thermodynamics - On an equality following from the Helmholtz free energy expression

Let's, for the sake of argument, agree that the thermodynamic relation for equilibrium in the form of zero change of the Helmholtz potential, is applicable for particles suspended in a liquid,

$$\delta F = \delta E - T \delta S = 0, \ \ \ \ \ \ \ \ \ \ (1)$$

where $F$ is the Helmholtz potential, $E$ is the energy, $T$ is the absolute temperature and $S$ is the entropy. Then, the following integral for $\delta E$ is written

$$\delta E = -\int_0^l K \nu \delta x dx \ \ \ \ \ \ \ \ \ \ (2)$$

where $K$ is the force per unit particle, acting only along the $x$-axis, $\nu$ is the number of particles per unit volume, $\delta x$ is a variation of $x$, $\delta x$ being also a function of $x$. The liquid is bounded by the planes $x = 0$ and $x = 1$ of unit area.

Now, it can be intuitively seen how, because $K$ is a force on a unit particle, that force has to be multiplied by the number of particles residing on a plane perpendicular to the $x$-axis and that has led to the expression for $\delta E$ in eq.(2)

Then, the variation of entropy, $\delta S$, is defined as

$$\delta S = \int_0^l k \nu \frac{\partial\delta x}{\partial x} dx, \ \ \ \ \ \ \ \ \ \ (3)$$

where $k$ is Boltzmann's constant.

Now, from eq.(2) and eq.(3), using eq.(1) we get

$$-\int_0^l K \nu \delta x dx = T\int_0^l k \nu \frac{\partial\delta x}{\partial x} dx. \ \ \ \ \ \ \ \ \ \ (4)$$

or, if the integrands are continuous functions

$$-K \delta x = kT \frac{\partial \delta x}{\partial x}. \ \ \ \ \ \ \ \ \ \ (5)$$

It seems to me that l.h.s of eq.(5) is not equal to its r.h.s. -- on the l.h.s. we have work with negative sign, while on the r.h.s. we have the expression of energy for two degrees of freedom, according to the equipartition theorem, multiplied by a positive non-zero factor. What do you think about this? Am I right or am I missing something?