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- Let's solve some rational equations.
- Let's say I had 5 over 3x minus 4 is equal to--
- actually, let me write it.
- Let me scoot it over a little bit.
- So let's say I have 5 over 3x minus 4 is equal to
- 2 over x plus 1.
- So your first reaction might be, geez, I've
- never seen this before.
- I have x's in the denominators, how do I
- actually solve this equation?
- And the easiest way to proceed, and there's other
- ways to do this, is to try to multiply both sides of this
- equation by expressions that will get rid of it these x's
- in the denominator.
- So if we multiply both sides of the equation by x plus 1,
- it's going to get rid of this x plus 1 in the denominator.
- Of course, we have to do it to both sides.
- We can't just do it to one side.
- But we also want to get rid of this expression right here.
- So let's multiply both sides of the equation by that, by 3x
- minus 4, so you're multiplying by 3x minus 4 on
- this side as well.
- So I'm just multiplying both sides of the equation times
- this and times this.
- So whatever I multiply on this side, I also have
- to do on this side.
- What is it going to simplify to?
- Well, this 3x minus 4 is going to cancel out with this 3x
- minus 4, so the left-hand side is just going to be x plus 1
- times 5, or 5 times x plus 1.
- And my right-hand side, this x plus 1 is going to cancel with
- that x plus 1.
- It's just going to equal 2 times 3x minus 4.
- And if you just look at it, if you didn't look at what we
- actually did, it looks like we did something called
- cross-multiplying.
- When you have something over something is equal to
- something else over something else, notice, the end product
- when we multiplied both sides by both of these expressions
- was 2 times 3x minus 4 is equal to 5 times x plus 1.
- So it looks like we cross-multiplied.
- 5 times x plus 1 is equal to 2 times 3x minus 4.
- But the reality is we didn't do anything new.
- We just multiplied both sides by this expression and that
- expression.
- But now we just have a straight-up linear equation.
- We can just simplify and solve.
- So the left-hand side becomes 5x plus 5, just distribute the
- 5, is equal to 6x minus 8.
- Now, let's say we subtract 5x from both sides.
- If we subtract 5x from both sides, the left-hand side just
- becomes a 5.
- The right-hand side, we are left with x minus 8, and now
- we could add 8 to both sides.
- The left-hand side becomes 13, the right-hand side is just an
- x, and we're done.
- x is equal to 13 solves this equation.
- You could even try it out.
- 5 over-- 3 times 13 is 39.
- 39 minus 4 is 35, and this should be equal to 2 over 13
- plus 1, or 14.
- And they both equal 1/7.
- So it checks out.
- Let's do a more involved one.
- Let's do a more involved problem.
- So let's say I had-- this one's pretty
- involved right here.
- Let's say I have negative x over x minus 2 plus-- I'm
- going to give it some space here-- 3x minus 1 over x plus
- 4 is equal to 1 over x squared plus 2x minus 8.
- So once again, if we want to get rid of all of these x
- terms in the denominator, we want to multiply essentially
- by the least common multiple of this expression, this
- expression, and this expression.
- So this one looks like it can be factored.
- So maybe it already has some of these guys in it.
- So let's try to figure that out.
- So this right here, this is what?
- This is x plus 4 times x minus 2, so it does, indeed, have
- both of these in it.
- So lets them multiply both sides of this equation by this
- thing or by x plus 4 and x minus 2.
- So if we multiplied-- so this thing, it could just be
- rewritten as that, so I just did that.
- Now let's multiply both sides of the equation by that.
- So let's see, we're going to multiply by x plus 4
- times x minus 2.
- If we do it on the right-hand side, we have to do it on the
- left-hand side.
- Scoot over a little bit.
- So this term, I'm going to have to multiply by x plus 4
- times x minus 2.
- Same thing for this term.
- That's why I left some space.
- X plus 4 times x minus 2.
- I'm just multiplying every term in this equation by this
- right there.
- Now, on the left-hand side, what do we get?
- This x minus 2 cancels with this x minus 2.
- We have nothing left in the denominator.
- This term right here will become negative x
- times x plus 4.
- That's that term right there.
- Now, this term and this term right here, this x plus 4 will
- cancel with that x plus 4.
- So we're a left with plus 3x minus 1 times x minus 2.
- And then the right-hand side, that cancels with that, that
- cancels with that.
- And we've essentially multiplied both sides by this
- denominator or by the inverse of this whole expression.
