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Assuming $f\neq 0$ I'm trying to rewrite $$0 = 2 \cdot g \cdot ((x-a)^2 + 1) - 2 \cdot d \cdot f \cdot (x - a)$$

into a system of equations of the form

$a = $(something not containing $d$)

$d = $(something with $a$ in it)

I'm pretty sure you need to introduce another variable but I'm just not seeing the "trick".

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You won't get a linear system, and I don't think you can decouple the variables, but here goes: divide by 2, put $df(x-a)$ on the left side, divide out the extra factor to get $d=\cdots$; write $y=x-a$ and solve the equation as a quadratic equation to get $y$ in terms of $g, f, d$, then write $a=x-y$. – anon Jul 13 '11 at 18:37
Thank you for the comments, but I really would like the variables decoupled. Eventually I need to generalize for more terms and the quadratic trick won't work for this. – Jon Jul 13 '11 at 18:57
Are you saying you want the upper equation not to contain $d$ and/or the lower equation not to contain $a$? That won't be possible. You only have one scalar equation, you can't turn it into two independent ones. Are you thinking of something like $xy=0\Rightarrow x=0\lor y=0$? There's an "or" in there, whereas between your two equations there's an implied "and". – joriki Jul 13 '11 at 19:07
If that isn't possible, maybe relax the condition so $d = $(something with $a$ in it) but $a = $(something not containing $d$) – Jon Jul 13 '11 at 19:15
Since you can solve the quadratic for $a$ and get an expression for $a$ that depends on $d$, it's impossible to also get an expression for $a$ that doesn't depend on $d$. – joriki Jul 13 '11 at 19:26

If you don't mind trig parametrizations, another quite simple looking solution is:

$a = x - \tan(\phi)$

$d = \dfrac{2 g}{f \sin(2 \phi)}$

Maybe $\phi$ was the extra variable you mentioned, although the expression for $d$ doesn't involve $a$ explicitly.

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Thanks for editing, J M - I tried "frac", but the result was too small. Must remember "dfrac" for future use ;-) – John R Ramsden Aug 13 '11 at 10:52

This is probably not what you want, but here goes: $$\displaylines{a=Q+17\cr d={((x-a)^2+1)g\over(x-a)f}\cr}$$ Maybe if you tell us what's wrong with this answer, we can make some progress toward understanding your question.

EDIT: I wonder if what you really want is

new variable equals something containing $a$ but not $d$,

$d$ equals something containing new variable but not $a$

such as $Q=x-a$, $d=(Q^2+1)g/(Qf)$.

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