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Hello, and welcome to Bay College's Intermediate Algebra. online lectures.
My name's Jim Helmer.
And today we're going to be talking about Section 5.4,
which is solving equations with rational expressions or
rational terms.
And what we're going to use here are some of the things we
covered in the last section, in Section
5-3 and Section 5-2.
We're going to determine what the LCD is and utilize this
when working with equations.
We can solve equations, but we can only simplify expressions,
which was the last few sections.
So now we're going to look at how we can apply LCDs to solve
in equations.
And the nice about equations is we can always check our
work and make sure our answer makes sense and
makes a true statement.
So the first thing I'm going to do here is I'm going to
determine what is the LCD of all these terms.
Well, we can do that by factoring, but since they're
just numbers, maybe we can look at it and say, well, 12
is divisible by all of these numbers so our LCD here is 12.
Now, what I'm going to do with this information, now that I
know that the LCD is 12, I'm going to multiply every single
term by this LCD.
So essentially, I'm going to set it up as a distribution.
I'm distributing it to the entire equation, which means I
have to distribute it to each term in the equation.
Now, there's two ways to do it, I can go 12 times 5 to
give me 60, divided by 12 is 5.
Or I can do the division first, 12/12 is 1, it
simplifies to 1.
1 times 5x is just 5x.
Now I'm going to distribute it to the next term.
And I'm going to do that division first.
12 divided by 3 is 4.
4 times negative 2 is negative 8.
12 times this term, the 12 simplify to 1, negative 7x.
12 times the last term, 12 divided by 6 is 2.
2 times this quantity is negative 10.
And now, if you assess the problem, there are no more
fractions and we have an equivalent equation, which
means we can go ahead and solve this.
I'm going to add 7x to both sides, and
add 8 to both sides.
I get 12x equals negative 2.
Divide both sides by 12 and simplify.
Negative 2/12 is negative 1/6.
Now, it's very important that we check our work, and I'm
going to do that here, but I just want to point out
sometimes checking our work can be way more tedious than
finding the answer.
But it is a good policy to always check your work.
Know that you're right, know that your work makes sense.
So I'm going to switch it up a little bit and go with another
color because I have that opportunity on
a whiteboard here.
What we're going to do is we're going to take 5/12 times
the x we found, negative 1/6, minus 2/3 equals negative 7/12
times our x, negative 1/6, minus 5/6.
Now, if we go ahead and simplify this, we can do
multiplication before any addition or subtraction.
I get negative 5/72 minus 2/3 equals positive 7/72--
a negative times a negative--
minus 5/6.
And even when we're dealing with equations that have
fractions in them, we can still use this policy.
I can multiply by the LCD instead of changing these to
have that LCD.
This one I'd have to multiply by 24/24 to get that
denominator to be 72, and this one I'd have to multiply by
12/12 to get that 72.
Or I can just multiply everything by 72.
That is my LCD here.
And if I do that, I get negative 5 minus negative 2
time 24 is 48.
Multiply this side by 72, I get 7, and this by 72, I get
negative 60.
Now, is this a true statement?
Negative 5 minus 48 is negative 53.
7 minus 60 is negative 53.
Negative 53 is negative 53.
That's a true statement, it is what it is.
I know that my solution here, x equals
negative 1/6, is correct.
So like I said, sometimes checking your work can be more
tedious than actually finding the solution,
but it's good policy.
And we'll see in future examples at times it is very
necessary to check your work.
When we talk about domain, we can never divide by 0, so we
got to watch out for that, especially with rational
expressions.
All right, let's move on to the next example here.
This one here is we see a 6 and a denominator and a 4 and
a denominator.
First thing we want to do is just assess any
problem before we start.
We look at it and say, oh, it's just a fraction equal to
a fraction.
And many of us might say, hey, I know that as a proportion,
something we'll talk about in the next section of--
or the next video of this section.
But we're going to use the policy of what we used here,
we're going to multiply by the LCD.
And I look at this and 6 and 4, well, they're LCD because
this has the factors of 2 and 3 and this has 2 and 2, so I
need the factors 2 twice and 3 once and 2 times 2 times 3 is
12, so my LCD here is 12.
Just by coincidence, it's the same as the last one.
So I'm going to multiply all the terms by 12.
Now, 12 divided by 6 is 2.
2 times this quantity.
Well, 2 times 4 is 8 times that quantity.
