# 4x+y+2z=100, x>y, y>z, x+y+z=j, 3x+z=k, j>k, j+k=100.

July 22, 2014 6:07 PM Subscribe

4x+y+2z=100, x>y, y>z, x+y+z=j, 3x+z=k, j>k, j+k=100. WolframAlpha can only give me alternative forms or a solution involving 'real and imaginary parts of z'. Is there a solution I can count on my fingers?

I wish to buy a thing. The thing I buy will have two desirable characteristics: j, and k.

j comprises three sub-characteristics, while k comprises four sub-characteristics.

I have three levels for ranking how important these sub-characteristics are to me - let's say 'super important', 'kinda important' and 'meh, whatever'. Obviously, they are of decreasing importance.

I have rated the three sub-characteristics for j 'super important', 'kinda important' and 'meh, whatever'.

I have rated three of the four sub-characteristics for k 'super important', and the fourth 'meh, whatever'.

This means I have a total of four 'super important', one 'important', and two 'meh, whatevers'.

Finally, j is a little bit more important to me than k. If further defining these variables help, then I think they'd be about a 60/40 split, though I'm flexible. They can be equal if that helps. At this stage I'd settle for k being less than j if it means a solution pops out.

I would like to assign positive numbers to x, y and z, because I'm a bureaucrat, that's why.

So,

x>y (must)

y>z (must)

j = x + y + z

k = 3x + z

4x + y + z = 100 (per cent)

j + k = 100 (per cent)

I can't change the number of characteristics, sub-characteristics, or levels of importance.

I could make k=2x+y+z, and so change the overall equation to 3x+2y+2z, but I'm not smart enough to know if that helps. It didn't help WolframAlpha.

I wish to buy a thing. The thing I buy will have two desirable characteristics: j, and k.

j comprises three sub-characteristics, while k comprises four sub-characteristics.

I have three levels for ranking how important these sub-characteristics are to me - let's say 'super important', 'kinda important' and 'meh, whatever'. Obviously, they are of decreasing importance.

I have rated the three sub-characteristics for j 'super important', 'kinda important' and 'meh, whatever'.

I have rated three of the four sub-characteristics for k 'super important', and the fourth 'meh, whatever'.

This means I have a total of four 'super important', one 'important', and two 'meh, whatevers'.

Finally, j is a little bit more important to me than k. If further defining these variables help, then I think they'd be about a 60/40 split, though I'm flexible. They can be equal if that helps. At this stage I'd settle for k being less than j if it means a solution pops out.

I would like to assign positive numbers to x, y and z, because I'm a bureaucrat, that's why.

So,

x>y (must)

y>z (must)

j = x + y + z

k = 3x + z

4x + y + z = 100 (per cent)

j + k = 100 (per cent)

I can't change the number of characteristics, sub-characteristics, or levels of importance.

I could make k=2x+y+z, and so change the overall equation to 3x+2y+2z, but I'm not smart enough to know if that helps. It didn't help WolframAlpha.

No, I don't see how those equations can possibly give a solution (given your requirement that they be positive).

You have x > y. You also have x + y + z > 3x + z. Therefore, y > 2x. That means x > 2x, which you can't have with positive numbers.

posted by UrineSoakedRube at 6:21 PM on July 22, 2014 [7 favorites]

You have x > y. You also have x + y + z > 3x + z. Therefore, y > 2x. That means x > 2x, which you can't have with positive numbers.

posted by UrineSoakedRube at 6:21 PM on July 22, 2014 [7 favorites]

If j>k, then substituting your formulas for j and k:

x + y + z > 3x + z

Simplifying, that means that

y > 2x

But because you've already stipulated that

x > y

You're saying that x is more than y but 2x is less than y. Assuming finger-counting numbers, I don't think it's possible.

posted by Paragon at 6:22 PM on July 22, 2014

x + y + z > 3x + z

Simplifying, that means that

y > 2x

But because you've already stipulated that

x > y

You're saying that x is more than y but 2x is less than y. Assuming finger-counting numbers, I don't think it's possible.

posted by Paragon at 6:22 PM on July 22, 2014

j = x + y + z

k = 3x + z

4x + y + z = 100 (per cent)

j + k = 100 (per cent)

Take j + k = 100 (per cent) and substitute x + y + z for j to get

x + y + z + k = 100 (per cent)

Take x + y + z + k = 100 (per cent) and substitute 3x + z for k to get

x + y + z + 3x + z = 100 (per cent)

Take a moment to simplify:

4x + y + 2z = 100

But you just said 4x + y + z = 100 (per cent)

Which implies that x and y are 0, and z = 100.

