Welcome to Georgia Highlands College Math 97 and Math 99 video tutorials.
In this video segment we’ll be answering the question, how do you solve rational
equations.
Well, the first step in this process goes back to an old video in that you've
got to list the restrictions on what you can plug into the equation. So you need
to go through and find out what will make any of the denominators zero, because
we know we don't zero in the denominator because dividing by zero is undefined.
So we want to identify those values that would make the denominator zero first
because if they show up as solutions, we want to discount them.
Secondly, you're going to clear fractions out of the equation and you're going
to do that by multiplying every term in the equation by the LCD or least common
denominator.
Then you’ll solve the resulting equation, now that may look different for
different types of rational equations. You might end up with a linear equation
or a quadratic equation, so you have to know how to solve all the types of
equations that we’ve talked about in previous chapters.
And then finally you'll either reject any of the solutions that show up that
Aren’t really solutions because of the restrictions placed on the input variable
or you're going to check and accept the other solutions that you are left with.
So let's take a look at an example.
All right, so we have a rational equation here, 5 over 2X equals 17 over 18
minus 3 over X, and the first thing that we need to do is figure out what we
absolutely cannot plug in for zero.
So if you look over to the side here. I have the situations set up where I've
taken each of my denominators and said these cannot equal zero. So if I solve
these equations here, I'll find out what X cannot be.
So to isolate X in this first I divide by 2 and zero divided by anything is just
zero, so in that case we have the restriction that X cannot be zero, so if we
run through all of our steps to solve this equation and we get zero out, we
simply cross that out because it cannot be zero because it will give us a zero
in the denominator if we plug it in.
Well our second denominator, 18, that's not zero and it never will be zero, so
we don't really have to worry about that denominator when it comes to
restrictions.
And in this last when one, 3X, well if we divide by 3, we also get here that X
cannot be equal to zero, so the same restriction shows up with that denominator
as well.
All right, so now that we've identified what X cannot be, let's go back and see
if we can figure out what X actually is.
So the first step in this process is to clear fractions and we do that by
multiplying every term. So we’ve got three terms here in this equation, one,
two, three; we’re going to multiply every term by the LCD.
Well, I’ve set off a little area to the bottom right to figure out what the LCD
is. So I just want to do the prime factorization for each of my denominators, 2
times X, 18 is 6 times 3 which is 2 times 3 times 3 and then 3X is 3 times X. So
my LCD, remember how we find the LCD, I list out the first factors of that first
expression there, and then I move to the second expression and add any factors
that haven't shown up yet. So I actually have two 3?s in that second one.
Since I already have a 2, I don't need to add it again.
Then I look down at 3X and I’ve already got a 3 and I’ve already got an X, so
I've got my LCD there, which is 2 times X times 3 times 3 or 18X. But I’m
actually going to leave it factored and you'll see why here in a minute. I find
it to be a little bit easier when it's in factored form.
All right, so the next step I’m going to do is just rewrite my problem and I’m
actually going to go ahead and factor all of the denominators within the
problem.
So I have 5 over 2 times X and that equals, I’m going to leave myself space and
you'll see here in a minute why am leaving this space, 17 over okay so 18 is 2
time 3 times 3 minus 1 over 3 times X.
Now I'm ready to multiply every term by the LCD, so I come back and multiply
each of these by a 2, an X, a 3, and a 3.
Now this is completely legal because this is an equation and as long as you do
the same thing to both sides of the equation, it remains an equation.
All right, so I’ve multiplied each of my terms by the LCD. Now here's the reason
that we did this. We want to get rid of these denominators. So look, anything,
and by the way, these were all living in the numerator. So let me just extend my
fraction bar there to show you that all of those factors are living in the
numerator right now which allows us to come back and say anything divided by
itself is one. Anything divided by itself is one. This is exactly why we had
to hit up every factor that showed up in the denominator because we wanted to be
able to cross out all of our denominators.
And now I'm just in a rewrite what we're left over with. So I have 3 times 3
times 5, and that's all over 1 now so I don't have a denominator left except for
1, which you know is understood to be there. And then I've got X times 17 and
I'm subtracting 1 times 2 times 3.
So no more fractions to worry about, now I just need to keep on with the
simplification process, 3 times 3 times 5 is 9 times 5 which is 45.
17 times X is just 17X. And 1 times 2 times 3 is six and we’re subtracting that
6 off. So I have a nice, neat, linear equation to solve.
So I begin unpacking everything that's hanging onto that X to isolate the
variable. So first I’m going to add 6 to both sides to get that negative 6 away
from the variable and this is 51 equals 17X and now I divide both sides by 17,
and I am left with X is 3.
So, whenever you get to the bottom and you have your potential solutions, and
I'm calling it potential on purpose, the first thing that you need to do is
check the solution you found against your restrictions. So the only restriction
that I have is zero for this particular problem. So right now 3 is legal, we
just need to make sure that it's actually the solution and makes our original
statement true. So I'm going to check it to make sure.
All right, and you always check in the original equations, so 2 times 3 equals
17 over 18 minus 1 over 3 times 3, all right. So just continuing to simplify I
have 5 over 6, let me just remind you at this point, whenever you're checking an
equation you simplify down either side of the equation. You don't move things
from side to side when you're checking, you just simplify down each side and see
if you equal out to the same thing on both sides. That’s the true statement
we’re looking for. So 17 over 18 minus 1 over 9, okay I need to get a common
denominator on the right-hand side, so I multiply by our clever form of 1, 2
over 2, and I have 5/6 equals 17/18 minus 2/18, oops, I’m running out of room,
let me come this way, sorry about that, so I have 5/6 equals, well 17 minus 2 is
15/18 and when I simplify 15/18, 3 goes into 15, 5 times and into 18, 6 times,
so I get 5/6 there.
So three is our solution to the equation, we’ve checked it, we made sure that it
wasn't a restrictive value, so we’re safe to say that 3 is the solution.
I hope that this has been helpful for you in beginning to understand how to
solve these rational equations. The big idea here is to not even deal with the
fractions, just clear them out by multiplying every term by the LCD. Now we’ll
have a couple more videos in this series solving more complex rational equations
each time just so that you can get a good feel for how to go about solving
these. I
If you have any other questions, please contact your Highlands instructor.
Thank you.
