Well, I didn't help you with the graphing, all I gave you was what intervals of the real line satisfy the inequality.

The way I learned how to graph the first problem (there are at least two conventions being used)

would be to put an open circle on x = -2, x = 0, and x = 1, then draw a line connecting x = -2 and x = 0 to indicate that these points are not part of the solution set, then draw a line to the right starting at x = 1 and continuing to

. (I put an arrow on the line from x = 1.)

For the second problem:

I put an open circle on -1 and a closed circle on 4 (to indicate that x = 4 is part of the solution) and connect x = -1 and x = 4.

Why can't x = -1 in the second solution? First of all, PLEASE use parenthesis!!

What you wanted was "(x - 4)/(x + 1)"

Anyway, plug x = -1 into the original inequality. Is this possible? What about x = 4?

-Dan