解有理式 1 (英)
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解有理式 1 (英) : 解有理式 1
- Solve the equation, 5 over x minus 1 is equal to
- 3 over x plus 3.
- And they give us two constraints here. x can't be
- negative 3 or 1.
- And that's because if x was negative 3, this expression
- right here, you'd be dividing by 0, it'd be undefined.
- If x was 1, this expression right here would be dividing
- by 0 and it would be undefined.
- So let's see what we can do here.
- So let me just rewrite it so we have some
- space to work with.
- So 5 over x minus 1 is equal to 3 over x plus 3.
- We're going to assume these constraints
- throughout this video.
- Now, I don't like having my x terms in the denominator, so
- let's see if we can get them out of the denominator.
- Well, a good place to start, if we don't want this x plus 3
- in the denominator right here, we can multiply both sides of
- this equation by x plus 3.
- Anything you do to one side you have to do the other.
- Now, by the same argument, I don't like having
- this x minus 1 here.
- So let's multiply both sides of the equation by x minus 1.
- Now, when we do this, what's going to happen?
- Well, the whole point of multiplying by x minus 1 is so
- that that cancels out with that.
- And the whole reason behind multiplying by x plus 3 is so
- that this cancels out with this.
- So we end up with 5 times x plus 3 is equal to 3
- times x minus 1.
- And all we did is we multiplied both sides of the
- equation by both denominators.
- Both sides of the equation by x plus 3 times x minus 1.
- And this is where the whole notion of cross multiplying
- comes from.
- When we did that, it looks like we just took 5 times x
- plus 3 is equal to 3 times x minus 1.
- Which is a legitimate thing to do, but it just comes from the
- idea of multiplying both sides by both denominators,
- essentially in one step.
- But now this is a pretty
- straightforward linear equation.
- We can just distribute the numbers and get the x's on one
- term-- on one side and just solve for things.
- So let's see, we have 5 times x plus 3.
- That's the same thing as 5x plus 15 when you
- distribute the 5.
- And that's going to be equal to 3x minus 3 when you
- distribute the 3.
- Now, let's get all the x's on the left-hand side.
- So let's subtract 3x from both sides.
- And we get 2x plus 15 is equal to-- those cancel out-- is
- equal to negative 3.
- And then we can subtract 15 from both sides.
- And we're left with 2x, this 2x, is equal to-- negative 3
- minus 15 is negative 18.
- Divide both sides by 2.
- Divide both sides of the equation by 2, and we're left
- with x is equal to negative 18 over 2, which is negative 9.
- We get x is equal to negative 9.
- And, of course, we say well, it's good it doesn't equal one
- of these things.
- That would have messed things up.
- And now let's test. Let's make sure that x does not equal
- negative 9 satisfies this equation.
- Let's make sure it satisfies it.
- So 5 over negative 9 minus 1 is equal to-- let's see, 5
- over negative 9 minus 1 is negative 10.
- So it's equal to 5 over negative 10,
- which is negative 1/2.
- That's where I substitute it into the
- left-hand side of the equation.
- Now, what happens when I substitute it into the
- right-hand side of the equation?
- 3 over negative 9 plus 3 is equal to 3 over negative 6,
- which is equal to negative 1/2.
- So the left-hand side does equal the right-hand side when
- x is equal to negative 9.
- And to some degree, they didn't even have to write this
- condition here, because the only x for which this is true
- is x is equal to negative 9.
- That's the only thing that you can legitimately put in here
- and this equality holding.