Algebra

$f(x) = \frac{2x + 5}{x - 4} + \frac{x^2 + 1}{4x - 3}$

$y = \frac{2x + 5}{x - 4} + \frac{x^2 + 1}{4x - 3}$

$x= \frac{2y + 5}{y - 4} + \frac{y^2 + 1}{4y - 3}$

$x(y-4)(4y-3)= (2y + 5)(4y-3) + (y^2 + 1)(y-4)$

$4xy^2-19xy+12x= y^3+4y^2+15y-19$

$y^2(4x-4)-y(19x+15)+12x= y^3-19$

Honestly, Idk what to do from here, but you have to solve for y in this equation. My idea would be to pile them into a cubic and try solving for x that way but.... I have doubts.