The first thing we have to see is that this equation is indeterminate when the value of its denominator is equal to zero. We call this value critical value
[tex]\begin{gathered} x-8=0 \\ x=8\to\text{critical value} \end{gathered}[/tex]Now we will solve the inequality
[tex]\begin{gathered} \frac{x+7}{x-8}\ge3 \\ x+7\ge3(x-8) \\ x+7\ge3x-3(8) \\ 7+24\ge3x-x \\ 31\ge2x \\ x\le\frac{31}{2} \end{gathered}[/tex]Using the value of the inequality and the critical value found earlier, we can determine the range of this
[tex]\begin{gathered} 8