Wikipedia:Reference desk/Archives/Mathematics/2016 October 11

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October 11

How to isolate a variable

X = AB

Y = AC

Z = C/B

I know the exact values of X, Y, and Z. Now I want to solve for one of the unknowns A, B, or C. Is it possible?98.20.64.142 (talk) 18:00, 11 October 2016 (UTC)

Yes it can be solved. Since you say that X, Y, and Z are knowns and not variables, then to solve for the unknowns, by a method known as the system of equations method can be employed. The basic principle is that any system of equations is soluble so long as you have at least as many equations as unknowns. This walks you through the methods. --Jayron32 18:27, 11 October 2016 (UTC)

Except these are not polynomial equations (the variables are not joined by addition or subtraction).98.20.64.142 (talk) 19:15, 11 October 2016 (UTC)

(ce) I assume that Jayron meant a system is solvable if you have at least as many unknowns as equations, not vice versa. (Otherwise, A=1, A=2 would be solvable.) But either way, not all such systems are solvable in the sense of having an analytic solution. For the one variable, one equation case, see Abel-Ruffini theorem or quintic equation. For multiple equations and variables, see system of polynomial equations. Regarding the particular system the OP is inquiring about, try solving the first equation for A, then substituting that expression for A into the other equations, then solving the next one for B and substituting into the last equation, then solving that one for C, then using that expression for C in the previous expressions, etc. Loraof (talk) 19:34, 11 October 2016 (UTC)

That's just it though. I only know values for X, Y, and Z! I need to isolate either A, B, or C from the first three. But how?98.20.64.142 (talk) 19:49, 11 October 2016 (UTC)

XZ = (AB)(C/B) = AC = Y. Your equations are not independent. For the values of X, Y, and Z you know, do they satisfy XZ = Y? If not, then there are no solutions. If so, then there is a family of solutions. -- ToE 20:06, 11 October 2016 (UTC)

Yes, they do satisfy that condition. Can you point me in the direction of finding that family of solutions?98.20.64.142 (talk) 20:15, 11 October 2016 (UTC)

If your X, Y, and Z, do satisfy XZ = Y, then you can characterize your family of solutions with any one of A, B, or C. For instance, {(A, X/A, Y/A)}. -- ToE 20:19, 11 October 2016 (UTC)

I see what you mean. That doesn't help much for my problem though.98.20.64.142 (talk) 20:35, 11 October 2016 (UTC)

If you are not interested in the family of solutions but instead just want any one solution, then you can pick any A ≠ 0 and generate a solution such as A = 1, B = X, C = Y. But if you are looking for a unique solution, then your system is insufficiently constrained. You need another equation. -- ToE 21:06, 11 October 2016 (UTC)

If you are more comfortable with systems of linear equations, you can take the logarithm of both sides of all three equations. For convenience, let the lower case variable be the logarithm of its upper case, for instance x = log X, etc. Then your three equations become:

x = a + b

y = a + c

z = c - b

It is now obvious that these equations are not independent because adding the first and third equations gives x + z = a + c, but y = a + c. So your constants must satisfy y = x + z, and you have only two independent equations in three variables. -- ToE 21:10, 11 October 2016 (UTC)

Interesting, using logarithms to reduce an equation to linear form. I need to remember that one! That still doesn't apply to this situation. I can take the logarithm of X and Y (discarding dependent variable Z) but still can't see how to set up a system that can be solved. 98.20.64.142 (talk) 22:09, 11 October 2016 (UTC)

Discarding the Z equation as dependent, we are left with three variables A,B,C and two known constants X and Y. In this case there is a one dimensional family of solutions. To find it, let's make A the independent variable and B and C the dependent variables. Then the first two (original) equations give C = Y/A and B = X/A for A nonzero. For every nonzero value of A, C and B are determined. --Mark viking (talk) 22:46, 11 October 2016 (UTC)

A family of solutions doesn't really tell you much about solving for specific cases though. Or does it? 98.20.64.142 (talk) 23:00, 11 October 2016 (UTC)

But it does. It tells you that without further constraints (i.e. a third independent equation), that is the best you can get.--Jasper Deng (talk) 23:04, 11 October 2016 (UTC)

And it tells you that if you know there really is only one solution to your problem, then you are missing an equation. -- ToE 23:16, 11 October 2016 (UTC)

Yeah, I figured it probably wasn't solvable as is. Just wanted to be sure though. Thanks. 98.20.64.142 (talk) 23:55, 11 October 2016 (UTC)

• To the OP: You may have missed that I corrected my typos (in which I wrote X, Y, Z when I meant A, B, C) before you posted your reply to me. My post above tells you how to solve your equations. Loraof (talk) 23:58, 11 October 2016 (UTC)

I still don't see how that can be done. Let's say we have

X = AB = 7.1094

Y = AC = 11.7834

How to begin to solve for A in the first place?98.20.64.142 (talk) 00:19, 12 October 2016 (UTC)

You don't need to solve for A. Your system is underdetermined, so any nonzero value of A will work. For example, take A=1. Then B = X/A = 7.1094 and C = Y/A = 11.7834. Or take A=2. Then B = 3.5547 and C = 5.8917. And so on. You can pick any arbitrary value for A and produce values for B and C that will satisfy the original equations. CodeTalker (talk) 01:59, 12 October 2016 (UTC)

I actually DO need to solve for A in this case, it just isn't possible. Not enough independent equations. I get that. Thank you.98.20.64.142 (talk) 03:58, 12 October 2016 (UTC)