>> This is part two of Simplifying Complex Fractions.
In part one, we went over method one.
Here's a reminder of how to do method one.
We simplify the numerator and denominator
of the complex fraction, so each is a single fraction.
And then we multiply the fraction in the numerator
of that complex fraction by the reciprocal of the fraction
in the denominator of that complex fraction.
And then we simplify it.
At the end of part one, I introduced another method,
and we worked, one
of the problems we did using method one, using a new method,
which I'll call method two.
In this method we multiply the numerator and denominator
of the complex fraction by the least common denominator
of the fractions in both the numerator and denominator.
Then simply.
Okay. That's kind of a mouthful,
but if we actually work some problems,
we'll see this isn't so bad.
All right.
Here's an example of a complex fraction
and we want to simplify it.
Remember complex fraction has a fraction either
in the numerator, the denominator or in both,
as it does in this case.
Method two is especially useful
if you don't have a single fraction in the numerator
and a single fraction in the denominator.
So like there's some plus signs up here
in the numerator or denominator.
So what's the LCD?
Method two says find the LCD of all the denominators.
So we've got like, for all the denominators there's an x
squared, an x, and x squared,
an x. The least common denominator is simply x squared.
So we want to multiply the numerator and the denominator
of this big complex fraction by the least common denominator
which is just x squared over x squared.
So we're multiplying the numerator by x squared,
the denominator by x squared,
that'll make an equivalent fraction,
because you're multiplying the numerator
and denominator by the same thing.
All right.
So you have to distribute.
So we've got in the big numerator here,
x squared times 2 over x squared.
So let's write that out.
x squared times 2 over x squared plus distributing,
I've got x squared times 1 over x. Now
if you want you can write x squared over 1, those of you
that like to see it written as a fraction.
All right.
I've done the numerator, by x squared.
Now I'm going to do the denominator by x squared.
Same reasoning.
I'm going to write that as x squared over 1 times 4
over x squared, minus distributing x squared
over 1 times 1 over x. Now what's great about this method,
you immediately get rid of any of the fractions
in the numerator and the denominator,
because things will cancel.
So my x squares cancel here.
Now what happens over here?
I've got x squared over x, that's going to cancel
to just give me one x in the numerator,
knowing your laws of exponents, right.
We've got an x squared over x squared here.
Those are going to cancel.
And let's see, x squared over x squared,
that'll give me just one x in the numerator.
So look what happens to my numerator.
What I'm left with is just 2 over 1
which is 2 plus just an x. And in the denominator,
I've got just a 4 minus an x. And those can't be,
neither of those can be factored,
so you're not going to be able to cancel.
We're done.
That's it.
Now you could have done this problem using method one
as well.
You would have to get a common denominator in the numerator,
add the fractions, get a common denominator in the denominator,
add the fractions, then you'd have to multiply
by the reciprocal and you should end up with the same result.
So you could, why don't you put this on pause and see
if you can do method one, and then I'll show the steps.
Right let's do this using method one.
The numerator, I need a common denominator
so I could add the fractions.
So I'll have to multiple this second fraction by x over x,
so that I have the same denominator as I have
for the first fraction here,
so they're going to both be x squared.
And in the denominator just happens to be x squared.
That might not happen.
In fact usually it doesn't,
but in this case that's how it written.
So what do I have in the numerator?
I've got a common denominator of x squared,
so I could add the numerators 2 plus x. Now the denominator,
all right, I'm going to have x squared for the denominator.
And in the numerator I'll have 4 minus x. All right.
So I've written it as a single fraction in the numerator,
a single fraction in the denominator.
So the next step is to take the numerator, which is 2 plus x
over x squared and multiply it by the reciprocal
of what's in the denominators.
So remember, taking the reciprocal of this, right.
So we're going to multiply it by x squared
over 4 minus x. The x squares cancel.
We end up exactly with the same answer we did using method one.
Okay. Both methods are completely valued, valid,
you just have to make sure you do them correctly.
Let's try this one.
3 over x plus 2, plus 5 over x minus 2,
all over 4 over x squared minus 4.
So you would need to determine the least common denominator
of all these denominators.
If you were going to use method two.
And hopefully you'll realize
that x plus 2 times x minus 2 actually equals x squared
minus 4.
So x squared minus 4 is your least common denominator.
But you might want to rewrite this x squared minus 4
as x plus 2 times x minus 2
when you actually do the multiplication.
So I'm going to move this over here.
All right.
So let's do this.
I'm going to take the whole numerator,
and the whole denominator, and we're going
to multiply it by x squared minus 4.
Now, remember, you can write x squared minus 4
as x plus 2 times x minus 2.
Right. Either way is fine.
It might help you in this to break it up into its factors.
So now I'm going to multiply
through using the distributive property.
So on the numerator here,
hopefully you can see what's going to happen.
Let's see if we could do this
without writing out all the steps.
When you multiply this x plus 2 times x minus 2,
times the first term here, the x plus 2's will cancel
and you'll only have 3 times the x minus 2.
You see how that works?
Because if you wrote the x plus 2 in the numerator here,
and the x plus 2, they're just going to cancel.
Plus, all right let's do that.
Now when you multiply it by the second fraction,
do you see how the x minus 2's will cancel?
So you'll only have 5 times x plus 2.
So you see how that eliminated all the fractions?
And in the denominator what happens?
When you multiply this x plus 2 times x minus 2,
its going to cancel with x squared minus 4
because that's exactly the same as x squared minus 4.
So you just end up with a 4 in the denominator.
So all we need to do is simply this numerator.
So we're going to have to do the distributive property,
3x minus 6, plus 5x plus 10, all over 4.
Are you with me?
And then let's add like terms.
We have 8x plus 4 all over 4.
All right.
Then what do you do?
You want to factor to see if anything reduces.
So let's factor a 4 out of the numerator.
And wow that cancels out that 4.
Maybe that's like 1.
So if you want to write this down.
So what's the final answer?
All of this ends up just equaling 2x plus 1.
So wow, that original complex fraction is a complicated way
of writing 2x plus 1.
So the answer is 2x plus 1.
This can also be done using the second method
where you get a common denominator
in the numerator up here.
The numerator of the complex fraction.
And get a single fraction first.
And then you could multiply by the reciprocal.
You get exactly the same answer.
My suggestion is if you have like an addition problem
with fractions up in the numerator denominator,
the second method is usually easier.
If you already have a single fraction in the numerator
or denominator, then usually it's just as easy to go ahead
and multiple by the reciprocal.
Let's do one more.
Let's end with one easy one.
Okay. This has a single fraction
in the numerator and denominator.
So you can just take the numerator,
multiply it by the reciprocal of the denominator.
And that's pretty easy, right?
Because it already had single fractions, 5 over 6.
Now let's just do method two.
What's the least common denominators?
Well you've got a 4 in the denominator down here and an 8.
So the least common denominator is 8.
So I'm just going to multiply the numerator by 8
and the denominator by 8.
And let's see, the 8's cancel here.
So we have a 5.
And of the denominator, we have 2 times 3 is 6.
I also get the same answer.
Both methods work.
This is just on a simple arithmetic problem that's the
same is true for algebra.