Algebra (all content)
- Reducing rational expressions to lowest terms
- Reducing rational expressions to lowest terms
- Simplifying rational expressions: common monomial factors
- Reduce rational expressions to lowest terms: Error analysis
- Simplifying rational expressions: common binomial factors
- Simplifying rational expressions: opposite common binomial factors
- Simplifying rational expressions (advanced)
- Reduce rational expressions to lowest terms
- Simplifying rational expressions: grouping
- Simplifying rational expressions: higher degree terms
- Simplifying rational expressions: two variables
- Simplify rational expressions (advanced)
- Simplifying rational expressions (old video)
Sal simplifies & states the domain of (5x²+20xy+20y²)/(x²-xy-6y²).
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- Would you also have to say that "y" cannot equal -1/2*x?(28 votes)
- Good observation!
The answer is yes. However, finding the domain of a multivariable function is not very common in Algebra II.(30 votes)
- what about x-3y isn't it supposed to not equal 3y also?(15 votes)
- Yes, that is true, and we can say that. The reason we don't emphasize the one that remains is that it is obvious that we have that constraint by looking at our answer, while the x + 2y factor has disappeared and is therefore more dangerous--it is easy to forget because is is no longer visible.(20 votes)
- At6:00, how come there are no restrictions on x-3y? How can we tell if there will be restrictions on a certain part or not?(4 votes)
- If a factor "disappears" from the denominator, then its restriction needs to be stated. Otherwise there'd be no way of knowing that it had ever existed in the original problem.
If a factor is still left in the denominator of the simplified fraction, then its restriction does not need to be stated because it is still visible and we all know that division by zero is not allowed.(6 votes)
- At1:55, why does Sal ignore the variable x? I understand that he thought of y as the coefficient to x, but why doesn't a+b have to equal -yx opposed to just -y?(2 votes)
- Remember, factoring is the reverse process to multiplication. When you multiply 2 binomials, the middle term created is a combination of all 4 terms from the original binomials.
Sal knows by looking at the lead term: x^2, that the 2 binomials start with "x" like this:
(x + _ ) (x + _ )
Those x's in create the "x" in the middle term. So, The rest of the middle term, the "-y" must come from the last terms in each binomial. This is why he is focused just on that portion of the middle term.
Hope this helps.(5 votes)
- I was wondering if the constraint that x cannot equal y equaling 0 would also have to be applied because if x and y both equal zero at the same time it's undefined? Also if that's true how would you state that?(4 votes)
- Well, I'm not certain about this, but I think that the answer is yes, they cannot both equal 0, and you would express that as x=y=0. Again, I'm not certain on either count, but it's my best guess.(1 vote)
- I keep messing up trying to find out when the equation is undefined. Is there a video that specifically addresses that ? thanks(2 votes)
- I don't know where the video is, but here's what you need to do.
The aim is to see what values of x make the denominator zero (since division by zero is never allowed)--- and then figure out which of those values needs to be listed separately along with your answer..
Factor the denominator and the numerator.
Look at any common factor that is completely divided out.
Find the value of x that makes it zero.
That value (or values) will have to be stated separately when you write the simplified version of the fraction.
So, in Sal's example, the ( x + 2y ) was divided out completely and did not appear in the simplified fraction. When x = -2y, that expression equals zero.
That's why he had to state that x cannot equal -2y.
But because the other factor ( x - 3 y ) in the denominator was still present in the reduced fraction, everyone could tell that x also couldn't equal 3y. It didn't need to be stated separately.(4 votes)
- What do I do if I can't factor the equation so the 1st term to 1? For example, 6m^2-5my-y^2.(2 votes)
- Use factoring by grouping.
