Expert: Josh - 4/4/2006

Question
Hello:

I want to thank you for the reply; however I'm confused! I'll try to make it simple as I can as to what I want. I am looking for a simple short solution for the calculation. Is there one or perhaps more possible?  Something not requiring algebra?

I thank you for your assistance and help regarding my question.  Please be advised that my math skills are not as advanced as yours; so when you explain mathematical situations please keep this in mind!
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Followup To
Question -
Hello:

I want to thank you for the follow-up reply.

In my calculation, I want it to be so that the missing factors total 20, 4 + 16. So, it is not a "just a coincidence."  From you answer, I have the understanding that there are more than one possible numbers that can be used. When I indicated that the two missing factors (4 and 16) must total 20 from the denominator and the sum for the numerator (2 + 4) must be 30% of this denominator, (4 + 16 = 20) and (2 + 4)/20 = 30% X 20 = 6.
Isn't the only solution the the calulation regarding the above stipulation only 4 and 16? I'm still somewhat confused.

I thank you once again for your help and assistance with my questions!
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Followup To
Question -
Hello:

I want to thank you for the reply and answer. I'll try to be less confusing with this follow-up question.

Yes, I want to fill in the blank spaces in the calculation. However, assume that we do not know what the two products are, 2 and 4 in (2 + 4)/20. I cannot use 2=0.5*a and 4=0.25*b to determine the answers because the 2 and 4 are not known yet.  I need to determine the two missing factors of 4 and 16 are first.

I thank you for your follow-up reply and your assistance with my question.
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Followup To
Question -
Hello:

By using the information provided from the following calculation, 1/2 , 1/4, 6, 20 and 30%, is it possible to determine the two products of 2 and 4, which become the numerators 2/20 + 4/20, if the other factors of 4 and 16 in ( 1/2 X 4) + (1/4 X 16) are unknown?
I am especially interested in a simple solution.

(1/2 X ____) + (1/4 X ____)/20 = (2 + 4)/20 = 6/20

Answers: (1/2 X 4) + (1/4 X 16)/20 = (2 + 4)/20 = 6/20

The two missing factors (4 and 16) must total 20 from the denominator and the sum for the numerator (2 + 4) must be 30% of this denominator, (4 + 16 = 20) and (2 + 4)/20 = 30% X 20 = 6.

I thank you for any helpful reply provided.

Answer -
I am a little confused by your question. You want to fill out the blanks...Is that the question you are asking?

Forget about "The two missing factors (4 and 16) must total 20 from the denominator" for the time being,

We have the fraction 6/20 and this is equivalent to (2+4)/20. Comparing the numerator with (1/2 X ____) + (1/4 X ____)/20, 2+4=(0.5*a)+(0.25*b)

One obvious solution is obtained by setting 2=0.5*a, 4=0.25*b. In this case, a=4, b=16.

Answer -
To solve (1/2 X ____) + (1/4 X ____)/20 = 6/20.

Focus on the numerator in the expression
[(1/2)*a + (1/4)*b]/20 = 6/20 ...[#1]
Equating the numerator in the LHS and RHS of the equation, writing (1/2) as (1/4)*2, we get
(1/4)*2*a+(1/4)*b = 6
(1/4)*[2a+b]=6
2a+b=24
We see that the solution is NOT unique...[#2]
An obvious solution is found by letting a=b. In this case, 2a+b=3a=24. It follows a=8 is a solution.

To verify this, put a=8, b=8 into [#1].

Another solution is obtained by inspection.
We can set a=4, b=24-2a=16 is the solution pair which you found last time.

Note: In your original question, you mentioned that (4+16)=20. This is just a coincidence; not a requirement for the solution. It just turned out that one feasible solution for the blanks happen to add up to twenty. It is by no means a constraint.

In today's reply, see [#2] from above, we see that the solution is not unique. Making the observations we did in my last reply, the general solution is not obvious.
Answer -
I had the answer written two days ago. Due to internal server problems, I had not been able to post this until today.

Although I don't see a good reason for stipulating the condition that the two unknowns ("a" and "b") should add up to 20, if you insist, an additional constraint "a+b=20" may be enforced.

Continuing on from line [#2] in yesterday's reply, we have
2a+b=24 ...call this [#3] and now we also have
a+b=20 ...call this [#4]

Since these two equations are linearly independent, from linear algebra, it can be shown that there is only one solution for "a" and "b".

Solving [#3] and [#4] gives a=4, b=16.

There are pros and cons to this:
- Insisting that a+b=20, you delve into greater depths in linear algebra, which I don't think was your original intention. But this is the proper way for understanding the problem, even though it started out as a "fill in the blanks" problem.
- Having "a+b=20", the general solution to the problem is overlooked. It is normally a good idea to begin with minimal restriction on the solution space.
- It's best not to make "ad-hoc" assumptions in the beginning. Once the general solution is found, you can add other constraints (see [#4] in addition to [#3]) to obtain an unique solution.  

Answer
As far as I can see, you have two choices. Solutions can be found by trial and error, this requires very little formal understanding of a problem and there is usually not a lot of intuition that we can talk about. For the problem that you have described, the algebraic techniques that I have introduced are in my opinion essential to understanding how the solutions are determined. If anything, trying to spot the answer by inspection is not going to advance your understanding of the problem. You need a concrete method that finds you the answer in less time. So, the short answer is "yes", algebra is required to determine the full set of solutions. Indeed, it is the simplest methodology for finding the answer.

Regarding "Is there one or perhaps more (solution) possible?" The answer is yes. If you do not enforce the condition that the two unknowns "a" and "b" add up to 20, then, there are multiple solutions (please refer to my second reply). If you insist that the two unknowns in the numberator "a" and "b" must add up to 20, as I have shown in my third reply (the last reply), there can be only one solution for "a" and "b" (the unknowns for the blanks).

Regarding "my math skills are not as advanced as yours", I think that you are getting the false impression that the algebraic method is more difficult than it really is.
The steps I have taken to determine the answer require nothing more than simple additions and subtractions. Perhaps you should mention what actually causes the confusion. I have made every effort to keep things simple, and I have taken a standard approach to your problem. The technique seems long, only because I have given ample explanation (at least I tried). If the answer is communicated between two people with a similar background or level of understanding, the solution can be written in a few lines without any explanation.

For instance, the solution may be condensed as:

To solve (1/2 X ____) + (1/4 X ____)/20 = 6/20.
Let the blanks be given by "a" and "b".
We have [(1/2)*a + (1/4)*b]/20 = 6/20 ......[#1]
Equating the numerator in the LHS and RHS of the equation, writing (1/2) as (1/4)*2, we get
(1/4)*2*a+(1/4)*b = 6
(1/4)*[2a+b]=6
2a+b=24 ...[#2]

From [#2], we see that the solution is not unique.
However, if the condition a+b=20 is imposed, an unique solution may be found, viz., a=4, b=16.

Sure, there might be things you may not be able to follow, but if you work through this, asking for clarification, I don't think this technique is unreasonable. It is a "systematic" way of finding the answer (all possible solutions). This, of course, requires a willingness on your part, to have a go at formulating and solving equations with two unknowns. Without this, if you rely on trial-and-error, your "guess" of the solution is as good as anyone's guess.