The easiest example is money. Let's say I have $100 USD, and I want to convert part of this money to CAD, at an exchange rate of 1.3. I want to have the same quantity of USD as CAD at the end of the process.
If you do rough napkin math for the $100 USD example above, the exchanged amount would be roughly 43.55. This leaves 56.45 USD, and results in 56.62 CAD.
Are there any formulas or functions that can determine the 43.55 amount?
On
Yeah, you just solve a system of 2 linear equations in 2 unknowns.
The best way to think about this is to systematically translate "pieces of information I need" into "unknowns", and "pieces of information I have" into "equations".
The pieces of information you need are $x =$ final amount of USD and $y =$ final amount of CAD.
The pieces of information you have are (1) the final amount of USD equals the final amount of CAD, and (2) the exchanged amount (or $100 minus the final amount) of USD times the exchange rate equals the final amount CAD.
So your equations can be $$ \begin{align} x &= y \\ 1.3(100 - x) &= y \end{align} $$ You can eliminate $y$ right away and set $x = 1.3(100 - x)$. Solve: $$ \begin{align} x &= 1.3(100 - x) \\ x &= 130 - 1.3x \\ 2.3x &= 130 \\ x &= 130 / 2.3 \\ x &\approx 56.521739 \\ 100-x &\approx 43.478261 \end{align} $$ So you can exchange USD ~43.48 to end up with USD ~56.52 and CAD ~56.52.
I will generalise the statement made in my comment. Suppose we have $z$ monetary units of type"$A$". We can split this into two parts - the part which is not exchanged and the part which is. Call these parts $x$ and $y$. We trivially have $x+y=z$ as the two parts should add to give the entire sum of money. If the exchange rate is $k$ monetary units of type "$B$" per unit of type "$A$" then we want the resulting exchanged amount $(ky)$ to be equal to the non-exchanged amount $(x)$. This is equivalent to $x=ky$ or $x-ky=0$. This system of equations can then be solved by various methods to give $$\begin{cases}x+y&=z\\x-ky&=0\end{cases}\iff(x,y)=\left(\frac{kz}{k+1},\frac{z}{k+1}\right)$$ in which case the exchanged amount is $z/(k+1)$. So, in your example, the exchanged amount is $100/(1.3+1)=1000/23\approx43.478$.