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Let's do a few more problems that bring together the

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concepts that we learned in the last two videos.

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So let's say we have the inequality 4x plus 3 is less

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than negative 1.

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So let's find all of the x's that satisfy this.

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So the first thing I'd like to do is get rid of this 3.

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So let's subtract 3 from both sides of this equation.

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So the left-hand side is just going to end up being 4x.

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These 3's cancel out.

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That just ends up with a zero.

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No reason to change the inequality just yet.

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We're just adding and subtracting from both sides,

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in this case, subtracting.

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That doesn't change the inequality as long as we're

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subtracting the same value.

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We have negative 1 minus 3.

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That is negative 4.

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Negative 1 minus 3 is negative 4.

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And then we'll want to-- let's see, we can divide both sides

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of this equation by 4.

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Once again, when you multiply or divide both sides of an

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inequality by a positive number, it doesn't change the

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inequality.

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So the left-hand side is just x.

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x is less than negative 4 divided by 4 is negative 1.

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x is less than negative 1.

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Or we could write this in interval notation.

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All of the x's from negative infinity to negative 1, but

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not including negative 1, so we put a

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parenthesis right there.

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Let's do a slightly harder one.

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Let's say we have 5x is greater than 8x plus 27.

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So let's get all our x's on the left-hand side, and the

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best way to do that is subtract 8x from both sides.

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So you subtract 8x from both sides.

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The left-hand side becomes 5x minus 8x.

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That's negative 3x.

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We still have a greater than sign.

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We're just adding or subtracting the same

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quantities on both sides.

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These 8x's cancel out and you're just left with a 27.

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So you have negative 3x is greater than 27.

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Now, to just turn this into an x, we want to divide both

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sides by negative 3.

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But remember, when you multiply or divide both sides

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of an inequality by a negative number, you swap the

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inequality.

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So if we divide both sides of this by negative 3, we have to

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swap this inequality.

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It will go from being a greater than sign

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to a less than sign.

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And just as a bit of a way that I remember greater than

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is that the left-hand side just looks bigger.

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This is greater than.

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If you just imagine this height, that height is greater

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than that height right there, which is just a point.

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I don't know if that confuses you or not.

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This is less than.

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This little point is less than the

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distance of that big opening.

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That's how I remember it.

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But anyway, 3x over negative 3.

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So now that we divided both sides by a negative number, by

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negative 3, we swapped the inequality from greater than

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to less than.

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And the left-hand side, the negative 3's cancel out.

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You get x is less than 27 over negative 3,

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which is negative 9.

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Or in interval notation, it would be everything from

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negative infinity to negative 9, not including negative 9.

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If you wanted to do it as a number line, it

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would look like this.

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This would be negative 9, maybe this would be negative

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8, maybe this would be negative 10.

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You would start at negative 9, not included, because we don't

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have an equal sign here, and you go everything less than

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that, all the way down, as we see, to negative infinity.

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Let's do a nice, hairy problem.

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So let's say we have 8x minus 5 times 4x plus 1 is greater

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than or equal to negative 1 plus 2 times 4x minus 3.

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Now, this might seem very daunting, but if we just

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simplify it step by step, you'll see it's no harder than

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any of the other problems we've tackled.

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So let's just simplify this.

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You get 8x minus-- let's distribute this negative 5.

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So let me say 8x, and then distribute the negative 5.

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Negative 5 times 4x is negative 20x.

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Negative 5-- when I say negative 5, I'm talking about

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this whole thing.

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Negative 5 times 1 is negative 5, and then that's going to be

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greater than or equal to negative 1 plus

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2 times 4x is 8x.

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2 times negative 3 is negative 6.

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And now we can merge these two terms. 8x minus 20x is

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negative 12x minus 5 is greater than or equal to-- we

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can merge these constant terms. Negative 1 minus 6,

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that's negative 7, and then we have this plus 8x left over.

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Now, I like to get all my x terms on the left-hand side,

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so let's subtract 8x from both sides of this equation.

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I'm subtracting 8x.

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This left-hand side, negative 12 minus 8,

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that's negative 20.

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Negative 20x minus 5.

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Once again, no reason to change the

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inequality just yet.

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All we're doing is simplifying the sides, or adding and

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subtracting from them.

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The right-hand side becomes-- this thing cancels out, 8x

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minus 8x, that's 0.

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So you're just left with a negative 7.

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And now I want to get rid of this negative 5.

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So let's add 5 to both sides of this equation.

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The left-hand side, you're just left with a negative 20x.

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These 5's cancel out.

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No reason to change the inequality just yet.

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Negative 7 plus 5, that's negative 2.

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Now, we're at an interesting point.

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We have negative 20x is greater than or equal to

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negative 2.

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If this was an equation, or really any type of an

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inequality, we want to divide both sides by negative 20.

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But we have to remember, when you multiply or divide both

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sides of an inequality by a negative number, you have to

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swap the inequality.

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So let's remember that.

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So if we divide this side by negative 20 and we divide this

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side by negative 20, all I did is took both of these sides

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divided by negative 20, we have to swap the inequality.

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The greater than or equal to has to become a less than or

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equal sign.

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And, of course, these cancel out, and you get x is less

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than or equal to-- the negatives cancel

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out-- 2/20 is 1/10.

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If we were writing it in interval notation, the upper

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bound would be 1/10.

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Notice, we're including it, because we have an equal sign,

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less than or equal, so we're including 1/10, and we're

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going to go all the way down to negative infinity,

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everything less than or equal to 1/10.

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This is just another way of writing that.

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And just for fun, let's draw the number line.

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Let's draw the number line right here.

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This is maybe 0, that is 1.

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1/10 might be over here.

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Everything less than or equal to 1/10.

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So we're going to include the 1/10 and everything less than

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that is included in the solution set.

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And you could try out any value less than 1/10 and

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verify that it will satisfy this inequality.

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