 So let's graph the solution to minus 29 less than 5x plus 6 less than or equal to 31. So remember that when we write a inequality this way, this actually corresponds to the compound inequality, negative 29 less than 5x plus 6, and also 5x plus 6 less than or equal to 31. So we'll solve this first inequality as an equality. Now, when we solve the equality, like a good math student or a good human being, we acknowledge the existence of the inequality. And since x equals minus 7 solves the equality, minus 29 equals 5x plus 6, but the inequality is strict, then x equals minus 7 is not part of the solution. And so we note that our solution has a critical value at x equals minus 7, which is not included. Now we'll solve the second as an equality, 5x plus 6 equals 31, so solving that gives us. And so that tells us that x equals 5 is a critical value. And again, a good math student or a good human being would recognize that the inequality exists and act accordingly. Now, x equals 5 solves the equality, 5x plus 6 equals 31, and the inequality allows for 5x plus 6 to equal 31, but this is a compound inequality with an AND, so in order for something to be part of the solution, it has to satisfy both of these inequalities. Since this is a compound inequality, we have to check that x equals 5 also solves the other inequality. Is it true that minus 29 is less than 5x plus 6 when x is equal to 5? And we find that minus 29 is in fact less than 31, and so this is true, and so that means that x equals 5 is going to be included. And so our solution has a critical value x equals 5, which is included. Let's go ahead and graph those critical values. So the critical value x equals minus 7, which is not included, we'll want to graph using an open circle. So we'll select our open circle tool and put an open circle at minus 7. The critical value x equals 5 is included, which means we'll use a closed circle, and so we'll select our closed circle tool and put a closed circle at x equals 5. And now we see that the critical values separate the number line into three parts. In the region to the left, we'll test x equals minus, oh, I don't know, how about minus 1000. And so we need to see if x equals minus 1000 satisfies this compound inequality. And remember that since this is an AND inequality, we have to satisfy both to be part of the solution. If x equals minus 1000, then 5x is a large negative number. 5x plus 6, well, if you add 6 to a large negative number, you still get a large negative number. But importantly, a large negative number will not be greater than minus 29. And so x equals minus 1000 will not be part of the solution, and we should exclude the region to the left. Now we might see that x equals 0 is in the central region, so we'll use that as our test point. If x equals 0, our left inequality minus 29, is it true that it's less than 5x plus 6? So we'll check it out, and this is true. But since this is a compound inequality, we also have to check out our right inequality is 5x plus 6 less than or equal to 31. And this is also true. And since both inequalities are true, x equals 0 solves a compound inequality, so we should include the central region. So I'll switch to the line tool and drag a line from the starting point of the interval to the ending point of the interval. And finally, in the region to the right, we'll test x equals, oh, I don't know, 1 million. And so we find that 5x plus 6 will be a large positive number, satisfying our first inequality minus 29 less than 5x plus 6, but a large positive number will not be less than or equal to 31. And since both inequalities must be satisfied, x equals 1 million is not a solution, so we exclude the region to the right. And again, it's useful to remember that these cross-outs just indicate that we have checked these outside intervals, so we don't really need them, and they're not really part of our answer. And now we're ready to submit our answer.