Adding rational expression: unlike denominators | High School Math | Khan Academy

Pause the video and try to add these two rational expressions. Okay, I'm assuming you've had a go at it. Now we can work through this together.

So, the first thing that you might have hit when you tried to do it is you realize that they have different denominators. It's hard to add fractions when they have different denominators. You need to rewrite them so that you have a common denominator.

The easiest way to get a common denominator is you can just multiply the two denominators, especially in a case like this where they don't seem to share any factors. Both of these have factors you can get, and they don't share anything in common.

So, let's set up a common denominator. This is going to be equal to something. Let's see, it's going to be equal to something over our common denominator. Let's make it 2x. Let me do this in another color. So, we're going to make it ( 2x - 3 ) times ( 3x + 1 ), and then plus something else over ( 2x - 3 ) times ( 3x + 1 ).

To go from just ( 2x - 3 ) here in the denominator to ( 2x \times (3x + 1) ), we multiply the denominator by ( 3x + 1 ). So, if we do that to the denominator, we don't want to change the value of the rational expression we have; we'd also have to do that to the numerator.

So, the original numerator was ( 5x ) (doing that in blue color), and now we're going to multiply it by the ( 3x + 1 ). So, times ( 3x + 1 ). Notice I didn't change the value of this expression; I multiplied by ( \frac{3x + 1}{3x + 1} ), which is 1 as long as ( 3x + 1 ) does not equal zero.

So, let's do the same thing over here. Over here, I have a denominator of ( 3x + 1 ). I multiply it by ( 2x - 3 ). So, I would take my numerator, which is ( -4x^2 ), and I would also multiply it by ( 2x - 3 ). Let me put parentheses around this so it doesn't look like I'm subtracting ( 4x^2 ).

So, then I can rewrite all of this as being equal to... Well, in the numerator, I’m going to have ( 5x \times 3x ), which is ( 15x^2 ) and ( 5x \times 1 ), which is ( + 5x ).

And then, over here (let me do this in green), let's see... I could do ( -4x^2 \times 2 ) which would be ( -8x^2 ), and then ( -4x^2 \times -3 ) which is ( +12x^2 ).

Did I do that right? Oh, let me be very careful. My spider sense could tell that I did something shady. In fact, if you want to pause the video, you could see and try to figure out what I just did that's wrong.

So, ( -4x^2 \times 2x ) is ( -8x^3 ), and then ( -4x^2 \times -3 ) is ( +12x^2 ). Now, our entire denominator, our entire denominator, we have a common denominator now, so we were able to just add everything.

It's ( (2x - 3)(3x + 1) ), and let's see how we can simplify this. So, this is all going to be equal to... Let me draw and make sure we recognize it's a rational expression.

And so let's see, we can look at, our highest degree term here is the ( -8x^3 ). So, it's ( -8x^3 ), and then we have ( 15x^2 ) and we also have ( +12x^2 ). We could add those two together to get ( 27x^2 ).

So, we've already taken care of this. We've taken care of those two, and we're just left with ( +5x ). So, all of that is over ( (2x - 3)(3x + 1) ), and we are all done.

It doesn't seem like there's any easy way to simplify this further. You could factor out an ( x ) out of the numerator, but that's not going to cancel out with anything in the denominator. And it looks like we are all done.