f` and `f+10` the 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 2f + 5 =  6f² + 60f 0 =  6f² + 60f - f - 50 =  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.