An algorithm to make a variable the sole term of one side

Question

Is it always possible to refactor an equation so that a desired variable appears as the sole term of one side?* If so, how?

Simple example:

$$xa = ya + b \\ xa - ya = b \\ a(x - y) = b \\ a = \frac{b}{x -y}$$

I'm inclined to say yes (for any non-infinitely recursing $a$) but it's hard to pull an algorithm out of how I intuitively solve equations.

Tagging with recursive-algorithms because I think the solution will be recursive.

*excluding $a=a$

Answer 1

It's not always possible without defining new functions. Some classic examples:

• $x^2=2$ okay we know how to solve this one but we can't just get $x$ alone we need two different functions: $x=\sqrt{2}$ and $x=-\sqrt{2}$ . In a sense the square root function is made up to help us solve these sorts of equations.
• $\cos x =x$. It's clear they intersect but we can't solve for $x$
• $xe^x=1$ Note people invented a new function to solve equations of this type it's called the Lambert W-function $x=W(xe^x)=W(1) \approx 0.567143290409784 $. But it's not a function that's taught in the usual mathematical canon https://en.wikipedia.org/wiki/Lambert_W_function
• $\sin x=\frac{1}{2}$. Sure we can solve it using the inverse sine function $x=\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}$. But that's only one of infinitely many soltutions so you'd need to know that all solutions have the form $\frac{\pi}{6}, \frac{\pi}{6}\pm 2\pi,\frac{\pi}{6}\pm 4\pi, \frac{\pi}{6}\pm 6\pi, \ldots$ and $\frac{5\pi}{6}, \frac{5\pi}{6}\pm 2\pi,\frac{5\pi}{6}\pm 4\pi, \frac{5\pi}{6}\pm 6\pi, \ldots$

So if the problem is interesting enough in a family, it may be worth defining a new function to solve. Then maybe dedicating your life to giving effecient numerical approximations ;)

It's also not even possible to solve for a variable for linear equations:

• $2x+1=2x+1$ Trying to solve for $x$ results in $1=1$ (or $0=0$). Actually every $x$ is a solution
• $x+1=x$ trying to solve for $x$ results in $1=0$ . This one has no solutions

Answer 2

$$ a + e^a = b + \sin b + e^{e^b} $$