Question

Solve $\frac{x+3}{x-1}>0,xϵR$

Answer 1

given equation is

$\frac{x+3}{x-1}>0$ as x-1 is in denominator it must not be equal to zero

therefore x not equals to 1

let x be >1

then the equation becomes

$x+3>0(x-1)$

$\Rightarrow x>-3$ and we assumed x>1 therefore the solution for this case is $x\in (1,\infty )$

now let x<1,

then the equation becomes

$x+3<0(x-1)$

$\Rightarrow x<-3$ and we assumed that x<1 therefore the solution for this case is $x\in (-\infty ,-3)$

therefore the overall solution is $x\in (-\infty ,-3)\cup (1,\infty )$