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2 equations, 3 unknowns - finding 2 ratios of unknowns

by uby
Tags: equations, ratios, unknowns
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Sep12-10, 08:11 PM
P: 176
1. The problem statement, all variables and given/known data
I have two equations with three unknowns. Obviously, this is indeterminate and not all three variables can be independently solved. However, I only need (want) to solve for two ratios of the three unknowns. I am writing greatly simplified versions of the equations below:

2. Relevant equations
K1 = Ax + Ay - Bz
K2 = Ax + (A+B)y

I wish to solve for x/y and z/(x+y) in terms of A, B, K1 and K2 only.
3. The attempt at a solution
I have been using excel spreadsheets to attempt numerical solutions and have quickly found the system to STILL be indeterminate even when trying to only solve for these ratios.

When solving for z/(x+y) using eqn. 1, you immediately find it varies with (x+y) as:
z/(x+y) = (A/B) - (K1/(B*(x+y)))

I suspect it is impossible to solve for both (x+y) and (x/y) since knowing both allows you to calculate both x and y. But, on the other hand, I don't want to know (x+y), only z/(x+y).

Any ideas?

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Sep12-10, 08:51 PM
HW Helper
P: 2,155
When solving a system for combinations of variables, if the solution isn't immediately obvious, a good technique to use is to define a new letter for each combination of variables you want to solve for. For example, you might set
[tex]p = \frac{z}{x+y}[/tex]
[tex]q = \frac{x}{y}[/tex]
Then use these definitions to eliminate two of the original variables, say z and y, from the original equations, and solve for p and q. If a solution exists for p and q in terms of A, B, K1, and K2 only, you should find that the remaining variable, x, cancels out of the equation.

Of course, it is possible that the ratios you're looking for are not constant but are inherently dependent on the (x,y,z) point at which they are computed. In that case, you won't be able to eliminate x from the equations, and the problem as given cannot be solved.
Sep12-10, 09:02 PM
P: 176
thanks diazona, your approach is very helpful

i choose to eliminate x and z instead using your definitions: x = qy and z = (x+y)p
then, substituting x = qy into K2 = Ax + (A+B)y yields y = K2 / (Aq + A + B)
substituting y (and the original substitutions for x and z) into K1 = Ax + Ay - Bz yields the following expression:
K1 = K2*(Aq + A - Bpq - Bp)/(Aq + A + B)
thus, I am left with an expression for p and q in terms of A, B, K1 and K2 only -- though the system remains indeterminate as I only have 1 equation with 2 unknowns (p and q).

so, it would seem as though i cannot solve for these ratios (though, i don't understand why from a degrees of freedom argument: any two pieces of information should be solvable so long as their combinations cannot yield all three unknowns).

is there some way to find the set of all ratios that are determinate in this system?

Sep12-10, 09:42 PM
HW Helper
P: 2,155
2 equations, 3 unknowns - finding 2 ratios of unknowns

Quote Quote by uby View Post
any two pieces of information should be solvable so long as their combinations cannot yield all three unknowns
Actually, that's not true. To take a simpler example, consider the equation
[tex]x = y - 1[/tex]
In this system, it's impossible to solve for the ratio [itex]x/y[/itex] - or rather, you can solve for it, but not in terms of constants only, because the ratio depends on which (x,y) point you choose. You can see this by taking a few sample points: at (3,4) the ratio is 0.75, at (9,10) it's 0.9, and so on.

What the concept of degrees of freedom (DOF) really tells you is that if you have an underdetermined system, the dimensionality of the solution space is equal to the number of variables minus the number of constraints. (At least for "normal" e.g. algebraic systems) In your example, you have 3 variables and 2 unknowns, thus the space of solutions to your system will be a 1-dimensional object, namely a line or curve. If you created a 3D plot of (x,y,z) points that satisfy your equations, you should be able to see that.

Now, if you think about it, hopefully you can see that in order for you to solve for the ratios you're looking for (in terms of A,B,K1,K2 only), the values of those ratios would have to be constant along the entire curve. But they're not. Again, if you had a 3D plot of the solution space, you should be able to calculate x/y and z/(x+y) at various points in that space and you would find an assortment of different values.

Generally speaking, you can't pick any arbitrary function of x, y, and z (such as one of your ratios) and expect it to be constant throughout the solution space (the 1D curve). In fact, most functions of x,y,z will not be constant throughout the solution space. Finding a function that is constant is actually an interesting problem of its own. This sort of problem is quite relevant in physics, and it sometimes takes some rather advanced tools to solve it, e.g. Noether's Theorem.

In principle, I think you could use Noether's Theorem to find quantities that are constant throughout the solution space of your system, although it may not be necessary. One thing you could try is to just combine the two equations by solving one equation for one variable and substituting into the other. In the resulting equation, make sure that the left side contains only constants, and then whatever appears on the right side should give you a quantity that will be constant throughout your solution space. It may be more complicated than a simple ratio, though.
Sep12-10, 09:46 PM
P: 176
Diazona, thanks again for your expertise!

So, an approach to this might take the form of finding a function for the line containing all solutions in (x,y,z) space, and then deducing from this function what ratios, if any, are constant?
Sep12-10, 10:12 PM
HW Helper
P: 2,155
Hmm... I suppose that would be one way to go about it.

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