This conflicts with the constraints:

x>y (must)

y>z (must)

Hence why you're having such trouble: you've set yourself up with an impossible problem.

posted by adipocere at 6:23 PM on July 22, 2014

k = 3x + z

4x + y + z = 100 (per cent)

j + k = 100 (per cent)

Take j + k = 100 (per cent) and substitute x + y + z for j to get

x + y + z + k = 100 (per cent)

Take x + y + z + k = 100 (per cent) and substitute 3x + z for k to get

x + y + z + 3x + z = 100 (per cent)

Take a moment to simplify:

4x + y + 2z = 100

But you just said 4x + y + z = 100 (per cent)

Which implies that x and y are 0, and z = 100.

This conflicts with the constraints:

x>y (must)

y>z (must)

Hence why you're having such trouble: you've set yourself up with an impossible problem.

posted by adipocere at 6:23 PM on July 22, 2014

And note that Wolfram Alpha is not smart enough to tell you that, which is a lesson about Wolfram Alpha.

posted by escabeche at 6:26 PM on July 22, 2014 [3 favorites]

posted by escabeche at 6:26 PM on July 22, 2014 [3 favorites]

Ok so really the issue, aside from nonsensical bounds, is that you only have two equations and 4 unknowns (x,y,z,k). So I solved the equations

x+y+z+k = 100 (i.e. j+k = 100)

4x+y+z = 100

And you get

k = 3x ; z = 100-4x-y

This is a bounded region; you can probably figure out some good solutions on your own from here. A start would be: x = 20, y=11, z = 9, k = 60, j = 40.

posted by Maecenas at 6:32 PM on July 22, 2014

x+y+z+k = 100 (i.e. j+k = 100)

4x+y+z = 100

And you get

k = 3x ; z = 100-4x-y

This is a bounded region; you can probably figure out some good solutions on your own from here. A start would be: x = 20, y=11, z = 9, k = 60, j = 40.

posted by Maecenas at 6:32 PM on July 22, 2014

*A start would be: x = 20, y=11, z = 9, k = 60, j = 40.*

This doesn't quite work because k doesn't equal x + 3z here.

*They can be equal if that helps. At this stage I'd settle for k being less than j if it means a solution pops out.*

Like Paragon said above, k < j is a necessary condition to have solutions to your system. For instance, if you let k = 40 and j = 60, your system

x + y + z = 40

3x + z = 60

(Which together imply that 4x + y + 2z = j + k = 100.)

x > y > z >= 0

has the positive integer solutions

x = 17, y = 14, z = 9

x = 18, y = 16, z = 6

x = 19, y = 18, z = 3.

posted by bassooner at 6:41 PM on July 22, 2014

Oh, right, there're a few typos around. The general solution is now

k = 50 + x - y/2

z = 50 - 2x - y/2

Which satisfies k = 3x+z.

Now, where to go from here? I'll assume you want z>0; therefore, y≥2 since it otherwise yield non-integral answers. Just pick an x for

Trial solution: x = 22, y = 6, z = 3. -> j = 31, k = 69. Indeed, j+k = 100, and k = 3x + z.

posted by Maecenas at 6:54 PM on July 22, 2014

k = 50 + x - y/2

z = 50 - 2x - y/2

Which satisfies k = 3x+z.

Now, where to go from here? I'll assume you want z>0; therefore, y≥2 since it otherwise yield non-integral answers. Just pick an x for

*24≥x≥15*, then pick a*y*, then everything is determined.Trial solution: x = 22, y = 6, z = 3. -> j = 31, k = 69. Indeed, j+k = 100, and k = 3x + z.

posted by Maecenas at 6:54 PM on July 22, 2014

If I am following your explanation here, then you can simplify (as in make this easier) your solution by just ignoring values for subtraits which you rate "meh, whatever." That gives you fewer variables to consider.

