[tex]\begin{gathered} (-\infty,-4)\cup(7,\infty) \\ \end{gathered}[/tex]
Explanation
[tex]\frac{x+4}{x-7}>0[/tex]
Step 1
Multiply both sides by (x-7)
[tex]\begin{gathered} \frac{x+4}{x-7}>0 \\ \frac{x+4}{x-7}\cdot(x-7)>0(x-7) \\ x+4>0 \\ \text{subtract 4 in both sides} \\ x+4-4>0-4 \\ x>-4 \end{gathered}[/tex]
Check possible critical points.
x=-4
[tex]\begin{gathered} \frac{x+4}{x-7}>0 \\ \frac{-4+4}{-4-7}>0 \\ \frac{0}{-11}>0 \\ 0>0 \end{gathered}[/tex]
so, test each interval for a positive or negative result, the divide ths igns
we need a result greater than zero ( so a positive number)
therefore, the answer is
[tex]\begin{gathered} (-\infty,-4)\cup(7,\infty) \\ \end{gathered}[/tex]
I hope this helps you
