>> In this video, we solve
this rational equation.
And when you eliminate
the denominators,
we get a quadratic
equation and solve it.
All right, we're going
to solve this equation,
and it's a rational equation
because it contains fractions.
So there's two methods for
getting rid of the fractions,
clearing the denominators,
you can multiply both sides
by the common denominator or
you can write all the terms
so that they all have the
same denominator and then
that means the numerators
will be equal.
That's how I did it on
problem 1 of this type.
And that's going to
take a lot less space
so that's how I'm
going to do it.
So I know I have a factor of 2x
minus 5 and a factor of x plus 3
and what about this 2x
square plus x minus 15?
Well, if you multiply
these two factors together,
you actually get 2x
squared plus x minus 15.
So, this is really
the same thing
as 2x minus 5 times x plus 3.
So the least common denominator
is 2x minus 5 times x plus 3.
So, let's see if we
could write everything
with the same denominator.
So this already has the
least common denominator,
so that's cool.
We're all right here.
All right.
Next term, on the right side
of the equal sign I've got 5
over 2x minus 5, but I don't
have my factor of x plus 3.
So I need to multiply the
top and bottom, by x plus 3,
to get the same denominator.
And then over here we have x
over x plus 3 and let's see.
What do I need to
multiply this by?
The 2x minus 5.
All right.
Now, all the denominators
are the same,
which means the numerators
are equal.
If you don't believe that,
the real reason it's true is
because if you multiply
both sides
by the least common denominator,
it would simply cancel
out every denominator
and you'd be left again
with the numerators.
So they end up being equal.
So, what do I have here?
In the numerator I've got an 11,
all right I've got
a 5 times x plus 3
and then I have a minus
x times 2x minus 5.
You've got to be careful
if there is something
like an x plus 2 up here, with
a minus sign in front of it.
You might want to distribute
your minus sign first.
Okay, so, we've eliminated
the fractions.
So the next step is to do the
distributive property here.
So I have 11 equals
and we're going
to distribute 5x plus 15
minus 2x squared and remember,
this is a minus x times a minus
5, that will be a plus 5x.
Now just take a look
at this for a minute.
Notice we now have an x squared,
we've got some x terms
and some constants.
We have a quadratic equation.
So in a quadratic equation,
we want to set the
equation equal to zero.
Let's just simplify one more
step here by adding the 5x
and the 5x, so we've got
negative 2x squared plus 10x
plus 15.
And I want to set
it equal to zero,
and I have a negative 2x
squared on the right side
so I always think it's easier
when the a term is positive,
although it's okay to just
subtract 11 from both sides.
In fact, just to be different,
I'm going to do it that way.
Let's subtract 11
from both sides.
We have negative 2x
squared plus 10x plus 4.
If you would have put everything
on the left hand side,
you would have 2x
squared minus 10x minus 4.
All your A's, B's and C's would
be opposite and when you put it
in the quadratic
formula, you know what?
You're still going to get the
same values for your solutions.
Okay. Now we can go ahead
and use the quadratic
formula right here,
but do you notice how
there's a common factor of 2
or of negative 2 here?
I'm going to go ahead
and divide both sides
by negative 2, so it's easier.
So if I divide the left by
negative 2 and every single term
over here, then I've divided
both sides of the equation
by negative 2, I'm going to
have easier numbers to plug
in the quadratic formula.
So I have zero equals, see,
I made a positive after all.
X squared minus 5x and minus 2.
Now at this point, you would
see if you could factor this,
because it's a quadratic.
But the only factors
of two are 2 and 1
and there's no way
you're going to get
that negative 5 for
your middle term.
So, we're going to use
the quadratic formula.
A is 1, B is negative
5, C is negative 2.
By the way, if you wouldn't
have divided by negative 2,
you would have had A is
negative 2, B is 10 and C is 4
and you can verify that if you
use those for your values of A,
B and C, when you're all
done, in the end, you're going
to get the same answer that
I get when I use these values
of A, B and C. It's
just that I'm going
to have easier arithmetic.
So let's do, B squared
minus 4AC.
I'd like to do that first.
So B squared will be
25 minus 4 times AC.
1 times negative
2 is negative 2.
So I've got 25 plus 8 or 33.
So that's what's going
to go underneath the
square root symbol,
in the quadratic formula.
So you're ready for
the quadratic formula?
X equals negative
B. B is negative 5
so negative B is going to
be the opposite of that
or positive 5 plus or
minus the square root of...
there's your 33.
That's what goes underneath the
square root, 33, all over 2A.
All right?
A is 1 so 2A is 2.
And so those are my solutions.
I have two of them.
5 plus the square
root of 33 over 2
and 5 minus the square
root of 33 over 2.
And there we go.
Let's go ahead and eliminate the
fractions the traditional way
by multiplying both
sides of the equation
by the least common denominator
of 2x minus 5 times x plus 3.
So I take my, you're going
to see how it just is hard
to get all on one page.
I have 11 over and
I'm going to write
that as 2x minus
5 times x plus 3.
And you have to multiply that
by 2x minus 5 times x plus 3.
Okay? And then that's
going to be equal
and I've got the next term.
5 over... 2x minus 5 and
you're going to multiply
that by 2x minus 5 times x
plus 3 minus the last term
on the right hand side and
that also gets multiplied
by 2x minus 5 times x plus 3.
Okay. So in other words you
could see it takes two lines
to get all this in here.
But now let's do our cancelling.
This 2x minus 5 is cancelled,
the x plus 3's cancel here,
the 2x minus 5's cancel here,
the x plus 3's cancel here,
so what do I end up with?
I've got the 11, it's
hard, be careful.
11 equals then I have 5 times x
plus 3 minus x times 2x minus 5.
So I have 11 equals 5x plus
15 minus 2x squared plus 5x
and that's exactly
what we got before
when we were eliminating
the fractions
by just making all the
denominators the same.
So this is the traditional
way of doing it, but you sort
of have to have a big
piece of paper in order
to write it all down,
but it should lead you
to the same basic equation.
At this point, you would,
you know, add like terms,
set the equation equal to zero.
You could divide again, both
sides by negative 2 like I did
in the previous example.
But the point is, that is the
same equation we had right here.
All right?
So just alternate ways to
eliminate the fractions,
and that only works when
you have an equation.
When you're just
adding fractions,
and you've got a denominator,
you can't just make
it disappear,
you can't just eliminate it.
