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Studying for a test? Prepare with these 5 lessons on Solving inequalities.
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Let's do a few more problems that bring together the concepts that we learned in the last two videos. So let's say we have the inequality 4x plus 3 is less than negative 1. So let's find all of the x's that satisfy this. So the first thing I'd like to do is get rid of this 3. So let's subtract 3 from both sides of this equation. So the left-hand side is just going to end up being 4x. These 3's cancel out. That just ends up with a zero. No reason to change the inequality just yet. We're just adding and subtracting from both sides, in this case, subtracting. That doesn't change the inequality as long as we're subtracting the same value. We have negative 1 minus 3. That is negative 4. Negative 1 minus 3 is negative 4. And then we'll want to-- let's see, we can divide both sides of this equation by 4. Once again, when you multiply or divide both sides of an inequality by a positive number, it doesn't change the inequality. So the left-hand side is just x. x is less than negative 4 divided by 4 is negative 1. x is less than negative 1. Or we could write this in interval notation. All of the x's from negative infinity to negative 1, but not including negative 1, so we put a parenthesis right there. Let's do a slightly harder one. Let's say we have 5x is greater than 8x plus 27. So let's get all our x's on the left-hand side, and the best way to do that is subtract 8x from both sides. So you subtract 8x from both sides. The left-hand side becomes 5x minus 8x. That's negative 3x. We still have a greater than sign. We're just adding or subtracting the same quantities on both sides. These 8x's cancel out and you're just left with a 27. So you have negative 3x is greater than 27. Now, to just turn this into an x, we want to divide both sides by negative 3. But remember, when you multiply or divide both sides of an inequality by a negative number, you swap the inequality. So if we divide both sides of this by negative 3, we have to swap this inequality. It will go from being a greater than sign to a less than sign. And just as a bit of a way that I remember greater than is that the left-hand side just looks bigger. This is greater than. If you just imagine this height, that height is greater than that height right there, which is just a point. I don't know if that confuses you or not. This is less than. This little point is less than the distance of that big opening. That's how I remember it. But anyway, 3x over negative 3. So now that we divided both sides by a negative number, by negative 3, we swapped the inequality from greater than to less than. And the left-hand side, the negative 3's cancel out. You get x is less than 27 over negative 3, which is negative 9. Or in interval notation, it would be everything from negative infinity to negative 9, not including negative 9. If you wanted to do it as a number line, it would look like this. This would be negative 9, maybe this would be negative 8, maybe this would be negative 10. You would start at negative 9, not included, because we don't have an equal sign here, and you go everything less than that, all the way down, as we see, to negative infinity. Let's do a nice, hairy problem. So let's say we have 8x minus 5 times 4x plus 1 is greater than or equal to negative 1 plus 2 times 4x minus 3. Now, this might seem very daunting, but if we just simplify it step by step, you'll see it's no harder than any of the other problems we've tackled. So let's just simplify this. You get 8x minus-- let's distribute this negative 5. So let me say 8x, and then distribute the negative 5. Negative 5 times 4x is negative 20x. Negative 5-- when I say negative 5, I'm talking about this whole thing. Negative 5 times 1 is negative 5, and then that's going to be greater than or equal to negative 1 plus 2 times 4x is 8x. 2 times negative 3 is negative 6. And now we can merge these two terms. 8x minus 20x is negative 12x minus 5 is greater than or equal to-- we can merge these constant terms. Negative 1 minus 6, that's negative 7, and then we have this plus 8x left over. Now, I like to get all my x terms on the left-hand side, so let's subtract 8x from both sides of this equation. I'm subtracting 8x. This left-hand side, negative 12 minus 8, that's negative 20. Negative 20x minus 5. Once again, no reason to change the inequality just yet. All we're doing is simplifying the sides, or adding and subtracting from them. The right-hand side becomes-- this thing cancels out, 8x minus 8x, that's 0. So you're just left with a negative 7. And now I want to get rid of this negative 5. So let's add 5 to both sides of this equation. The left-hand side, you're just left with a negative 20x. These 5's cancel out. No reason to change the inequality just yet. Negative 7 plus 5, that's negative 2. Now, we're at an interesting point. We have negative 20x is greater than or equal to negative 2. If this was an equation, or really any type of an inequality, we want to divide both sides by negative 20. But we have to remember, when you multiply or divide both sides of an inequality by a negative number, you have to swap the inequality. So let's remember that. So if we divide this side by negative 20 and we divide this side by negative 20, all I did is took both of these sides divided by negative 20, we have to swap the inequality. The greater than or equal to has to become a less than or equal sign. And, of course, these cancel out, and you get x is less than or equal to-- the negatives cancel out-- 2/20 is 1/10. If we were writing it in interval notation, the upper bound would be 1/10. Notice, we're including it, because we have an equal sign, less than or equal, so we're including 1/10, and we're going to go all the way down to negative infinity, everything less than or equal to 1/10. This is just another way of writing that. And just for fun, let's draw the number line. Let's draw the number line right here. This is maybe 0, that is 1. 1/10 might be over here. Everything less than or equal to 1/10. So we're going to include the 1/10 and everything less than that is included in the solution set. And you could try out any value less than 1/10 and verify that it will satisfy this inequality.