 Alright, let's solve inequalities. Remember that when we solved inequalities before, we found out that if we multiply or divide by a negative, then that will cause us to switch the inequality, but otherwise the inequality stays the same. All we need to do is add or subtract across the inequality and then multiply and divide across the inequality. So here we have ten that I can take to the other side so that I'll have my variable, negative six B term, on one side and my constant, which is now thirty, on the other side. Now notice, in this very first example, I purposely gave you a negative coefficient so that you would remember that we're going to have to switch the inequality. This is going to have to become less than or equal to. And now when I divide by negative six, I get B on this side and thirty divided by negative six will be negative five and B is less than or equal to negative five. If we wanted to check it, we could pick a number that was less than negative five. Let's say B is going to be equal to say negative six. So if I come in here and I say negative six and then times my B, which is negative six right here, and plus ten, that should be greater than forty. Negative six times negative six is positive thirty six plus ten, which is greater than forty because forty six is greater than forty. It worked. Here I have variables on both sides of the inequality and I also have constants on both sides of the inequality. So I'm going to have to move across my inequality several times. And it doesn't matter if I move x or I move constant first. So I'm purposely going to move the x first, trying to keep it positive. Now I see that eight x is bigger than negative two x, so I want to move everything over to this eight x side. So I add two x to both sides and that'll give me ten x. And I still have one on this side and that's greater than fifty one. Now I have to move the constant. So I subtract the one and ten x will be greater than fifty. And if I divide by ten, it's a positive ten, so I don't have to change my inequality at all. So that x is going to be greater than five. Over here on this last example, we have, again, variables on both sides of the inequality and constants on both sides. So I want to move my variable first. Now I see that this is a negative three and this is a negative five. Negative five is smaller, so I want to move it over here toward that negative three, which is larger, because then nice things will happen. If I add five x to both sides, then I get a positive two x over here. Positive five x minus three x is positive two x, plus five. The center equal to negative seven. Then I subtract the five from both sides and two x is less than or equal to negative twelve, subtracting five. And I divide by positive two, so my inequality stays the same. On this side I have x and negative twelve divided by positive two will be negative six. What happens if we look at a table? This is the second example that we had earlier. Eight x plus one is greater than negative two x plus fifty one. Well, it's going to ask us, what is f of x equal to g of x? Where are they the same? And they are the same when x is equal to, right here at five, they're both 41. So when x is equal to five, I have 41 equal to 41, f of x equal to g of x. f of three, remember we come to three in the x and f of three is going to be 23. And g of three is going to be 45. So we would say f of three is less than g of three. And when I look at f of six, I see that f of six is 49 and g of six is 29. So f of six is bigger or greater than g of six. So what interval satisfies the inequality? Well, remember that this one is my f of x. And it's asking me for f of x is greater than g of x. Well, f of x is greater than g of x. And here at f of six, it's also at f of seven. And anything larger than x equaling five will give us f of x greater than g of x. It can't be five, but where they're equal will always tell us where to begin. And it can be greater than five. My inequality won't let it be equal to five. Let's look at a graph. The same graph of the same equation we did now symbolically and with a table, eight x plus one, negative two x plus 51 on both sides. That's g of x is negative two x plus 51, f of x is eight x plus one. And we want to compare f of negative five and g of negative five. Remember that's an x is equal to negative five. So down here, f of negative five looks like it's going to be approximately negative 40. And up here for x equal negative five, g of negative five looks like it's going to be approximately 60. So we could say that this would be negative 40 and positive 60. So f of negative five is less than g of negative five. If I do f of ten, doing the same thing, I'm going to let x be ten. Now remember, this point right here is on my g line. So g of ten looks to be approximately 30. And my f of ten, because this is my f line, f of ten looks to be approximately 80. We don't know that for sure, but it's approximate. So f of ten is now going to be greater than g of ten. Where are they the same? It looks like they're the same right about here where x equal five. So does the interval we found before seem to satisfy this inequality from the graph? And in our inequality before, we said that x had to be greater than five. And you can see that over here where x is greater than five to the right of five on the x-axis, if I go to my graph, all the points on my f graph are above the ones on my g graph. So yes, it satisfies.