# Math Help - Could this be solved?

1. ## Could this be solved?

Hi Everyone,

I am sorry, but I can't think of a good descriptive title.

I am writing here hoping that I can obtain some advice from the mathematicians. I work in analytical chemistry, and my math knowledge is very dusty, to say the least.

Here are two equations:

T-F = G*K*F / (1+K*F+H*E)

U-E = G*H*E / (1+H*E+K*F)

I would like to find a solution for F as a function of all the above parameters, except for E. I tryied to rearrange, but I got totally lost at this stage:

K*F^2+{G*K-K*T-0.5(G*H-H*U+K*F+1)+0.5*sqrt[(G*H-H*U+K*F+1)^2+4*H(U+U*K*F)]+1}*F-T+0.5*T*(G*H-H*U+K*F+1)-0.5*T* sqrt[(G*H-H*U+K*F+1)^2+4*H(U+U*K*F)]=0

I know the interval of the parameters, it that would help with some simplifications.
K ~25
H ~ 50
G ~ 0.4
U = [0.003 - 0.5]
T = [0.0043 - 0.77]

I don't even know what to read, to get an idea how to approach this problem. Unfortunatelly I don't have any friends who are good in math, and the couple "shots in the dark" sent to a couple of people had no result. Hopefully, a good Samaritan can help me on this forum. Thanks so much!

2. ## Re: Could this be solved?

WHOA! I'm confused! Are you saying you are trying to solve the equations for F and you have to use both equations? And what exactly do you mean by "except for E"? And are these the equations:

$T-F = \frac{GHF}{(1 + KF + HE)}$
$U-E = \frac{GHE}{(1 + KF + HE)}$

If so, do you see that dividing these two equations will eliminate part of the equations allowing for you to solve for F?

3. ## Re: Could this be solved?

Sorry for the confusion. I am looking to express F as a function of K, H, G, U and T, because that is what I measure. I don't measure E, hence it should not be in the equation for F. If I divide, E is still present, but I lose G, K and H, which I don't want to lose, because F depends on them.

4. ## Re: Could this be solved?

Originally Posted by amix
Sorry for the confusion. I am looking to express F as a function of K, H, G, U and T, because that is what I measure. I don't measure E, hence it should not be in the equation for F. If I divide, E is still present, but I lose G, K and H, which I don't want to lose, because F depends on them.
So, you want to eliminate E, and solve for F?

5. ## Re: Could this be solved?

Seems impossible...

Take very simple case:
a = uv + w [a=10, u=2, v=3, w=4]

Try solve that for w "without the v"; to be blunt: makes no sense!

6. ## Re: Could this be solved?

Originally Posted by Youkla
WHOA! I'm confused! Are you saying you are trying to solve the equations for F and you have to use both equations? And what exactly do you mean by "except for E"? And are these the equations:

$T-F = \frac{GHF}{(1 + KF + HE)}$
$U-E = \frac{GHE}{(1 + KF + HE)}$

If so, do you see that dividing these two equations will eliminate part of the equations allowing for you to solve for F?
Believe me when I say that I did not want to do this.

We basically have to start from equation 2 because starting from equation 1 gives something too messy to substitute 3 times.

So:

$U-E = \frac{GHE}{(1 + KF + HE)}$

$(U-E)(1+KF+HE)=GHE$

$U+KFU+HEU-E-KFE-HE^2-GHE=0$

This gives us a quadratic to solve for $E$:

$HE^2+E(GH+KF-HU+1)-KFU-U=0$

$E=\frac{-(GH+KF-HU+1)\pm\sqrt{(GH+KF-HU+1)^2-4H(-KHU-U)}}{2H}$

You'd then have to substitute this into the first equation:

$T-F=\frac{GHF}{1+KF+H(\frac{-(GH+KF-HU+1)\pm\sqrt{(GH+KF-HU+1)^2-4H(-KHU-U)}}{2H})}$

$T-F=\frac{GHF}{1+KF+(\frac{-(GH+KF-HU+1)\pm\sqrt{(GH+KF-HU+1)^2-4H(-KHU-U)}}{2})}$

...and good luck solving that for $F$!

$T-F = \frac{GHF}{(1 + KF + HE)}$

...which starts off lovely.

$(T-F)(1+KF+HE)=GHF$

$1+KF+HE=\frac{GHF}{T-F}$

$HE=\frac{GHF}{T-F}-(1+KF)$

$E=\frac{GF}{T-F}-\frac{1+KF}{H}$

But now you have to substitute that into the second equation:

$U-E = \frac{GHE}{(1 + KF + HE)}$

$U-(\frac{GF}{T-F}-\frac{1+KF}{H}) = \frac{GH(\frac{GF}{T-F}-\frac{1+KF}{H})}{(1 + KF + H(\frac{GF}{T-F}-\frac{1+KF}{H}))}$

Again, good luck solving that for F.