On this side, if I multiply this by 12, here it is, 12
times this quantity, 12 over 4 is 3.
3 times this would be 3 times 2 times this, it would be 6
times the quantity x minus 5.
Now you notice, by multiplying by the LCD and just doing a
little bit of simplification, I now have an equation with no
fractions in it.
So we can eliminate these grouping symbols by just
distribution.
8x minus 24 equals 6x minus 30.
And now I can just get my x's on one side, my number's on
the other, and solve.
So I'm going to subtract 6x from both sides and add 24,
and I get 2x equals negative 6, divide by 2, x equals
negative 3.
Let's go back and check our work.
We'll do it the quick way instead of
showing all the work.
And if you want to work this out, you can.
But I'm just going to, for time's sake-- negative 3 minus
3 is negative 6.
Negative 6 over 6 we can reduce to negative 1 times 4.
So we get negative 4 on the left side when we plug in that
value, negative 3.
Here, if I put in negative 3, negative 3 minus
5 is negative 8.
Negative 8 over our denominator of
4 is negative 2.
Negative 2 times this 2 is, also, negative 4.
This is a true statement, it doesn't get
any truer than this.
Negative 4 is negative 4, that is a true statement.
So let's review the tools that we used to solve these
equations containing rational terms or rational expressions.
Here we have--
the first thing we want to do is clear those
fractions using the LCD.
Determine what your LCD is, multiply all the terms by it,
and now we have something without any fractions.
We can remove any grouping symbols whether they be
braces, or brackets, or parentheses by distribution
using the distributive property to eliminate those.
We can combine any terms and solve for x.
What we'll see in one example is sometimes our x's are not a
single value, sometimes they're at a higher power such
as x squared.
Well, hopefully we remember when we see an x squared term,
we want to set that equation equal to zero, get all your
terms on one side, and maybe try to solve it by factoring.
Hopefully factoring will work and we won't have
to use other methods.
That's something we will explore more in this class,
you might be somewhat familiar with that.
And then of course, check your solution.
In some of the examples over here you're going to see when
we check our solutions one of them actually doesn't work.
So it's very important to check your solutions.
All right, let's look at some more examples.
And we'll utilize this because we know that math is only
learned through repetition.
And doing it yourself, of course, so hopefully you
worked on the homework.
Let's do this one first, we're going to identify the LCD.
Well, we have the factors of x and 6 and 6 and x, so our LCD,
in this case, to include all those factors would be 6x.
So I'm going to multiply every term by 6x.
Well, here if I multiply, the x is reduced to 1 and I have 6
times 5, which is 30.
6x times 1/6, well, just the 6 is canceled and I have 6 times
1, which is-- or excuse me, x times 1 because 6 is reduced
to 1, which just gives me x.
And then I have 6x times this last term, well, both the 6
and the x cancel or reduce to 1, leaving me with 23.
Now this, hopefully, you recognize and say, hey, that's
a very simple equation.
I can just subtract 30 from both sides and I'm done.
So if I subtract 30, I get x equals negative 7.
And I can plug in negative 7 and work it
out, but it does work.
I'm going to have you trust me on this one,
just for time's sake.
All right, let's look at this one.
It looks a little bit different.
We notice, hey, I have an x in the denominator as well as in
the numerator.
This is one I'm going to be sure to check my answer
because I can never divide by 0.
So maybe you can identify what value needs to be excluded and
hope that doesn't show up in your answers
when you're done working.
But it's always important to check your work, so we will
check it on this one.
Here, we identify the LCD.
Well, there's only two terms and they have the same
denominator.
So this is my LCD, x minus 5 is the only factor in any
denominator.
One thing you could do, you could just combine this to a
single fraction, but you're going to end up multiplying by
x minus 5 either way.
So let's do it right away.
Let's just get into the habit of doing this and if we're
consistent, we'll eliminate the fractions.
x minus 5 times this first term-- the x minus 5 is
reduced to 1, leaving me with 3x.
Multiply this by the second term, same
thing's going to happen.
The x minus 5 will reduce to 1, leaving me
with a negative 2.
Here, x minus 5 times 2, well, I'm going to actually write
this out because 2 times x minus 5 requires a little bit
more work than just reducing or canceling, we have to
distribute.
So let's distribute it.
3x minus 2 equals 2x minus 10 when we distribute that 2.
And now I can subtract 2x's from both sides and add 2.
And I have x equals negative 8.
Now it's time to check our work.
