Answer
The solution is $(-\infty,-1)\cup[3,\infty-)$
The graph is:
Work Step by Step
$\dfrac{x-3}{x+1}\ge0$
Find the intervals. The factor of the numerator is $x-3$ and the factor of the denominator is $x+1$. Set them equal to $0$ and solve for $x$:
$x-3=0$
$x=3$
$x+1=0$
$x=-1$
The factors are zero when $x=-1,3$. These two numbers divide the real line into the following intervals:
$(-\infty,-1)$ $,$ $(-1,3)$ $,$ $(3,\infty)$
Elaborate a diagram, using test points to determine the sign of each factor in each interval: (refer to the attached image below)
It can be seen from the diagram that the intervals $(-\infty,-1)$ and $(3,\infty)$ satisfy the inequality. The endpoint $-1$ does not satisfy the inequality because it makes the denominator equal to $0$ but the endpoint $3$ does satisfy the inequality, because of the symbol $\ge$
The solution is $(-\infty,-1)\cup[3,\infty)$
