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Multiplying rational expressions: multiple variables

Sal multiplies and simplifies (3x²y)/(2ab) X (14a²b)/(18xy²). Created by Sal Khan and Monterey Institute for Technology and Education.

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  • marcimus pink style avatar for user princessshree85
    Is there a video that covers domains... I think I may have missed something
    (30 votes)
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    • old spice man blue style avatar for user ahenzinger
      Domains are constraints you have to set up, so that the denominator does not equal 0. You have to do this, because if the denominator equal 0, the number will be undefined. So for example if you have (a+34) as a denominator in your expression, the domain is all real numbers except for a= -34. A can be all possible real numbers, except for (-34), because (-34+34)=0, and so the number would be undefined. I hope this helps you ...
      (51 votes)
  • blobby green style avatar for user Michael Bils
    At , when he's canceling out the y factors, why doesn't this produce y^-1 ?

    I thought y^1 / y^2 = y^-1
    (7 votes)
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  • male robot johnny style avatar for user Adi
    I still don't get what domain is - can anybody tell me?
    (6 votes)
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  • male robot donald style avatar for user Sean Bailey
    Instead of saying a, b, x, or y =/= 0, could you state it as a*b*x*y =/= 0?
    (6 votes)
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    • spunky sam blue style avatar for user Niema Moshiri
      You definitely could. If a, b, x, or y WAS equal to 0, then a*b*x*y would equal 0, and that is the only way that we could end up with 0 on the right hand side. Therefore, yes, we can assume that, if we can ONLY end up with a result of 0 if one of the 4 variables is 0, that if none of them are 0, the result will not be 0.

      In short, yes you could.
      (8 votes)
  • leaf grey style avatar for user ARodMCMXICIX
    In the skill after this there have been a few errors I think or I am making a mistake. For example: there was an expression -8n+8 and the hints said to factor out a -4 which should give you
    -4(2n-2) while the hints said it gives you -4(2n-1). Am I the only one with issues here or am I missing something? Thanks for any help.
    Note it wasn't the skill after this video but rather this skill "Multiply and divide rational expressions with polynomial numerators and denominators"
    (5 votes)
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  • blobby green style avatar for user Grace Gall
    My teacher has taught us to factor the numerator and denominator, and then cross out matches. Although the way you are teaching it seems ten times easier, would I still get the same solution?
    (2 votes)
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    • hopper cool style avatar for user Chuck Towle
      Yes, both reach the same answer.
      In the video Sal took the 14/2 and divided both by 2.
      You could also factor the 14 in 2*7 and cancel the 2s.
      Both are ways of doing the same thing.

      Using your teachers method you would have had a 7 left in the numerator and a 2*3 left in the denominator which would have turned back into a 6. So, in this case, your teachers method added a couple extra steps.

      But, the method your teacher taught is a good way to always get the correct answer everytime because you are sure to find all the common factors.

      If you had a 35/21 you might carelessly miss the fact that you can divide both by 7.
      But if you use your teachers method and factor everthing, you would see they both had a factor of 7 which could be cancelled.

      So, use which ever one clicks best for you, but be careful not to miss something when you don't factor everything down to the prime mumbers.
      (6 votes)
  • blobby green style avatar for user Josh Treitz
    dont you have to common factor the equation?
    (5 votes)
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  • piceratops sapling style avatar for user Jennah
    im still confused...i searched this and i didnt understand wat number or key word they were equaling ,im having bad grades for math because of this and i need alot of help...and wat is tht 2 on the top of a and x? im seriously confused!!please help!! can u help!
    (0 votes)
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  • blobby green style avatar for user faith32363
    So you cant divide by zero right?
    (2 votes)
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  • blobby green style avatar for user ausslatt
    When stating the domain of a multi-variable rational expression. Sal doesnt expressly say there are any specific rules for finding it. He just says it can only be that a,b,x,y are not equal to zero. Based on this, can I assume for all multi-variable rational expressions. The domain will simply be the variables not equal to zero? Or, will there sometimes be a different case? Ex. "a,b,x,y" can not equal zero and "a" cannot be equal to -3
    (2 votes)
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    • female robot grace style avatar for user MsPhillips
      The rule for determining domain is that you can't divide by 0. Also, the domain must be determined using the original expression; that's why Sal said a,b,x,y can't equal 0 even though there's only a y in the denominator in the simplified expression. So if your original denominator was x-2 then your domain would exclude any values of x that make x-2=0 (domain is x cannot equal 2). This would be the domain even if the denominator eventually simplified to 1.
      (1 vote)

Video transcript

Multiply and express as a simplified rational. State the domain. We'll start with the domain. The only numbers that will make this expression undefined are the ones that would make the denominator equal to 0, and those are the situation, or that situation would occur, when either a, b, x, or y is equal to 0. If any of those are equal to 0, then we have an undefined expression. We could say the domain is all real a's, b's, x, and y's, except 0. Or we could be specific: except a, b, x, and y can't equal to 0. Or you could write this: given that a, b, x, and y does not equal 0, that none of them can be equal to 0. These are just multiple ways of stating the same thing. With that stated, let's actually multiply and simplify this rational expression. So when we multiply, you just multiply the numerator and multiply the denominator, so you have 3x squared y times 14a squared b in the numerator. Then in the denominator, we have 2ab times 18xy squared. Let's see where we can simplify this thing. We can divide the 14 by 2, and the 2 by 2, and we get 14 divided by 2 is 7, and 2 divided by 2 is 1. We could divide the 3 by 3 and get 1, and divide the 18 by 3, and get 6. Every time we divided the numerator and the denominator by 2, now the numerator and the denominator by 3, so we're not changing the expression. Then we can divide a squared divided by a, so you're just left with an a in the numerator, and a divided by a is just 1. You have a b over a b. Those guys cancel each other out. You have an x squared divided by an x, so x squared divided by x is x, and x divided by x is just a 1, so this becomes an x over 1, or just an x. Finally, you have a y over a y squared. if. You divide the numerator by y, you get 1. If you divide the denominator by y, you just get a y, and so what are we left with? We're left with in the numerator, these 1's we can ignore. That doesn't really change the number. We have a 7 times a times x. That's what we have in the numerator. In the denominator, we just have a 6y. And we have to add the constraint that a, b, x, and y cannot equal 0. When you just look at this expression right here, you're like, hey, what's wrong with x? There isn't even any b here, so it's a little bit of a bizarre statement, but you say, hey, why can't x or a be equal to 0 over here? It doesn't make it undefined. But in order for these to really be the same expressions, they have to have the same domains. Or actually, if these were functions that equal them, in order for that f of x to be equal to this f of x right over here, you'd have to constrain the domain in a similar way. This is a fundamentally different expression if you allow x's and a's. In this one, you can't allow x's and a's. For them to be really the same, you have to put the same constraints on it.