Write out the combination terms.

The given equation is $^x C_2 + ^x C_3 - 4x = 0$.
The formula for combinations is $^n C_r = \frac{n!}{r!(n-r)!}$.
For $^x C_2$:

$$^x C_2 = \frac{x!}{2!(x-2)!} = \frac{x(x-1)(x-2)!}{2 \cdot 1 \cdot (x-2)!} = \frac{x(x-1)}{2}$$

For $^x C_3$:

$$^x C_3 = \frac{x!}{3!(x-3)!} = \frac{x(x-1)(x-2)(x-3)!}{3 \cdot 2 \cdot 1 \cdot (x-3)!} = \frac{x(x-1)(x-2)}{6}$$

For these combinations to be defined, $x$ must be an integer and $x \ge 3$.

Since $x$ is a positive integer and $x \ge 3$, $x \neq 0$. We can divide the entire equation by $x$:

$$\frac{x-1}{2} + \frac{(x-1)(x-2)}{6} - 4 = 0$$

Multiply the entire equation by 6 to eliminate the denominators:

$$6\left(\frac{x-1}{2}\right) + 6\left(\frac{(x-1)(x-2)}{6}\right) - 6(4) = 0$$ $$3(x-1) + (x-1)(x-2) - 24 = 0$$

Expand the terms:

$$3x - 3 + (x^2 - 2x - x + 2) - 24 = 0$$ $$3x - 3 + x^2 - 3x + 2 - 24 = 0$$

Combine like terms:

$$x^2 + (3x - 3x) + (-3 + 2 - 24) = 0$$ $$x^2 - 25 = 0$$

$$x^2 = 25$$

Take the square root of both sides:

$$x = \pm\sqrt{25}$$ $$x = \pm 5$$

The problem states that $x$ is a positive integer. Therefore, we choose the positive value.

$$x = 5$$

This value satisfies the condition $x \ge 3$.