Question

Solve the inequality $\frac{x-1}{x^2-4x+3}<1$.

Answer 1

$\frac{x-1}{x^2-3x-x+3}<1$

$\frac{(x-1)}{(x-3)(x-1)}<1\{x\ne 1\}$

$\frac{1}{x-3}-1<0$

$\frac{1-x+3}{x-3}<0$

$\frac{x-4}{x-3}>0$

$x\epsilon (-\infty ,3)\bigcup (4,\infty )$

$butx\ne 1$

$x\epsilon (-\infty ,1)\bigcup (1,3)\bigcup (4,\infty )$