Answer
The inequality is valid for values between -7 and -1 and for values more than 3 (not including them) i.e. $(-7,-1)\cap (3,\infty)$
Work Step by Step
First, we are going to move everything to the left side and simplify:
$\dfrac{5}{x-3}>\dfrac{3}{x+1}$
$\dfrac{5}{x-3}-\dfrac{3}{x+1}>0$
$\dfrac{5(x+1)}{(x-3)(x+1)}-\dfrac{3(x-3)}{(x-3)(x+1)}>0$
$\dfrac{5x+5-(3x-9)}{(x-3)(x+1)}>0$
$\dfrac{2x+14}{(x-3)(x+1)}>0$
Now, we find critical points by equating the numerator and denominator to zero:
$2x+14=0$
$(x-3)(x+1)=0$
There are three critical points:
$2x_1+14=0\rightarrow x_1=-\frac{14}{2}=-7$
$x_2+1=0\rightarrow x_2=-1$
$x_3-3=0\rightarrow x_3=3$
First test with a value less than -7:
$\dfrac{5}{-10-3}>\dfrac{3}{-10+1}$
$\dfrac{5}{-13}>\dfrac{3}{-9}$
$-\dfrac{5}{13}>-\dfrac{1}{3} \rightarrow \text{ FALSE}$
Second test with a value between -7 and -1:
$\dfrac{5}{-2-3}>\dfrac{3}{-2+1}$
$\dfrac{5}{-5}>\dfrac{3}{-1}$
$-1>-3 \rightarrow \text{ TRUE}$
Third test with a value between -1 and 3:
$\dfrac{5}{2-3}>\dfrac{3}{2+1}$
$\dfrac{5}{-1}>\dfrac{3}{3}$
$-5>1 \rightarrow \text{ FALSE}$
Fourth test with a value more than 3:
$\dfrac{5}{5-3}>\dfrac{3}{5+1}$
$\dfrac{5}{2}>\dfrac{3}{6}$
$2.5>0.5 \rightarrow \text{ TRUE}$
The inequality is valid for values between -7 and -1 and for values more than 3 (not including them) i.e. $(-7,-1)\cap (3,\infty)$.
