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Adding and Subtracting Rational Expression - Problem 3

Teacher/Instructor Alissa Fong
Alissa Fong

MA, Stanford University
Teaching in the San Francisco Bay Area

Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts

Any time you’re adding fractions or rational expressions, it’s not too bad if they have the same denominators, so look out for problems like this. I’m bummed because I have to add them, but they have totally different denominators. So let me show you what a lot of my students try to do. A lot of my students see problems like these and they say okay well I know this has to be a common denominator so that’s on the right track and so they’ll say okay I need to add three more X's to top and bottom and I need, because that way I’d have 4x and I also need an x² so they’ll just go ahead and like add stuff, and that’s kind of the right idea because now these are common denominators if I combine like terms. They’re on the right track, but you guys remember that I can’t just add stuff to top and bottom of a fraction; I can multiply stuff by top and bottom of a fraction. That’s a huge distinction that’s really, really important.

So instead of just adding stuff to fractions, what you need to do is look at factoring. Look for things that can multiply the fraction by. For example look at this second denominator. This guy could be easily factored into x plus 3 and x plus 1. Now my denominators look kind of similar. This one has the x plus 3 piece, but it needs the multiplied by x plus 1. So this is how I can find the common denominator. Multiply this fraction by x plus 1 over x plus 1.

Now those are fractions that have the same denominators. I’m going to rewrite it so it looks a little better, but the most important piece if you don’t remember anything else from this video, I hope that you guys remember that you can’t just add staff to top and bottom of the fraction. You can multiply stuff top and bottom of the fraction.

Okay so rewriting this first fraction on top now I have 4 times the quantity x plus 1 over x plus 3 times x plus 1. Then my second fraction still looks the same x² plus x only after the bottom. Okay now I can see clearly these have the same denominator so I know the denominator of my sum is going to be x plus 3 times x plus 1. I’m just going to write it once.

Then I’m going to look at the tops. I want to combine these, but first I’m going to distribute this 4 so I can combine them a little more effectively. If I distribute the 4, then I can combine these and write them in Standard form. X² will become first because that’s my highest exponent. Then I have 5 X's and then plus 4 at the end. I’m still not done, a lot of students think they’re done here because they added them together yeah you’re on the right track, but we need to simplify. Simplify means I need to factor the top and then the bottom I’m sorry done and I need to cross out any factors that are the same.

So my next step is going to be to factor the top. The bottom is already done. On top I need numbers that multiply to 4 and add up to 5 so that will be 4 and 1. I want to look for any factors that are the same on top and bottom, cross those guys out so my final answer is going to look like x+4 over x+3.

Another mistake I see a lot in my classes and please don’t make this mistake is students look at this and they think oh well the x is cancelled out right? So my answer should just be 4 over 3. A lot of students want to go like that, sound effects optional, but a lot of students want to cancel out those X's and just write 4/3 as the answer, but be really careful you guys this cancelling process is mathematically accurate if I have things being multiplied. Here things are being added together so I can’t cancel out.

Remember cancelling things out is really like writing them as 1. X plus 1 over x plus 1 is like the fraction 1, so I can’t just go cancelling out whenever I want to. I can only cancel out things that are being multiplied together. Okay so that’s my most simplified answer an A+ student would go back and look for excluded values it’s up to whether your textbook or your teacher asks you to do that, it’s not too hard. Excluded values comes from looking at the denominators and looking think about x values that would make the denominator equal to 0.

In our case the only x values that would make this denominator equal to 0 would be -3, or -1. So I would write from my excluded values x cannot be equal to -3, or -1 in this answer because if they were, I would have my denominator equal to 0 and that’s undefined.

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