Question

$\frac{41x-12}{x^2-16}=\frac{4x+3}{x-4}.$

Answer 1

$\frac{41x-12}{(x+4)(x-4)}=\frac{4x+3}{(x-4)}$

$\Rightarrow 41x-12=(4x+3)(x+4)$

$\Rightarrow 41x-12={4x}^2+19x+12$

$\Rightarrow {4x}^2-22x+24=0$

$\Rightarrow x=3/2,4$

But at $x=4$, $x-4=0$

which is not valid

$\therefore$ $x=3/2$