 So again, we'll use our knowledge of solving E equations to allow us to solve in equations, inequalities, in other words. We'll solve rational inequalities The same way we solved other inequalities. First, we'll ignore the inequality and then rewrite it as an equation. We'll find the critical values and then, like a good human being or a good mathematician, we'll acknowledge the existence of the inequality and identify the solution intervals. The one important difference between rational inequalities and other sorts is that the forbidden values are critical values. So let's solve this inequality. First, we'll identify the forbidden values. Those will be x equals negative three and x equals negative one. Next, we'll solve our equation. We'll multiply through by our denominators x plus three and x plus one. We'll get a new equation, which we can solve, and that gives a solution x equals one-third and three critical values. The two forbidden values, negative three and negative one, and the actual solution to the equality one-third. So let's begin by graphing our inequality. So first, we'll locate our critical points. Since x equals negative three and x equals negative one are forbidden values, we're not allowed to include them in our solution set, and so we use open circles at these points. Meanwhile, since x equals one-third actually solves the equation and the inequality allows for equality, we'll use a closed circle at x equals one-third. And so our critical points have partitioned the interval into four. Count them four, one, two, three, four intervals, and we have to test a point at each of the intervals. In the first interval, it will test negative one million, and we find this inequality is false, so we exclude the interval. In the second interval, we might use, oh, I don't know, negative two as a test value, and we find this is true, so we include this interval. In the third interval, we can use x equals zero as a test point, and the inequality is false in this interval, so we exclude this interval. And in the last interval, which includes everything to the way we write, we can use x equals one million as our test point, and this is true, so we include this last interval. And finally, let's write down our solution. Notice that the problem is given to us in algebraic form, not as a picture, and so we want to write our final answer in algebraic form, which means we should probably give the solution in interval notation, and so translating our picture of the inequality solution into the interval notation we get, or we could try another inequality. The forbidden values are those that make the denominator equal to zero, so we'll set denominator equal to zero and solve. We can save ourselves a little bit of effort by noting that when we calculate the discriminant, we get a negative number, so this equation has no real solutions, and so there are no forbidden values. We'll solve the corresponding equation. We'll multiply through by our denominator. That gives us a quadratic equation, which we'll solve using our favorite method. Hint, it's the quadratic formula, and we get our two solutions. When we graph, it'll be useful to recognize that our critical values are 5 halves plus some amount and 5 halves minus some amount. So it's useful to remember to plot first and label, so 5 halves is somewhere, and we're going to go 5 halves plus a little bit and 5 halves minus a little bit. And because these critical values solve the equation and our inequality allows for equality, we'll go ahead and represent these using closed circles. And the critical values form three intervals on the real number line, so we'll test a point in each interval. So our first interval includes everything to the way, way left, and so we might test x equals minus 1 million. And we note that minus 1 million squared will be much larger than 8 times minus 1 million, so our denominator is going to be much larger than the numerator, and so our fraction is going to be significantly less than 1, and so this inequality is false, and we exclude this interval to the way left. Because of the way we've written our critical values, we know that x equals 5 halves is in the middle interval, and so we'll use x equals 5 halves as our test point. So testing x equals 5 halves, and after all the test settles, we see that this is true, so we include the middle part. And finally we have this part to the way, way right, so we'll try x equals 1 million for a point in this last interval. And again the same argument applies, since 1 million squared will be much larger than 8 times 1 million, the denominator is much larger than the numerator, and this inequality is false, and we'll exclude this way, way right. Again the graph is a geometric picture of our solution, but because the question was asked algebraically, we should write our answer algebraically and give our solution in interval notation.