Solve for x. (5x^2 - 12x) / (x^2 + 5x - 24) <= 0
x(5x-12) / (x+8)(x-3) = 0
We have several points where the graphs crosses the x-axis or has an asymptote. Since there are no double roots in the denominator, the function will change sign at the asymptotes, as well as at the zeros.
So, we need to check the sign of the function at x =
-8, 0, 12/5, 3
Since for large x, both top and bottom are positive, y>0 for x < -8. After that it changes sign at each point of interest. So, y < 0 in
(-8,0)U(12/5,3)
Now, we can't include the interval endpoints where they lie at an asymptote, so y <= 0 in
(-8,0] U [12/5,3)