Question

$\frac{14x}{x+1}-\frac{9x-30}{x-4}<0$

Answer 1

$\left(\frac{14x}{x+1}\right)-\left(\frac{9x-30}{x-4}\right)<0=\left(\frac{14x(x-4)-(9x-30)(x+1)}{(x+1)(x-4)}\right)<0=\left(\frac{14x^2-56-(9x^2+9x-30x-30)}{(x+1)(x-4)}\right)<0=\left(\frac{14x^2-56-9x^2+21x+30}{(x+1)(x-4)}\right)<0=\left(\frac{5x^2+21x-26}{(x+1)(x-4)}\right)<05x^2+21x-26$

will be always greater than zero (as D < 0)$\therefore x\in (-1,4)$