Answer
$(-\infty,3)\cup[7,\infty)$
Work Step by Step
$\frac{x+1}{x-3}\leq2\\\frac{x+1}{x-3}-2\leq0\\\frac{7-x}{x-3}\leq0$
The zeros from $\frac{7-x}{x-3}=0$: $7-x=0\\x=7$. The values for which the function is undefined: $x-3=0\\x=3$.
I use the real zeros and the values for which the function is undefined to separate the real number line into different intervals:
$(-\infty,3)$, $(3,7),$ $(7,\infty)$.
I now select a test number in each interval found in the step above and evaluate the function on the left side of the inequality at each number to determine if the function is positive or negative. Refer to the table for this.
According to the table the solution set: $(-\infty,3)\cup[7,\infty)$ because the function is negative or $0$ in this interval.
