Solve each of the following in equations and represent the solution set on

Solution:

Given:

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

Signs of $x-3$

$x-3=0 \rightarrow x=0$ (Adding both the sides by 3 )

$x-3<0 \rightarrow x<3$ (Adding both the sides by 3 )

$x-3>0 \rightarrow x>3$ (Adding both the sides by 3 )

Signs of x + 1

$x+1=0 \rightarrow x=-1$ (Subtracting both the sides by 1 )

$x+1<0 \rightarrow x<-1$ (Subtracting both the sides by 1 )

$x+1>0 \rightarrow x>-1$ (Subtracting both the sides by 1 )

$x-3$

$\overline{x+1}$ is not defined when $\mathrm{x}=-1$

The interval that satisfies the condition that $\frac{x-3}{x+1}<0$ is $-1<\mathrm{x}<3$

Therefore,

$x \in(-1,3)$