解有理式 1 (英)

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解有理式 1 (英): 解有理式 1

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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.

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