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Current time:0:00Total duration:6:59

CCSS Math: HSA.APR.D

- [Voiceover] So I have a
rational expression here and my goal is to simplify
it, but while I simplify it, I wanna make the simplified expression be algebraically equivalent. So, if there are certain x values that would make this thing undefined, that I have to restrict
my simplified expression by those x values. So, you could pause this
video and take a go at it and then we'll do it together. All right. So let's just think real quick. What x values would make
this expression undefined? Well, it's undefined if
we try to divide by zero. So if x is zero, 14 times zero is zero, it's going to be undefined. And so, we could say this is, we could say x does not,
cannot be equal to zero. For any other x, we can
evaluate this expression. Now let's actually try to simplify it. So, when you look at the
numerator and the denominator, every term we see is divisible by x and every term is divisible by seven. So it looks like we can factor
seven x out of the numerator and seven x out of the denominator. So the numerator, we can rewrite as seven x times, we factor seven x out of 14 x squared, we're gonna be left with two x and then you factor
seven x out of seven x, you're gonna be left with a one. And one way to think about it is we did the distributive
property in reverse, if we were to do it again, seven x times two x is 14 x squared, seven x times one is seven x. All right. And now let's factor seven x out of the denominator. So, 14 x could be rewritten as seven x times two, times two, and remember, I wanna keep
this algebraically equivalent, so I wanna keep the constraint that x cannot be equal to zero. And so we divide the numerator and the denominator by seven x
or one way to think about it, you can divide seven x by
seven x and just get one and we are left with two x plus one over two. Now this was the original expression, x could take on any value, but if we want it to be
algebraically equivalent to our original expression, it has to have the same constraints on. So we're going to have to say x, x does not equal zero, and it's a very subtle but
really important thing. For example, if you defined a function by this right over here, the domain of the function
could not include zero. And so, if you simplified how you defined that function to this, if you want that function to be the same, it needs to have the same domain. It has to be defined for the same inputs. And so that's why we're putting the exact same constraints
for them to be equivalent. If you got rid of this constraint, these two would be equivalent everywhere except for x equal zero. This one would have been
defined for x equal zero, this one would not have, and so they wouldn't have
been algebraically equivalent. This makes them algebraically equivalent. And of course you can write
this in different ways. You can divide each of these
terms by two, if you like. So you could divide two x by two and get x and then divide one by two and get 1/2. Once again, we would wanna keep the x cannot be equal to zero. Let's do another one of these. So it's a slightly hairier expression, but let's do the same drill. See if we could simplify it. But as we simplify it, we
really wanna be conscientious of restricting the Zs here so that we get an algebraically
equivalent expression. So, let's think about
where this is undefined. We can think about where it's undefined by factoring the denominator here. Actually, let me just... So, this is going to be equal to, actually, let me just... Let me just do the first
step where I could say, "Well, what's a common factor "in the numerator and the denominator?" Every term here is divided by z squared and every term is also divided by 17, so it looks like 17 z
squared can be factored out. So, 17 z squared can be
factored out of the numerator and then we would be left with, we factor out of 17 z squared
out of 17 z to the third and we're gonna be left with just a z. 17 z squared, you factor
17 z squared out of that and you're gonna be left with just a one. And once again, you can
distribute the 17 z squared, multiply it times z, you
get 17 z to the third. 17 z squared times one is 17 z squared. All right. All of that
is going to be over, we wanna factor out of 17 z squared out of the denominator, 17 z squared times, and so 34 z to the third
divided by 17 z squared, 34 divided by 17 is two, and z to the third
divided by z squared is z and then we have minus
51 divided 17 is three and z squared divided by
z squared is just one. So we'll just leave it like that. And so over here, it
becomes a little bit clear of, well one, how we're gonna simplify it. We're just gonna divided 17
z squared by 17 z squared, but let's be careful to
restrict the domain here. So, we can tell if z is equal to zero, then this, 17 z squared
is gonna be equal to zero and we make the denominator equal to zero and we could see that even looking here. So we could say z cannot be equal to zero. Now what else? Z cannot be equal to whatever makes this two z
minus three equals zero. So let's think about what makes two z minus three equal to zero. Two z minus three is equal to zero. You can add three to both sides and you would get two z is equal to three. Divide both sides by two and you would get z is equal to 3/2. So z cannot be equal to zero and z cannot be equal to 3/2. So that's how we're gonna
restrict our domain. But now let's simplify. So, if we simplify it,
these two cancel out and we are going to be left with, we're gonna be left with z plus one over two z minus three and we wanna keep that constraint. z cannot be equal to zero. And actually, this second
constraint is redundant because we still have the
two z minus three here. If someone were to just
look at this expression, like, well, the denominator
can't be equal to zero and so, z cannot be equal to 3/2. So this is still, if we
just left it the way it is, we don't even have to rewrite this, that would be redundant since
looking at this expression is clearly not defined for z equals 3/2. So, there you go. And so I'm gonna ask you for what values does this
expression not defined, well then you would also include that. Actually, let me just write it. It doesn't hurt to be redundant. z does not equal 3/2. But this constraint right
here is really important because it's not obvious by
looking at this expression. This expression by itself would be defined for z equals zero, but if we want it to be
algebraically equivalent to this one and that one, it has to be constrained in the same way.