- So we're just left with this 1.
- And now, we have to simplify it, or we have to multiply
- things out.
- So negative x times negative x is negative x squared.
- Negative x times 4 is minus 4x.
- And then we have 3x times x is plus 3x squared.
- 3x times negative 2 is minus 6x.
- Negative 1 times x is minus x or negative x.
- Negative 1 times negative 2 is plus 2.
- All of that's going to be equal to 1.
- Now let's add the same degree terms. So we have a second
- degree term and a second degree term.
- 3x squared minus x squared.
- That gives us 2x squared, and you have a negative 4x, a
- negative 6x and a negative x.
- So what is that going to be equal to?
- Negative 4 and a negative 6 is negative 10.
- Minus another 1 is a negative 11x.
- And then finally, we just have that constant term out here,
- plus 2 is equal to 1.
- Let me make sure I got all the terms. Yeah, I got all the
- terms. Now we could subtract 1 from both sides and the
- equation becomes 2x squared minus 11x plus
- 1 is equal to 0.
- So it just becomes a straight-up, traditional
- quadratic equation.
- And if we're just looking for the roots, we set it equal to
- zero, we're just looking for the x's that satisfy this, we
- can use the quadratic formula.
- So the solutions are x is going to be equal to negative
- b. b is negative 11.
- So negative negative 11 is positive 11 plus or minus the
- square root of b squared.
- Negative 11 squared is 121 minus 4 times a, which is 2,
- times c, which is 1, all of that over 2 times a, all of
- that over 4.
- So x is equal to 11, plus or minus the square root of-- 4
- times 2 times 1 is 8.
- So 121 minus 8.
- So what is that?
- That's 113; is that right?
- The square root of 113, all of that over 4.
- Can I factor 113 at all?
- Did I get that right?
- 4 times 2 times 1.
- Yeah, that looks right.
- Let's get our calculator out to actually figure out what
- these values are.
- Let me see, so we want to turn the calculator on.
- Go to my main screen, and we want the square root
- of 121 minus 8.
- 10.63.
- So this is equal to 11 plus or minus 10.63 over 4.
- Those are two answers.
- And if we want to find the particular answers, it would
- be 11 plus 10.63 over 4, which would be what?
- That would be 21.63/4.
- We could try to calculate that in a second.
- And then if you subtract it, 11 minus 10.63 is-- so 11
- minus 10.63 over 4, this is equal to what?
- Like 0.37, 0.37/4, if I'm doing my math correctly.
- So what do we get?
- So let's see, 21.63 divided by 4 is 5.51, so
- this is equal to 5.41.
- And then the other solution, 0.37 divided by 4 is going to
- be like 0.09 something.
- So then we have the other solution is 11 minus 10.63.
- Yep, we got our 0.37 divided by 4 is equal to 0.0925 So
- this is equal to 0.0925.
- So I lost some precision.
- This isn't the exact, because this wasn't 10.63.
- It was 6301.
- It just kept going.
- It's an irrational number, but this should get close.
- So let's check.
- Let's verify that these actually work.
- So let's try the 5.41 solution first. So if this is true, if
- we take-- so let me put 5.4-- let me just do it.
- So if we do-- so we're going to start off with 5.41.
- So our original equation was negative x, so negative 5.41
- divided by x minus 2.
- So 5.41 minus 2 is 3.41.
- So that's that term.
- And then we have plus 3 times 5.41 minus 1 divided by-- 5.41
- plus 4 is 9.41-- 9.41.
- So the left-hand side gives us 0.0319.
- So that's what the left-hand side equals.
- Now let's see what the right-hand side equals when we
- substitute x equals 5.41.
- And of course, we've lost some precision here.
- We've lost some of the zeroes, but we
- should get pretty close.
- So the right-hand side, if we take 1 divided by 5.41 squared
- plus 2 times 5.41 plus-- this is a minus 8, I think.
- I lost that.
- That's a negative 8 right there, so minus 8.
- That is that, and they are very, very, very close, at
- least up to three digits, 0.31.
- So it does work out.
- And these other things, they don't equal past that because
- we weren't precise enough.
- If we added the trailing zeroes here when we took the
- square root of 113, we would have gotten the right answer.
- If you actually kept it in this form you would have
- gotten the exact right answer.
- So I'll leave it up to you to verify that this is also
- another solution.
- But, hopefully, you found that slightly instructive.

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