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# Simplifying rational expressions: two variables

Sal simplifies & states the domain of (5x²+20xy+20y²)/(x²-xy-6y²).

## Want to join the conversation?

• Would you also have to say that "y" cannot equal -1/2*x?
• Good observation!
The answer is yes. However, finding the domain of a multivariable function is not very common in Algebra II.
• what about x-3y isn't it supposed to not equal 3y also?
• 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.
• At , how come there are no restrictions on x-3y? How can we tell if there will be restrictions on a certain part or not?
• 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.
• At , 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?
• 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.
• 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?
• 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
• 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.
• What do I do if I can't factor the equation so the 1st term to 1? For example, 6m^2-5my-y^2.
• Use factoring by grouping.
A=6; C=-1
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:
6m^2-6my+my-y^2
Find GCF for each pair of terms
6m(m-y)+y(m-y)
Then factor out the GCF of (m-y)
(m-y)(6m+y)

Hope this helps.