If you are being literal about your question

posted by Michele in California at 7:02 PM on July 22, 2014

If you are being literal about your question

*Is there a solution I can count on my fingers?*, then I wonder if counting in binary on your fingers would help. You can count up well above 100 that way and keep track of large numbers (with a little bit of occasional finger cramping) pretty easily.posted by Michele in California at 7:02 PM on July 22, 2014

If I understand correctly what you're trying to do, you're looking for values

where

If that is indeed what you're after, and you're starting from the

The actual values for

posted by flabdablet at 9:09 PM on July 22, 2014 [1 favorite]

**x**,**y**and**z**that correspond to "super important", "kinda important" and "meh", which you could subsequently plug into an equation of the form**rating = 60*(x*j**_{0}+ y*j_{1}+ z*j_{2}) + 40*(x*k_{0}+ x*k_{1}+ x*k_{2}+ z*k_{3})where

**j**,_{n}**k**= 0 means that a particular sub-characteristic is fully absent and 1 means that it is fully present?_{n}If that is indeed what you're after, and you're starting from the

**j**and_{n}**k**that apply to the particular thing you're initially considering, then where you need to start is by picking arbitrary numbers for any three of_{n}**rating**,**x**,**y**and**z**and solving for the missing one.The actual values for

**x**,**y**and**z**won't matter as much as the ratio**x**:**y**:**z**, which will end up being a numerical representation of the ratio**super**:**kinda**:**meh**; a clarification of*how many times as important*"super important" is, compared to "kinda important" and "meh". And really, that's completely up to you - it's a personal values thing, not something Wolfram Alpha could be expected to help you with.posted by flabdablet at 9:09 PM on July 22, 2014 [1 favorite]

Others have already well covered why the equations as you've written them have no solution.

Here's something to consider: when you say one of the three sub-characteristics for j is "super-important" and three of the four sub-characteristics for k are "super-important," does that mean that the super-important sub-characteristic of j is equally as important as a super-important sub-characteristic of k? Because that's what the equations you've written imply, and there's no solution.

If, on the other hand, the super-important sub-characteristic of j is super-important compared to the other sub-characteristics of j; and the super-important sub-characteristics of k are super-important compared to the other sub-characteristic of k,

An analogy: suppose you have two initially empty boxes, j and k. You also have an amount of paper currency in three different denominations, not necessarily corresponding to real-world denominations, but they are positive. Let's call the three denominations large, medium, and small.

With the equations as you've written them, you're saying, "I would like to put one of each of the three denominations into box j. And I would like to put three large bills and one small bill into box k. And then I would like the value of the currency in box j to be greater than that in box k." There's no way to do it. The three large bills alone in box k will be worth more than the one large, one medium, and one small bill in box j.

However, your explanation suggests that maybe you don't need the types of bills that go into box j be the same as the types of bills that go into box k. Maybe you have a set of three types of bills that go into box j (large-j, medium-j, and small-j), and a different set of two types of bills that go into box k (large-k and small-k; no "medium-k" is necessary since you have no "kinda important" sub-characteristics of k), and within each set the "large," "medium," and "small" designations only apply relative to the other bills within that set (e.g., small-j might be greater than large-k). Then there are certainly many possible solutions, but your equations would look something like this:

j=x+y+z

k=3u+v

x>y>z>0

u>v>0

j>k

j+k=100

You might even add the constraint that the

x/z = u/v

You could even set j and k to your original desired values, j=60, k=40.

One possible solution is:

x=30

y=20

z=10

u=12

v=4

posted by DevilsAdvocate at 9:56 PM on July 22, 2014

Here's something to consider: when you say one of the three sub-characteristics for j is "super-important" and three of the four sub-characteristics for k are "super-important," does that mean that the super-important sub-characteristic of j is equally as important as a super-important sub-characteristic of k? Because that's what the equations you've written imply, and there's no solution.

If, on the other hand, the super-important sub-characteristic of j is super-important compared to the other sub-characteristics of j; and the super-important sub-characteristics of k are super-important compared to the other sub-characteristic of k,

**but the super-important sub-characteristics of j and of k are not necessarily equally important**, then that's certainly doable. But then you can't use a single number to represent both the importance of the super-important sub-characteristic of j and the importance of the super-important sub-characteristics of k.An analogy: suppose you have two initially empty boxes, j and k. You also have an amount of paper currency in three different denominations, not necessarily corresponding to real-world denominations, but they are positive. Let's call the three denominations large, medium, and small.

