If $f(x) = a\log|x| + \dfrac{b}{x} + x$ has its extreme values at $x = -1$ and $x = 3$, then $(a, b)$ is equal to:

Differentiate $f(x)$:

$f'(x) = \frac{a}{x} - \frac{b}{x^2} + 1$

At an extreme value, $f'(x) = 0$. Multiply through by $x^2$:

$x^2 + ax - b = 0$

The roots of this quadratic are $x = -1$ and $x = 3$.

Sum of roots:

$-a = -1 + 3 = 2 \implies a = -2$

Product of roots:

$-b = (-1)(3) = -3 \implies b = 3$

Verification:

$f'(x) = -\frac{2}{x} - \frac{3}{x^2} + 1 = \frac{x^2 - 2x - 3}{x^2} = \frac{(x-3)(x+1)}{x^2}$

This equals $0$ at $x = -1$ and $x = 3$. ✓

$\therefore (a, b) = (-2, 3)$