Equations with rational expressions
Sal solves (x²-10x+21)/(3x-12)=(x-5)/(x-4), which has one real solution and one extraneous solution.
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- Why can't the denominator be zero?(6 votes)
- Division by zero is an undefined operation.
Let 𝑦 = 𝑥∕0 ⇒ 𝑥 = 0 ∙ 𝑦
For 𝑥 ≠ 0, there is no finite value 𝑦 that satisfies these equations, so 𝑦 is undefined.
For 𝑥 = 0, any finite value 𝑦 satisfies the equations, so again 𝑦 is undefined.(5 votes)
- Near the end of the video, Sal says that x cannot equal 4 because it would have both sides undefined. But wouldn't 4 actually work? Think about it: if x is 4, the left side has division by zero (which is undefined), and the right side has division by zero (which is undefined). If both sides are undefined, that means that both sides are equal, because "undefined" is equal to "undefined". So shouldn't 4 technically be a solution?(3 votes)
- When something is "undefined" it has no answer. It doesn't matter that the 2 sides are equal. You can't find a numeric result when you have an undefined condition.(18 votes)
- why is this so complicated?(8 votes)
- can any letter be a variable? I have been wondering this forever!(3 votes)
- Any symbol at all can be a variable, if you can write it consistently. Latin, Greek, and Hebrew letters all appear as variables in some parts of math. You can use any of those, draw your own symbols, use emoji, or whatever you like. They're all valid variables.(8 votes)
- So my teacher has been teaching this, and he says we have to find the LCD. Is that what is happening in this video?(4 votes)
- No. The LCD is the Least Common Denominator. Sal found the answer in a different form. There is a video on this, but I'm not sure how to link it.(2 votes)
- why cant we just multiply the denominator right away before simplifying it(3 votes)
- You can, but simplifying makes it easier to understand.(3 votes)
- Can the numerator equal 0? I know the denominator can’t equal 0 but can the numerator?(2 votes)
- Yes, its ok for the numerator to = 0. The one exception to this is if you are dividing by a fraction with a numerator of 0. Once it flips, you would have a denominator of 0 which would not be good.
Hope this makes sense.(6 votes)
- what happens when the denominator looks like (x+4)(x-1) or something like that?(2 votes)
- If the denominator were to have two binomials like that, do the same thing as you would in the video. Expectedly, it's a little bit more annoying as now as you have a third degree polynomial, but otherwise the same procedure as in the video.
Hopefully that helps !(2 votes)
- Is there no easy way to solve this via cross multiplication? Cross multiplying I get:
x^3-14x^2+61x-84 = 3x^2-27x+60
subtracting both sides:
and I'm stuck here.(1 vote)
- The problem with cross multiplication is that you end up with a much more difficult problem. Your polynomial takes a lot more work to try and find the factors.
The technique shown in the video eliminates some factors in the denominators making the problem more manageable and simple to solve.
Use cross multiplication when the denominators are simple monomials. If they are binomials or larger, then use the method in this video.(3 votes)
- Is Sal's technique essentially cross-multiplying in multiple steps? Thanks in advance!(2 votes)
- [Voiceover] So we have a nice little equation here dealing with rational expressions. And I encourage you to pause the video and see if you can figure out what values of x satisfy this equation. Alright, let's work through this together. So the first thing I'd like to do is just see if I can simplify this at all, and maybe by finding some common factors between numerators and denominators, or common factors on either side of the equal sign. So let's factor all of the numerators and the denominators. All the ones on the right-hand side are already done. So this thing up here, I could rewrite this as, let's see. What's product is 21? What two numbers, when I take their product, is 21? Positive 21, so they're going to have the same sign. And when I add them, I get negative 10. Well, negative seven and negative three, so this could be rewritten as x minus seven times x minus three. This over here, both are divisible by three. I could rewrite this as three times x minus four. And these are already factored. So the one thing that jumps out at me is I have x minus four in the denominator on the left-hand side and on the right-hand side. And so, if I were to multiply both sides by x minus four. So actually, let me just... Let me formally replace this with that. And up here, it's not so obvious that it's going to be valuable for me to keep this factored form. So I'm just going to keep it in this yellow form. In the expanded out form. So let me just scratch that out for now, because once I, well, let me multiply by x minus four. So, if we multiply both sides by x minus four, and once again, why am I doing this? Just so I get rid of the x minus four in the denominators. x minus four, and then x minus four. That and that cancels. That and that cancels, and then we're left with, in the numerator, we're left with our x squared minus 10 x plus 21, and, let's see, divided by three. Divided by three is equal to x minus five. Let's see, now, what we can do, and actually I could have done it in the last step, is I can multiply both sides by three. Multiply both sides. Do that in another color just so it sticks out a little more. So I can multiply both sides by three. So multiply both sides by three. On the left-hand side, that and that cancels, and I'll just be left with x squared minus 10 x plus 21, and on, and I don't have a denominator any more. My denominator is one, so I don't need to write it. Is going to be equal to three times, let's just distribute the three. Three times x is three x. Three times negative five is negative 15. And now I can get this into standard quadratic form by getting all of these terms onto the left-hand side. The best way to do that, let's subtract three x from the right, but I can't just do it from the right otherwise the equality won't held. I have to do it from both sides if I want equality to hold. And, I want to get rid of this negative 15, so I can add 15 to both sides. So, let's do that. And what we are left with, scroll down a little bit, so let's have a little more space. What we are going to be left with is x squared minus 13 x and then plus, what is this, plus 36. Plus 36. Did I do that right? Yep, plus 36 is equal to, is equal to zero. Alright, now let's see, we have it as a quadratic in the standard form. How can we solve this? So first thing, can we factor this. Product of two numbers, 36. If I add them, I get negative 13. They're both going to be negative, since they have to have the same sign to get their product to be positive. And, let's see, nine and four seem to do the trick. So, or negative nine and negative four. So, x minus four times x minus nine is equal to zero. Well, that's going to happen if either x minus four is equal to zero or x minus nine is equal to zero. We'll add four to both sides of this. This happens when x is equal to four. Add nine to both side of this. This happens when x is equal to nine. So we could say that the solutions are x equals four or x equals nine. So, x is equal to four, or x equals nine. But we need to be careful. Because we have to remember, in our original expression, x minus four was a factor of both denominators. And so if we actually try to test x minus four in the original equation, not one of these intermediary steps, the original equation, I would end up dividing by zero right over here, and actually, I'd end up by zero right over there as well. So the original equations, if I tried to substitute four, they don't make sense. So this is actually an extraneous solution. It's not going to be a solution to the original equation. The only solution is x is equal to 9.