With the equations as you've written them, you're saying, "I would like to put one of each of the three denominations into box j. And I would like to put three large bills and one small bill into box k. And then I would like the value of the currency in box j to be greater than that in box k." There's no way to do it. The three large bills alone in box k will be worth more than the one large, one medium, and one small bill in box j.

However, your explanation suggests that maybe you don't need the types of bills that go into box j be the same as the types of bills that go into box k. Maybe you have a set of three types of bills that go into box j (large-j, medium-j, and small-j), and a different set of two types of bills that go into box k (large-k and small-k; no "medium-k" is necessary since you have no "kinda important" sub-characteristics of k), and within each set the "large," "medium," and "small" designations only apply relative to the other bills within that set (e.g., small-j might be greater than large-k). Then there are certainly many possible solutions, but your equations would look something like this:

j=x+y+z

k=3u+v

x>y>z>0

u>v>0

j>k

j+k=100

You might even add the constraint that the

*relative*importance of "super-important" vs. "meh" within each set be the same, which would be represented as:x/z = u/v

You could even set j and k to your original desired values, j=60, k=40.

One possible solution is:

x=30

y=20

z=10

u=12

v=4

posted by DevilsAdvocate at 9:56 PM on July 22, 2014

Thanks all. You've given a very clear and concise explanation for why this isn't possible, and what can be done about it.

We had actually done what DevilsAdvocate has suggested - 'super important' for j doesn't need to be the same as 'super important' for k (and why would they be? They're nothing alike anyway, and concern two very different and distinct aspects of the thing I want to buy).

Alas, this hasn't gone down well with some, who insist there's a mathemamagical way to make everything the same across j and k, as well as make sense within j and k. You've shown (embarrassingly) quickly that this is a logical impossibility.

posted by obiwanwasabi at 6:01 PM on July 23, 2014

We had actually done what DevilsAdvocate has suggested - 'super important' for j doesn't need to be the same as 'super important' for k (and why would they be? They're nothing alike anyway, and concern two very different and distinct aspects of the thing I want to buy).

Alas, this hasn't gone down well with some, who insist there's a mathemamagical way to make everything the same across j and k, as well as make sense within j and k. You've shown (embarrassingly) quickly that this is a logical impossibility.

posted by obiwanwasabi at 6:01 PM on July 23, 2014

*'super important' for j doesn't need to be the same as 'super important' for k (and why would they be? They're nothing alike anyway, and concern two very different and distinct aspects of the thing I want to buy).*

If you're trying to develop a rating formula to help you choose between alternatives, then at some point you

*do*need to put some kind of value on how strongly you care about the various aspects of those alternatives.

The most direct way I can think of to do this in any semi-reasonable way is by using a simple weighted sum:

Rating = W

_{0}*C

_{0}+ W

_{1}*C

_{1}+ W

_{2}*C

_{2}+ ...

where the C

_{n}terms are a number between 0 and 1 inclusive that specifies the extent to which a certain characteristic is present in the product (could be just integers 0 and 1 for features that are fully absent or fully present) and the W

_{n}terms are arbitrary weightings that express how much you care about that characteristic (use positive numbers for desirable characteristics, negative for undesirable).

If certain characteristics are complete dealbreakers, assigning them a disproportionately large negative weighting will ensure that products that have those will always lose a ratings comparison regardless of how desirable their other characteristics might be. Similarly, you can use disproportionately high positive weightings to guarantee that products with killer features win hands down.

The kind of grouping you initially started with (the 60/40 j/k split) gives you some flexibility to tweak ratings in the direction you think they should go, while still being able to use a very restricted choice of weighting numbers to apply to individual sub-characteristics. That might work well on a web form, but I think it's probably a false economy. If you do care more about certain things than about certain other things, you

*do*need to think about that enough to be able to quantify the extent to which it is so, and having done that, using the results directly as weightings seems to me to be the most straightforward approach.

Which is basically the young rope-rider's point in the very first answer to this question.

posted by flabdablet at 7:28 PM on July 23, 2014 [1 favorite]

This thread is closed to new comments.

posted by the young rope-rider at 6:20 PM on July 22, 2014