When I plug negative 8 in, I can see, well, negative 8
minus 5 is negative 13.
Good thing this isn't 0, if that value made
this 0, we're done.
The whole problem would be no solution, but that's
not the case here.
So we get 3 times negative 8 is negative 24.
Negative 8 minus 5 is negative 13.
Minus 2 over negative 8 minus 5 is negative 13.
And 2 is 2.
Let's see if this is a true statement.
Negative 24 over negative 13 minus 2 over negative 13,
well, that's the same as negative 26 over negative 13.
A negative over a negative is positive.
26 over 13 is 2, so this quantity is equal to 2.
This is a true statement, I know that this value, x equals
negative 8, works.
All right, let's move on to here.
This is x over x minus 6 equals x plus 6 over nine.
And maybe you recognize that as a proportion and you want
to cross-multiply, that is something we'll look at in our
next video.
But if you don't do it that way, you can
still use this method--
what is my LCD?
I have the factor of x minus 6, I have the factor of 9, and
those are my two factors.
So if I combine them and multiply both sides by my LCD
of nine and x minus 6, well, when I multiply the first
term, the whole factor of x minus 6 can reduce to
1 and I have 9x.
When I multiply this term by this LCD, the 9's will reduce
to one, and I'm left with x minus 6 times x plus 6.
Now that's something that maybe we can use FOIL or maybe
we recognize it as a special product.
Or multiplying the difference of terms here.
And if we multiply that, we get x squared minus 36.
So I just FOILed this times that.
Now, here we have an x squared term.
So when we have an x squared term, well, we want to set the
equation equal to 0, we want to maybe try
and factor it first.
So I'm going to subtract 9x from both sides.
So I'm going to get x squared minus that 9x minus 36 equals
0, and then we're going to try and factor it.
Well, what are the factors of negative 36 that have a
difference of 9?
Well, if you don't see it right away, you could say,
well, I know the factors of negative 36 are 6 and 6, but
they don't have a difference of 9.
9 and 4 are factors of 36, they don't have a
difference of 9.
Well, what about 12 and 3?
12 and 3 are two factors of 36 that have a difference of 9.
So we can go ahead and factor that to x minus 12 x
plus 3 equals 0.
And if you don't recall, when it comes to factoring, this
sign here--
because we're summing the negative 12x and the positive
3x to give me negative 9x--
this sign determines the sign of a larger value.
So now we can just use the zero value theorem.
Essentially what that says is 0 times anything is 0.
Well, what value of x would make this factor 0?
We can see 12.
So I'm going to write x equals 12 here.
And what would make this 0?
Take negative 3.
And then I would test those.
And both of them do work, just for time's sake.
But I'm going to just check, 12 minus 6 is not 0.
That's good news.
Negative 3 minus 6 is not 0, so that's good news.
No 0's in the denominator.
So that's one way of-- if you're crunched for time maybe
on a quiz or a test, that's what you want to at least try.
Make sure it's not a value that's
going to make it undefined.
And if we did plug it in, like I said, sometimes checking
your work can be more tedious than finding the answer, those
solutions do work.
Now, I'm just going to clean this up a little bit because I
crowded my other problem here.
And if you need to see that work, well, one nice thing
about the technological age is you can move that little
cursor bar at the bottom of the video-- maybe it's right
about here, right now.
You can back that up and you can check that out.
So if you ever miss anything, don't
hesitate to move it back.
Or if you say, hey, how did he go from Step A to Step B, I
didn't catch that, you can always back up the video and
take a look at it.
All right, now this one here.
This one's going to take a little bit more work because
if we recall in the last videos when we were
simplifying rational expressions, we always
factored and reduced.
Well, in the other examples, there was no need to really
factor our denominators, but here, we have lots of x's,
lots of terms.
We should check to see if anything factors.
And if we look at this one here, maybe we recognize it as
the difference of two squares.
9x squared minus 1 is a difference of squares, of 3x
and 1, so those are my squares.
So I'm just going to factor it here.
3x minus 1, 3x plus 1.
Big difference of squares.
And if I FOILed 3x minus 1 times 3x plus 1, I'd get 9x
squared minus 1, so I did factor it correctly.
And now if we look at what are our factors in the
denominator, let's find that LCD.
I have a factor of 3x plus 1 here and here.
And I have a factor of 3x minus 1 here and here.
So the unique factors that I have are 3x
plus 1, 3x minus 1.
These are my LCDs--
or the factors of my LCDs.
