Solve for :
Recognise that the fraction is linear in , not a new unknown. The term has only in the numerator, so it is simply times a constant. That means the whole equation is linear and can be solved by collecting like terms — no cross-multiplication is needed.
Simplify the constant denominator.
so the fee term is . Dividing:
Keep the exact fraction to avoid rounding drift.
Move the constant across and collect the terms. Subtract from both sides:
since .
Divide to isolate .
This is exact, not rounded — the fraction terminates.
Check by substituting back. With : the fee term is , and . The equation balances exactly, confirming .
Need to solve a different problem like this? Open the solver →