7. ## Re: Could this be solved?

i think your 2nd approach shows promise. multiply the left by the denominator on the right, and then get all the terms with T-F in the denominator on the same side. i believe you wind up with a quartic in F, which is solvable, in theory (the formula is horrendously ugly, though).

8. ## Re: Could this be solved?

Originally Posted by Quacky
$E=\frac{GF}{T-F}-\frac{1+KF}{H}$

But now you have to substitute that into the second equation:

$U-E = \frac{GHE}{(1 + KF + HE)}$

$U-(\frac{GF}{T-F}-\frac{1+KF}{H}) = \frac{GH(\frac{GF}{T-F}-\frac{1+KF}{H})}{(1 + KF + H(\frac{GF}{T-F}-\frac{1+KF}{H}))}$
Fine...BUT you are still "using E";
you've ended up with a value for E in that equation;
so by substituting, you're actually using E's value...am I missing something?

9. ## Re: Could this be solved?

Going against any better judgement on my part, I tried it. This was my attempt. I decided I wanted to do it for the thrill of victory:

$U-(\frac{GF}{T-F}-\frac{1+KF}{H}) = \frac{GH(\frac{GF}{T-F}-\frac{1+KF}{H})}{(1 + KF + H(\frac{GF}{T-F}-\frac{1+KF}{H}))}$

I'd noticed the denominator of the fraction on the right simplifies considerably to give:

$U-(\frac{GF}{T-F}-\frac{1+KF}{H}) = \frac{GH(\frac{GF}{T-F}-\frac{1+KF}{H})}{(1 + KF + \frac{GHF}{T-F}-(1+KF))}$

Which then simplifies to:

$U-(\frac{GF}{T-F}-\frac{1+KF}{H}) = \frac{GH(\frac{GF}{T-F}-\frac{1+KF}{H})}{(\frac{GHF}{T-F})}$

To:

$U-(\frac{GF}{T-F}-\frac{1+KF}{H}) = \frac{(\frac{GF}{T-F}-\frac{1+KF}{H})}{(\frac{F}{T-F})}$

Putting the left over a common denominator:

$\frac{UH(T-F)}{H(T-F)}-(\frac{GHF}{H(T-F)}-\frac{(T-F)(1+KF)}{H(T-F)}) = \frac{(\frac{GF}{T-F}-\frac{1+KF}{H})}{(\frac{F}{T-F})}$

Now multiplying through by T-F:

$\frac{UH(T-F)}{H}-(\frac{GHF}{H}-\frac{(T-F)(1+KF)}{H}) = \frac{(GF-\frac{(1+KF)(T-F)}{H})}{(\frac{F}{T-F})}$

Combining the left:

$\frac{UH(T-F)-GHF+(T-F)(1+KF)}{H}) = \frac{(GF-\frac{(1+KF)(T-F)}{H})}{(\frac{F}{T-F})}$

Multiplying through by H:

$UH(T-F)-GHF+(T-F)(1+KF)) = \frac{(GFH-(1+KF)(T-F))}{(\frac{F}{T-F})}$

Rewriting:

$UH(T-F)-GHF+(T-F)(1+KF)) = \frac{(T-F)(GFH)-(1+KF)(T-F)^2}{F}$

And, finally for now, multiplying through by F:

$F(UH(T-F)-GHF+(T-F)(1+KF))) = (T-F)(GFH)-(1+KF)(T-F)^2$

Expanding brackets:

$F(UH(T-F)-GHF+(T-F)(1+KF))) = (T-F)(GFH)-(1+KF)(T-F)^2$

$F(UHT-UHF-GHF+T+KFT-F-KF^2)=(T-F)(GFH)-(1+KF)(T-F)^2$

$FUHT-UHF^2-GHF^2+FT+KF^2T-F^2-KF^3=TGFH-GF^2H-(1+KF)(T-F)(T-F)$

$FUHT-UHF^2-GHF^2+FT+KF^2T-F^2-KF^3=TGFH-GF^2H-(1+KF)(T^2-2FT+F^2)$

$FUHT-UHF^2-GHF^2+FT+KF^2T-F^2-KF^3=TGFH-GF^2H-(T^2-2FT+F^2+KFT^2-2KFT+KF^3)$

$FUHT-UHF^2-GHF^2+FT+KF^2T-F^2-KF^3=TGFH-GF^2H-T^2+2FT-F^2-KFT^2+2KFT-KF^3)$

Cancelling like terms:

$FUHT-UHF^2+FT+KF^2T-F^2=TGFH-T^2+2FT-F^2-KFT^2+2KFT)$

which, praying that I'm right so far, leads to:

$FUHT-UHF^2+FT+KF^2T-F^2-TGFH+T^2-2FT+F^2+KFT^2-2KFT=0$

Which is a quadratic to solve for F:

$KF^2T-UHF^2+FUHT+FT-TGFH-2FT+KFT^2-2KFT+T^2=0$ (Here, the +FT and -2FT can be simplified; note this below - my first error was putting +FT instead of -FT)

$F^2(KT-UH)+F(UHT-T-TGH-2KT^2-2KT)+T^2=0$

Then solve $F=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

or...

