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>> Julie Harland:
This is your math gal,
Julie Harland.
Please visit my website
at yourmathgal dot com
where all of my videos are
organized by topic.
This is an introduction
to solving rational
inequalities
where we only have polynomials
in the numerator
and denominator.
Make sure you already know how
to solve non-linear
polynomial inequalities.
Alright. So, first of all,
examples of rational
inequalities are shown here.
Basically,
rational means you're going
to have a fraction,
and you only have polynomials
in the numerator
and denominator.
Remember 2 it can be written
as 2 over 1.
So that's really still just a
fraction, and a constant could
also be a polynomial.
You don't have
to have an x value.
Now solving these is very
similar to solving the
non-linear polynomial
inequalities except these
fractions we have
to pay a little bit more
attention to.
For instance,
you can never have the
denominator equal to 0.
So if you look
at the denominator,
which is x minus 3,
x minus 3 can never be 0
because then
that means this fraction would
be undefined.
So, in other words,
what you could determine off
the bat are the values
that x cannot be.
So that's one thing that's
going to be a little bit
different here.
OK. Here are the steps.
Pretty similar
to solving the other ones
on the
polynomial inequalities.
Alright. First, we're going
to write the inequality with 0
on one side,
and then we're going to write
as one fraction
on the other side
and factor the numerator
and denominator, OK.
Next, we're going
to set each factor both
in the numerator
and denominator equal to 0
and solve the equation
to find the critical points,
and we're going to mark those
on a number line.
Now, here's the difference.
We're going to put a circle
on the number line
for any critical points
from the denominator
because we know those are not
solutions because it would
make the fraction undefined.
If we have,
for the critical points
from the numerator,
we'll put a circle
if we've got greater than
or just less than,
but if it's less than
or greater than or equal to,
right, less than or equal
or greater than or equal,
those will be solutions,
and we'll put a dot,
and then we'll go ahead
and test the intervals
like we've done before.
So here's the first problem.
X plus 2 over x minus 3 is
less than or equal to 0.
So our first step was
to make sure we have 0
on one side, and we have that.
Second thing is to make sure
that we have a single fraction
on the left,
on the other side,
and we have that.
Third is to make sure we have
the numerator
and denominator factored,
and we have that as well.
Alright. So those are the
first couple steps.
Now we want to set all
of the factors equal to 0.
So x plus 2 equals 0,
or x minus 3 equals 0,
gives us our critical points.
Alright. So our critical
points are negative 2 and 3,
and so we're going
to mark those
on a number line.
[ Background Sounds ]
Now we have to be careful.
The x minus 3 came
from the denominator.
So I sometimes
like to write it this way.
I know it can't be 3.
So I know I must not include 3
in my solution.
X is going to equal negative 2
because I have the less than
or equal symbols.
So look at this.
If you put a negative 2
into this original inequality,
you would have 0
over negative 5, OK.
So, in other words,
plug in negative 2.
You'd have negative 2 plus 2
over negative 2 minus 3 less
than or equal to 0.
Is that true?
Well, 0 over negative 5 is
less than or equal to 0
because you have 0 is equal
to 0.
So that's fine.
That means negative 2 is
a solution.
So because it had this equal
symbol, that critical point is
a solution.
That's similar to what we did
with the
polynomial inequalities.
Alright. Now
if this was less than,
strictly less than,
then both of these would have
circles, right,
but in all cases,
the critical point that came
from the denominator must have
a circle on it.
And now we do our test points
just like we did
with the
polynomial inequalities.
So let's pick a number less
than negative 2.
So how about we try
negative 10?
Alright. So we're going
to take the fraction and plug
in negative 10.
So in the numerator,
if I plug in negative 10,
negative 10 plus 2 is a
negative number.
So I have a negative number,
and in the denominator,
if I plug in negative 2,
I also have a negative number
because negative 2 minus,
I'm sorry,
if I plug in negative 10.
Said that backwards.
Negative 10 minus 3 is also a
negative number,
and what's a negative divided
by a negative?
It's a positive number.
Alright. Now let's test a
point between negative 2
and 3.
Zeros in there.
I always do 0
if it's an interval.
So try 0 for x.
So in the numerator,
if I put in 0 for x,
I get a positive number
at the top, right.
Zero plus 2 is positive,
and in the denominator,
0 minus 3 is negative,
and a positive number divided
by a negative number
is negative.
And then we'll pick something
in the other interval bigger
than 3.
So how about we try 5.
Alright. So in the numerator,
5 plus 2, that's positive,
and in the denominator,
5 minus 3 is also positive,
and a positive number divided
by a positive number
is positive.
Now, what do we want
for our solution?
So we want where it's going
to be negative, less than
or equal to 0.
So we want this part
in between negative 2 and 3.
So we put that in here.
And if we're going
to write our solution
in interval notation.
Alright. So we've got all the
numbers starting
at negative 2.
So it includes negative 2,
and it goes up to 3.
And that's our solution.
This is your math gal,
Julie Harland.
Please visit my website
at yourmathgal dot com
where all of my videos are
organized by topic.