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# 多步驟的不等式 (英) : 多步驟的不等式

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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.

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