Question

Solve the following inequality:
$\frac{x^2+1}{4x-3}>2$

Answer 1

$\frac{x^2+1}{4x-3}>2\frac{x^2+1}{4x-3}-2>0\frac{x^2-8x+7}{4x-3}>0\frac{(x-7)(x-1)}{(x-\frac{3}{4})}>0x\in (\frac{3}{4},1)\cup (7,\infty )$