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

Solution:

Given:

$\frac{x-3}{x+4}<0, x \in R$

Signs of $x-3$

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

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

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

Signs of x + 4

$x+4=0 \rightarrow x=-4$ (Subtracting both the sides by 4 )

$x+4<0 \rightarrow x<-4$ (Subtracting both the sides by 4 )

$x+4>0 \rightarrow x>-4$ (Subtracting both the sides by 4 )

$\frac{x-3}{x+4}$ $x+4$ is not defined when $\mathrm{x}=-4$

The interval that satisfies the condition that $\frac{x-3}{x+4}<0$ is $-4<\mathrm{x}<3$

Therefore,

$x \in(-4,3)$