So now I'm going to multiply every term by this LCD.
Now if I take this and I multiply it by the first term,
the 3x plus 1, as a factor, reduces to 1 and I have 1
times the remaining factor, 3x minus 1.
So I'm just going to write it here, 3x minus 1.
If I multiply this times this next term, well, this value is
the same as the LCD.
So both factors will reduce to 1.
1 times 2 is 2.
And then the LCD times the last term, the x minus 1 terms
will reduce to 1, and I'll have 3x plus 1 times 1.
Now, let's be careful because this is subtraction, so we
don't want to lose that sign.
It's negative 1 times 3x plus 1.
So I get negative 3x minus 1.
I distributed a negative 1 to that term.
So if you missed that, go back and look and try to see that.
Work it out yourself so you can see these examples.
I try to cut them out due to time's sake.
Now, I'm going to get my x's on one side.
Add 3x to both sides so I get 6x.
Here I can combine like terms, 2 minus 1, and then I'm going
to add one more to get my constants on the right side.
So I'm going to get 6x equals 2.
If I divide both sides by 6 and reduce, 2/6 is 1/3.
And then I can go ahead and plug this in.
This is so crucial that we check this.
Because if I plug in a 1/3 here, 3 times 1/3 is 1.
1 plus 1 is 2.
So here, when I'm checking it, I get 1/2.
When I plug it into here, 1/3 squared is 1/9.
1/9 times 9 is 1, 1 minus 1 is 0.
2/0.
Well, let's write that down for a moment, and hopefully we
say, whoa, hold up on the car wash.
When we divide by 0, that's undefined.
If any portion of an equation is undefined, the whole
equation is undefined.
We're stopping right there, we're saying this is not a
valid solution.
But let's continue anyways, maybe we don't see it there,
maybe we make some math error or sign error or something.
If I put in 1/3 here, 3 times 1/3 is 1, 1 minus 1 is 0.
2/0 minus 1/0 is not going to give me 1/2 no matter how I
look at it.
So this is undefined.
OK?
This is not a true statement.
That means the solution we found doesn't work.
Well, what does that mean about the whole problem?
It means there is no solution.
We can write null set or we can write out the term "no
solution." All right?
So nothing will solve this, it's not going to be true.
All right, let's look at one last example.
Here we have two different variables.
But all we're going to be asked to do for a problem like
this is solve for one of the variables.
Let's solve for x.
So what do I do to get just the single variable by itself?
Well, let's use these properties that we just used.
We're going to find the LCD.
If we look, we have three fractions
in these three terms.
And our LCD, we have a factor of 2, x, and y.
So my LCD--
I'll write it right here-- the LCD is 2, x, and y, all three
of these factors.
I'm going to use that to eliminate the fractions.
Distribute 2, x, y to each of these terms.
Well, if I multiply it by the first term, the 2's are going
to reduce to 1 and I get xy times 1.
The next term, 2xy times 1/x, the x term reduces to 1 and I
get 2y times a negative 1.
So I get xy minus 2y equals 2xy times the last term, the
y's reduce to 1 and I have 2x times 1, which is just 2x.
Now, our goal is to solve for x.
We eliminated the fractions using the method of the LCD,
but now I got to get my x's alone.
So what I have to do is I have to move my x's, group them on
one side of the table--
or equation sign.
Here we have to subtract 2x from both sides and add 2y.
So essentially, I'm just moving these across the equals
sign and changing their signs.
But now what can I do?
I have x in two separate terms.
Well, I can always factor.
Try to factor it.
All semester long in these videos or if you have me in
class, you're going to hear me say, when in
doubt, factor it out.
Intermediate algebra is all about factoring.
If you're weak in your factoring skills, definitely
work, work, work to improve your factoring skills.
All right.
And in the text, you can go back to Chapter Four and that
deals with factoring and rules of exponents.
Very important skills for this class.
All right, so let's factor out an x.
And that leaves me with y minus 2 equals 2y.
Well now, once we factored out that x, now that we've
recognized that step, we can say, you know what?
To get x by itself, all I have to do is divide by x minus 2.
So if I divide both sides by x minus 2, I'm going to get x
equals 2y over y minus 2.
Now this doesn't look much improved from what we started
with, but we did solve for x, and that was the instructions
given, let's solve for x.
This has been Video One of Section 5.4.
We'll continue the video with proportions in
the very next video.
Thank you for watching.