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# Reducing rational expressions to lowest terms

Learn how to simplify rational expressions, which are fractions with variable expressions in the numerator and denominator. This video also shows how to factor the expressions and cancel out common factors, just like with rational numbers. Don't forget to exclude values of x that would make the denominator zero, since those would make the expression undefined. Created by Sal Khan and CK-12 Foundation.

## Want to join the conversation?

• do you always have to add the condition? never heard of it from my teacher.
• Sometimes in an Algebra 1 course/text/curriculum, teachers will just teach the simplifying piece, and leave the restrictions for Algebra 2. This is because this is one of the most challenging type of problems in Algebra 1. Brains often melt solving rationals because many students can barely factor and simplify, let alone consider restrictions on the denominator. Even Sal makes mistakes in his examples, he forgets restrictions.
• At , Sal said that x cannot equal -1. However, x also cannot be equal to 2, based off of the answer he got. Why wouldn't you put x cannot equal negative 2 in the final answer, as that would also make the expression undefined?
• Now that the expression has been simplified to (x+5)(x-2), it is obvious that x cannot be equal to 2, but the fact that x could also not be equal to -1 (otherwise division by 0 resulted) in the original non simplified expression has been lost, so we add the condition as a reminder.
You see, we can set x=-1 into (x+5)(x-2) without a problem, but we could not set x=-1 in the original expression - and this simplified expression is based on that.
• Near couldn't you just factor out a 3 and get 3(x^2+x-6) and then factor the rest of it using the quadratic formula or other methods?
• That would work if I'm not mistaken, so if you're comfortable with that, go ahead! I personally prefer the method Sal uses.
• why is it that all math problems have numbers in them?
• I guess an answer would be that numbers can represent an almost unlimited amount of things. I wouldn't be surprised if there were some math problems without numbers, but it is a very essential part of the subject and so it shows up in lots of places. Something to think about would be, if not numbers, what else would be in them?
• in the final example, is it not necessary to also say that x ≠ 2, (ie. x ≠ 2, -1 ) because the fact that x - 2 is still in the denominator means that it's still present? i.e. would it only be necessary to specify x ≠ 2, -1 if somewhere later in the calculation x - 2 was somehow canceled out?
• Good question! The reason is that x ≠ 2 is still clear from our given expression. However, x ≠ -1 isn't (as that information is lost when we cancel the terms). That's why we mention the -1 and not the 2.

However, for the same function, if we were to define the domain, both 2 and -1 would have to be excluded.
• how do you get to the practice problems for this
• at , i still do not get how he got x cannot equal -1 over 3
• I've come across problems in my homework where I don't know what the condition should be. Is it the x value that makes what you cancel equal to zero, that makes the original expression's denominator equal to zero, or that makes the new expression's denominator equal to zero?
• In problems like those in the video you are expected to explicitly state the condition for the x value that makes what you cancel equal to zero.

However, since you're cancelling (and so there is such a factor in the original denominator), it implies the same thing makes the original expression's denominator equal to zero, or maybe—is ONE OF THE THINGS that make the original expression's denominator equal to zero.

If you think about it, the value that makes the new expression's denominator equal to zero is the reason for saying "one of the things..." in the previous sentence. Because THE OTHER THING is what makes the new expression's denominator equal to zero. It's the restriction (or several restrictions) shared by the old and the new expressions, therefore you don't have to explicitly state it in a task like this.
• I understand that we have to add the restrictions or conditions for the equation, but what if you forget it? Would it affect your graph or something?
• Any restrictions on a rational expression are a consequence of the expression not being defined at one or more values of the variable. At such values,
the graph of the corresponding rational expression function can have a vertical asymptote or a removable discontinuity . Both affect the graph.
• Ramey
a minute ago
Posted a minute ago. Direct link to Ramey's post “Can someone please walk me through this STEP BY ST...”
Can someone please walk me through this STEP BY STEP, regardless of how ridiculously minor the step is? I have tried and failed so many times it's not even funny. I have looked at countless videos, and asked several people for help. I know it is a simple problem, but I am really struggling. Any help would be greatly appreciated.

Two airplanes start at the same time from airports 500 km apart. Each one flies with an airspeed of 200 km/hr directly towards the other airport. But one reaches its airport half an hour before the other plane reaches the other airport. How fast is the wind blowing?

If you can include what laws you use along the way, it would really benefit my understanding. But at the least, i just want to see how to do it. *I AM NOT AFTER THE ANSWER. I AM AFTER THE PROCESS.* I really just want to understand and I am desperate. Thank you in advance!!
• Because you say that you don't want to know the answer, I'm guessing that you want the process to lead you to the answer. I would like to say, though, that this is a simple problem, but I get why you are struggling with it because it requires a lot of work to get to the answer.

Anyways, if both planes, which we will call the first one Plane A and the second one Plane B, are at opposite airports, and the airports are 500 km apart, that means that if the airplanes are going 200 km/h, then the flight of the planes should be about 2.5 hours, give or take.

Now, one of the airplanes is going a bit faster and the other is going a bit slower because the wind is blowing with plane A and against plane B.

Because Plane A arrives 30 minutes earlier than Plane B, Plane A took 15 minutes less time to get to the opposite airport, and Plane B took 15 minutes more time to get to the opposite airport.

Now, let's say that the airplanes took off at PM. Remember, the flight should have been around 2 hrs 30 min. Because of the wind factor, Plane A landed at PM, and Plane B landed at ; a 30 minute difference.

Alright. Now that we know that, we can solve for the wind. If the planes were going 200 km/h, and one arrived 15 minutes early and one arrived 15 minutes late, we will use simple subtraction to find the wind speed:
200(km/h the planes traveled at) - 15(the amount of time the flights were changed) = 175.

Therefore, the wind was going 175 km/h.

If you are confused with a step, all you need to do is ask me.

Hope that helps!