Welcome to Georgia Highlands College Math 97 and Math 99 instructional videos.
In this video segment we’ll also be answering the question, how do you solve
rational equations. Notice there's a 2 there meaning this is the second
example in this video series.
So the first thing we want to do, I’ll just remind you of the steps, is to list
the restrictions because the denominator cannot be equal to zero.
Then we’re going to clear all the fractions out by multiplying each and every
term by the LCD.
Finally we’ll solve whatever resulting equation we have, and then we'll either
reject or check and accept any solutions that show up.
Let's take a look at our example.
All right, so we have X over, I'm sorry X plus 6 over X equals negative 5.
And the first thing I’m going to do is make this a little easier on the eyes and
come back and show the denominators of 1 under the X and 1 under the negative 5
just to have everything looking uniform here before we get started.
So the first step in this process is to note any restrictions on the variable
that we have. Well we have two denominators that are 1, and so 1 does not equal
zero; we don't even have to worry about either of those.
However X could be equal to zero in this case, so we need to note that X cannot
be equal to zero because if we plugged zero into that second term on the
left-hand side that would have zero in our denominator, creating an undefined
situation for us.
Okay, so let's go ahead and move on to the second step which is to clear these
fractions out. To do that we need to note what the LCD is and the LCD is quite
simple for this particular problem. I have all my denominators written over here
and the next step is just prime factorize each of these and that just has a
factor of 1X is simply X and 1 is simply 1. So our LCD is 1 time X which is just
X.
All right, so I'm just going to rewrite my original equation and I'm going to
leave room to come back and multiply every term by the LCD. And the LCD is X, so
I just want to multiply each term by X. Let me remind you, this is happening in
the numerator and we do this so that we can come back and say well, anything
divided by itself is one. Our two other denominators are one, so they'll just go
away and be understood to be there.
X times X leaves X squared. We don't have anything left to multiply the 6 with,
so it's just plus 6 equals negative 5X. So actually this has morphed into a
quadratic equation, and you should remember from earlier studies in the textbook
that to solve quadratic equations, if you can factor them and use the zero
product property, that's how you should solve it.
So to do that we have to first have our equation equal to zero, so I’m just
going to add 5X to both sides to give X squared plus 5X plus 6 equals 0. We
factor the left-hand side there which gives X plus 3 times X plus 2. If you
don't remember how I just factored that, you can go back and watch the video
tutorial on factoring trinomials whose leading coefficient is one, to help you
understand how I just did that. And now we apply the zero product principal by
setting each of the factors equal to 0 and solve. So I'll subtract 3 here and I
get X is negative 3 and I subtract 2 with this one and I get X is negative 2.
So I have two potential solutions here of negative 3 and negative 2. The first
thing I want to do is check them against my restricted, my restrictive values
for the variable.
Well, neither of those values are zero. So at this point, they’re legal and we
just need to check them in the original equation to make sure that they give a
true statement.
So the first one I’ll check is X is negative 3. Once again, you always go back
to the original equation, so negative 3 plus 6 over negative 3 should equal out
to negative 5 if we have a solution here. So negative 3 plus, well six divided
by negative 3 is negative 2, and sure enough, when I combine negative 3 and
negative 2 I get negative 5 on both sides giving a true statement.
Now I also need to check negative 2, so I’m just going to plug it into the
original equation just like I did with negative 3 and simplify down each side,
so I have negative 2 plus negative 3 is negative 5 and negative 5 equals
negative 5, check. So you can write this one of two ways, you can say X is equal
to negative 3 and negative 2, or you can use set notation and show that this can
either be negative 3 or negative 2, either of these are acceptable. You may want
to check with your instructor to see which they prefer you to answer in.
So I hope that this has been helpful to take it one step further and actually
end up solving a quadratic in this rational equation. So it's possible to have
one, two, ten, however many solutions show up, and in this particular one we had
two solutions. So just remember your steps. Make sure that you check your
restrictions. You don't want to call something a solution that’s going to create
zero in your denominator.
And then next you clear all the fractions out by multiplying by the LCD and then
solve the resulting equation.
I hope that this has been helpful. If you have any other questions about this
topic, please contact your Highlands instructor.
Thank you.