AC = -6
You need 2 factors of -6 that add to -5 (ceofficient of middle term). Thef actors are -6 and +1
Use the factors to split the middle term:
Find GCF for each pair of terms
Then factor out the GCF of (m-y)
Hope this helps.(2 votes)
- At the very end, does it matter whether or not you distribute the 5 to (x+2y) or can you simply leave it in factored form?(1 vote)
- Unless the problem says otherwise, leaving it in factored form is usually the preferred choice --- less work to do and therefore less chance of mistake.(3 votes)
- what if x= -6, and y=3? wouldn't that also yield a zero in the denominator? You could also say that x=4, and y= -2. There seems to be an infinite number of possible x, y values that could make this rational expression undefined. Does saying x is not equal to -2y cover all of those solutions?(2 votes)
- [Voiceover] Let's see if we can simplify this expression and like always, pause the video and have a go at it. Now this one is interesting because it involves two variables, but it's really the same ideas that we've done when we factored things with one variable. So, for example, up here in the numerator, while I never liked having a non-one coefficient on the second degree term here, I mean, sometimes you have to, but it looks like every term here is divisible by five, so let's factor out of five first. So the numerator, I can rewrite as five times, five times, you factor out of five here, you get x squared. Factor out of five here, you get plus four. Actually, I'm gonna rewrite it as four y x and you'll see in a second why I'm doing that. Actually, I'll tell you why. I'll tell you why I'm doing that right now, why I'm writing the y there is that this way, it seems to, it seems to hit the pattern of how we're used to seeing quadratics. So let's see. So you have x squared plus four y x, you can view the four y's a coefficient on the first degree in x term, on the x term right over there, plus four y squared and it's gonna be over, over. Now, the denominator here, can we factor this out? Well, let's just think about it. Do we know two numbers, or I guess we would say, do we know two expressions that when you multiply, you get negative six y squared and then when you add them, you get negative x y. Sorry, when you add them, you get negative y. That's actually why I like to write and get like this. Actually, let me rewrite this. This is the same thing as negative y x. And so you can view the coefficient here as negative y. And so when you think of two numbers or two expressions, a times b, that is equal to negative six y squared and when I add them, a plus b, that is equal to negative y. And so you can imagine, both of them are going to be expressions, a and b are gonna be expressions that involve y. And so, let's see, if this was just a negative one and if this was just a negative six, what we would do, we would do negative three and positive two. Now let's see. If we did negative three y and positive two y, that indeed is going to be equal to negative six y squared and negative three y plus two y does indeed equal negative y. So that's our a and b right there. And it seems a little mysterious. How did Sal just all of a sudden get negative two y, or negative three y and positive two y? Let me write an analogous quadratic here that only has one variable. If I were to write x squared minus x minus six and I were to ask you to factor that out, you say, "Oh, okay." I have negative two. I have negative three times negative two is negative six and if I add them, well that's going to be negative one. So you'd say, "Well that's going to be "x minus three times x plus two." And so the only difference between this and that is instead of having just a negative one, you have a negative one y. Instead of having just a negative six here, you have a negative six y squared. And so you could just think of this instead of just negative three and positive two, negative three y and positive two y. Hopefully that makes sense and if it doesn't, I encourage you to kind of play around with this, multiply these out a little bit, get a little bit more familiar with this. But now that we know that it can factored like this, let's rewrite this. This is going to be x minus three y times x plus two y. And nothing seems to simplify out just yet, but it looks like what we have in magenta here can be simplified further. We're gonna do a very similar exercise to what we did just now. What two expressions if I multiply, I'm gonna get four y squared and if I add them, I get four y? It looks like two y would do the trick. So it seems like we can rewrite the numerator. This is going to be. So let me draw a little line here to make it clear that this is, this is going to be equal to five times x plus two y times, I could say this is x plus two y squared or I could just say x plus two y times x plus two y. Once again, two y times two y is four y squared. Two y plus two y is four y. And that's all going to be over, that is all going to be over x minus three y times x plus two y. And so now, I have a common factor, x plus two y in both the numerator and the denominator, so I can handle x plus two y divided by x plus two y, well that's just going to be one if we assume that x plus two y does not equal zero. And that's actually an important constraint because once we cancel this out, you lose that information. If you want this to be algebraically equivalent, we could say that x plus two y cannot be equal to zero or another way you could say it is that x cannot be equal to, cannot be equal to negative two y. I just subtracted two y from both sides there. And so what you're left with, and we can redistribute this five if we wanna write it out in expanded form. We could rewrite it as, the numerator would be five x. Let me write it over here. Five x plus 10 y. And the denominator is x minus three y. But once again, if we want it to be algebraically equivalent, we would have to say x cannot be equal to, x cannot be equal to negative two y. And now this is algebraically equivalent to what we had up here and you can argue that it's a little bit simpler.