Real Life Rounding Phenomena When Solving for Variables I have a question that I've been thinking a long time about without being able to come up with an answer and would appreciate some help:
I am attempting to subtract two distinct fees from a total transaction, depending on transaction price.
Fee #1 = 2.9% of transaction price
Fee #2 = 10% of transaction price
Let T = transaction price.  Therefore :
Let F = Total Fees
F = T(0.029) + T(0.1)
F = T(0.129)
F / 0.129 = T
This seems to look ok, HOWEVER, in the real world, each fee is rounded to the nearest cent.  So, for Fee #1, assuming a transaction price of 10.99, the fee would be 0.31871 and thus rounded to 0.32.  This would give a slightly different result from the algebraic result, given this rounding phenomenon.
My question is, in equations such as the one in the example, is there a way to account for  discrete rounding of terms before solving for a variable?
 A: In most generality: Trial and error. Most notably, since $0.129<1$, there are many (about eight) $T$ leading to the same rounded $F$. If you are given $T+F$ instead, i.e. a factor that should equal $1.129>1$, you can determine $T$ uniquely (and there are some values of $T+F$ that cannot legally be obtained): Compute $\operatorname{round}((T+F)/1.129)$ and try this value (i.e. compute $F$ from it); if it is too high/low, try one cent less/more.
In your original problem, all you can do is compute $\operatorname{round}(F/0.129)$ and try several cents up and down until the backwards calculation produces a differnt $F$.
A: How about multiplying by a factor $c$ first, then dividing the final answer by the same factor again?
For $f_1=$ fee #1, $f_2=$ fee #2, $c = 10$.
$$F\cdot c = T(f_1 \cdot c) + T(f_2 \cdot c) = T((f_1+f_2)\cdot c)$$
$$F = \frac{T((f_1+f_2)\cdot c)}{c}$$
Of course you solve the parentheses and numerator first. You can use bigger values of $c$ if you need more accuracy. e.g. $c=100, 1000, \ldots$.
