Get rid of B for (A-B)/(C-A+B)

Question

Is it possible to get rid of B: $$\frac{A-B}{C-A+B} $$ A, B, C are all non-negative.

Answer 1

There is no way to get rid of $B$ in $$\frac {A-B}{C-A+B}$$ and get an equivalent expression.

We can simplify a fraction if there is a common factor in the top and bottom of the fraction.

Here $B$ is not a factor and stays as it is in both top and bottom of your fraction.

Answer 2

$$ \frac{A-B}{C-(A-B)} = \left(\frac{C -(A-B)}{A-B}\right)^{-1}= \left(\frac{C}{A-B} - 1\right)^{-1} $$

so you only need the quantities $C$ and $A-B$. This obviously still depends on $B$ even if only through $A-B$.

Diffferent problem: Consider the possible values as $A,B,C$ range over all nonnegative integers.

If $A$ and $B$ are arbitrary nonnegative integers, then $A-B$ will be an arbitrary integer.

So you get $\left(\dfrac{C}{Z} - 1\right)^{-1}$ where $C$ is an arbitrary nonnegative integer and $Z$ is an arbitrary integer.

You can put $\dfrac{-C}{-Z}$ without changing the result so in fact you can get the same possible results by allowing all integers for both $C$ and $Z$.

$$ \left(\frac{C-Z}{Z}\right)^{-1} = \frac{Z}{C-Z} $$

By shifting $C$ by $Z$ we get arbitrary rational numbers because we have arbitrary integers on both the top and bottom.