At what substrate concentration will an enzyme work at 40% of its maximum rate if its Km = 0.095M?

Question

The question involves the Michaelis-Menten kinetics equation.

Here is what I did: I took the Michaelis-Menten equation, replaced v with 0.4Vmax, canceled the Vmaxes (one on each side), and solved for [S].

I got -0.056 which is obviously wrong, because [S] cannot be negative.

Answer 1

To be honest I know very few about enzymes and absolutely nothing about Michaelis–Menten.

However, when I "took the Michaelis-Menten equation, replaced v with 0.4Vmax, canceled the Vmaxes (one on each side), and solved for [S]", my result is positive:

0.4 * Vmax = S * Vmax / ( Km + S )
0.4 = S / ( Km + S )
S = 0.4 * Km + 0.4 * S
0.6 * S = 0.4 * Km
S = Km * 2 / 3.