$F=\frac{-(UHT-T-TGH-2KT^2-2KT)\pm\sqrt{(UHT-T-TGH-2KT^2-2KT)^2-4(KT-UH)(T^2)}}{2(KT-UH)}$

However, any attempt at checking it has so far lead me off the rails entirely. I don't think I'll be able to spot any mistakes (and I don't think anyone would want to) but if anyone else does want to have a go, feel free! I'll check my working tomorrow: I'm accepting defeat for now

Edit2: I noticed an error which I have corrected. However, I'm still dubious. I don't think the error I noticed would justify the depth of the incongruency when checking my result.

10. ## Re: Could this be solved?

Originally Posted by Wilmer
Fine...BUT you are still "using E";
you've ended up with a value for E in that equation;
so by substituting, you're actually using E's value...am I missing something?
I think this is what the OP is asking. I appreciate what you are saying, however.

11. ## Re: Could this be solved?

Originally Posted by Quacky
$FUHT-UHF^2-GHF^2+FT+KF^2T-F^2-KF^3=TGFH-GF^2H-T^2+2FT-F^2-KFT^2+2KFT-KF^3)$

Cancelling like terms:

$FUHT-UHF^2+FT+KF^2T-F^2=TGFH-T^2+2FT-F^2-KFT^2+2KFT)$
Hmmm...I'm probably wrong too...but I end up with K^2F^3 and -KHF^3
as my ^3 terms...so I guess "worse" than you...unless you cheated in
order to cancel 'em out!

12. ## Re: Could this be solved?

Originally Posted by Quacky

....Expanding brackets:

$F(UH(T-F)-GHF+(T-F)(1+KF))) = (T-F)(GFH)-(1+KF)(T-F)^2$

$F(UHT-UHF-GHF+T+KFT-F-KF^2)=(T-F)(GFH)-(1+KF)(T-F)^2$

$FUHT-UHF^2-GHF^2+FT+KF^2T-F^2-KF^3=TGFH-GF^2H-(1+KF)(T-F)(T-F)$

$FUHT-UHF^2-GHF^2+FT+KF^2T-F^2-KF^3=TGFH-GF^2H-(1+KF)(T^2-2FT+F^2)$

$FUHT-UHF^2-GHF^2+FT+KF^2T-F^2-KF^3=TGFH-GF^2H-(T^2-2FT+F^2+KFT^2-{\color{red}2KF^2T}+KF^3)$

$FUHT-UHF^2-GHF^2+FT+KF^2T-F^2-KF^3=TGFH-GF^2H-T^2+2FT-F^2-KFT^2+{\color{red}2KF^2T}-KF^3$

Cancelling like terms:....
i believe the terms high-lighted in red were mistakes on your part, changing the $F^2$ term in the final equation....

13. ## Re: Could this be solved?

Originally Posted by Wilmer
Hmmm...I'm probably wrong too...but I end up with K^2F^3 and -KHF^3
as my ^3 terms...so I guess "worse" than you...unless you cheated in
order to cancel 'em out!
Probably. I just don't have the energy to conquer it yet, though.

14. ## Re: Could this be solved?

Originally Posted by Deveno
i believe the terms high-lighted in red were mistakes on your part, changing the $F^2$ term in the final equation....
It's typical. I get through the whole process of removing the fractions, and fail at arranging the terms of a polynomial.

15. ## Re: Could this be solved?

Originally Posted by amix
Here are two equations:
T-F = G*K*F / (1+K*F+H*E)

U-E = G*H*E / (1+H*E+K*F)

I would like to find a solution for F as a function of all the above parameters, except for E.
OK guys, let's suppose this was the problem:
.................................................. .................................................. ................
Here are two equations:
A - B = CB / (1 + BX) [1]

D - X = CX / (1 + BX) [2]

I would like to find a solution for B as a function of all the above parameters, except for X.
.................................................. .................................................. ................
So we have: CB / (A - B) = CX / (D - X)
which leads to: X = BD / A [3]

So, subs. [3] in [1]:
A - B = CB / [1 + B(BD / A)]
DB^3 - ADB^2 + (A + AC)B - A^2 = 0

A solution is: A = 6, B = 3, C = 7, D = 4 ; which makes X = 2

So : I end up with a cubic...how can you guys end up with a quadratic
using the OP's much more complicated equations?

Also, OP stated in one post:
"Sorry for the confusion. I am looking to express F as a function of K, H, G, U and T, because that is what I measure. I don't measure E, hence it should not be in the equation for F. If I divide, E is still present, but I lose G, K and H, which I don't want to lose, because F depends on them."

I completely disagree with that; nothing will be "lost";
and I think my example shows that....

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