Troubleshooting Inverse Algebra Equations: Solving for A in a Complex Equation

[Jeff Cook]

All,

Hey, it's been a while since I've posted, but I have a problem that I am having a hard time moving forward with. I know there are some really intelligent algebra folks in here, and it would be great if someone could help me with this. The equation is kinda large, but there are other difficulties other than the size that keep me from solving it. I keep getting turned around with the inverse on the opposite side.

Here's the problem...

How would one go about solving for A in the following:

(C/(.5 (2^.5 + 1)(A^-1 - B)) + 2.5^-1)^-1 = (18A-F)^-1 (14A - F/V)

Any help with this would be great.

Thanks,

Jeff

 

[Vid]

Most of that is really ambiguous...

 

[Jeff Cook]

Ambiguous? What do you mean? It's just an equation. I am trying to solve for A. What's ambiguous about that?

 

[rock.freak667]

What exactly are F,V,C,A and B?

 

[Vid]

Mostly just the exponents, but assuming you only meant the -1 to be in the explonent. Mathematica gives two solutions though I don't know how helpful they are..

{A -> 1/((2.88989*10^9 B + 2.70298*10^9 C) V)(-4.66112*10^7 B F +
9.65349*10^7 C F + 1.44495*10^9 V + 1.16528*10^8 B F V -
0.5 \[Sqrt]((-1.9307*10^8 C F +
B F (9.32224*10^7 - 2.33056*10^8 V) - 2.88989*10^9 V)^2 -
4. (2.88989*10^9 B + 2.70298*10^9 C) F V (-9.32224*10^7 +
2.33056*10^8 V)))}

{A -> 1/((2.88989*10^9 B + 2.70298*10^9 C) V)
0.5 (-9.32224*10^7 B F + 1.9307*10^8 C F + 2.88989*10^9 V +
2.33056*10^8 B F V + \[Sqrt]((-1.9307*10^8 C F +
B F (9.32224*10^7 - 2.33056*10^8 V) - 2.88989*10^9 V)^2 -
4. (2.88989*10^9 B + 2.70298*10^9 C) F V (-9.32224*10^7 +
2.33056*10^8 V)))}

Note that mathematica uses spaces to imply multiplication.

 

[Jeff Cook]

What exactly are F,V,C,A and B?

They're variables

 

[D H]

How would one go about solving for A in the following:

(C/(.5 (2^.5 + 1)(A^-1 - B)) + 2.5^-1)^-1 = (18A-F)^-1 (14A - F/V)

This is your equation:

\left(\frac{C}{.5 (\surd 2 + 1)(A^{-1} - B)} + 2/5\right)^{-1} = \frac{14A - F/V}{18A-F}

The only thing I did was replace 2^{.5} with \surd 2, replace 2.5^{-1} with 2/5, and express the right-hand side in fractional form.

What exactly are F,V,C,A and B?

They're variables

Only A is a variable; this is what you want to solve for. The rest (F,V,C, and B) are unspecified constants. The first thing you should do is to clear the fraction on the left hand side -- this one:

C/(.5 (\surd 2 + 1)(A^-1 - B))

To do this, multiply by A/A. You might want to multiply by 2/2 at the same time. Then add 2/5, but form the result as a fraction so you can invert it. You will eventually end up with something like

\frac {k_1 A + b_1}{k_2 A + b_2} = \frac {k_3 A + b_3}{k_4 A + b_4}

which is obviously a quadratic.

 

[Jeff Cook]

Only A is a variable; this is what you want to solve for. The rest (F,V,C, and B) are unspecified constants. QUOTE]

In both Computer Science and Mathematics, the term, "variable," "is a symbolic representation used to denote a quantity or expression. In mathematics, a variable often represents an "unknown" quantity that has the potential to change; in computer science, it represents a place where a quantity can be stored. Variables are often contrasted with constants, which are known and unchanging." ...at least according to Wikipedia. But these semantics are aside from the point.

In any case, DH, thanks greatly for rewriting my equation in a more readable way. I am not up to speed yet on the proper conventions for preparing math in this forum. Thanks!

Is this a fourth degree polynomial when you say quadratic?

 

[Gib Z]

Is this a fourth degree polynomial when you say quadratic?

No, It is a second degree polynomial. It should be obvious if you cross-multiply the final form DH gave.

 

[Jeff Cook]

DH and all,

Thanks for your help. I was able to solve the equation finally due to your comments. However, no matter how hard I try, I cannot bring myself to think like the schooled mathematician; this is, of course, due to the fact that I am self-taught. What may seem obvious to you and the rest of the mathematicians in the world is not something that I can see all the time. In the same, what seems obvious to me often makes no sense to others. But such is my experiences.

In any case, this is how I went about solving the equation with your help. It’s kind of like going around one’s elbow to get to their x-hole, but this is sometimes the way I prefer to work…never know what’s going to turn up along the way. True it may take me 10^10 times as long, but it’s the only way I seem to be able to manage my work.

I took the equation and multiplied the left-hand side by 2A/2A, as you suggested. However, it is unclear to me what the addition of the 2/5 would do, so I ignored it.


(C / (.5 (2^.5 + 1) (A^-1 – B) + (1/2.5))^-1 = (14A / (18A – F)) – (F / (V (18A – F)))


2CA / (.5 Sqrt {2} + .5) (1/A – B) + (2A/2.5) = 2A / ((14A / (18A – F)) – (F / (V (18A – F))))


2CA / ((.5 Sqrt {2})/A + .5/A - .5 Sqrt {2} B + (2A/2.5) = 2A (18A – F) / (14A – F/V)


(14A – F/V)(2CA / (((.5 Sqrt {2}) / A + .5/A - .5 Sqrt {2} B - .5B) + (2A / 2.5) = 36A^2 – 2FA


Set: y = .5 Sqrt {2} / A + 5 / A - .5 Sqrt {2} B - .5B


…for the time being. Rearrange to get…


A (28C / y – 36 + 28/2.5) – 2CF / yV = -2F + 2F / 2.5V


Set: D = 36 – 28 / 2.5
And set: G = -2F + 2F / 2.5V


A (28C / y – D) = G + 2CF / yV


Open up y again and rearrange to get…


A ((28CVA – 2CF) / (GV + DVA) + .5B + .5 Sqrt {2} B) = .5 Sqrt {2} + .5


Set H = .5 Sqrt {2} + .5
And set M = B (.5 + .5 Sqrt {2})


To look at it as:

A ((28CVA – 2CF) / (GV + DVA) + M) = H

Rearrange and bring everything over to the right hand side:

0 = HG / A^2 + ((1 / A) (-MG + 2CF/V + HD)) + (-28C - MD)


x = 1 / A
a = HG
b = (-MG + 2CF/V + HD)
c = (-28C – MD)

x = (-b +- Sqrt {b^2 – 4ac}) / 2a


A = 1 / x

That’s the way I solved it, but yes, still stuck on the inverse. I just always feel better looking at things upside down. :)

Thanks for all your help! I wasn’t able to do it without you.

Jeff