yego.me
💡 Stop wasting time. Read Youtube instead of watch. Download Chrome Extension

Subtracting rational expressions: unlike denominators | High School Math | Khan Academy


3m read
·Nov 11, 2024

So right over here we have one rational expression being subtracted from another rational expression. I encourage you to pause the video and see what this would result in, so actually do the subtraction.

Alright, now let's do this together. If we're subtracting two rational expressions, we'd like to have them have the same denominator, and they clearly don't have the same denominator. So we need to find a common denominator, and a common denominator is one that is going to be divisible by either of these. Then we can multiply them by an appropriate expression or a number so that it becomes the common denominator.

The easiest common denominator I can think of, especially because these factors, these two expressions have no factors in common, would just be their product. So this is going to be equal to... So we could just multiply these two. This is going to be... actually, let me do this one right over here in magenta. So this is going to be equal to the common denominator. If I say, if I want to just multiply those two denominators for this one, I'll have my 8x + 7, and now I'm going to multiply it by 3x + 1.

I'm multiplying it by the other denominator, and I had negative 5x in the numerator. But if I'm going to multiply the denominator by 3x + 1, and I don't want to change the value of the expression, we'll have to multiply the numerator by 3x + 1 as well. Notice 3x + 1 divided by 3x + 1 is just 1, and you'd be left with what we started with.

From that, we are going to subtract all of this. Now, there's a couple of ways you could think of the subtraction. I could just write a minus sign right over here and do the same thing that I just did for the first term. Or another way to think about it, and actually for this particular case, I like thinking about it better this way, is to just add the negative of this.

So if I just multiplied negative 1 times this expression, I'd get negative 6x^3 over 3x + 1. If I had more terms up here in the numerator, I would have to be careful to distribute that negative sign. But here, I only have one term, so I just made it negative. I could say this is going to be plus... and let me do this in a new color, this in green. Our common denominator, we already established, is just the product of our two denominators.

So it is going to be 8x + 7 times 3x + 1. Now, if we multiply the denominator here, it was 3x + 1. We're multiplying it by 8x + 7, so that means we have to multiply the numerator by 8x + 7 as well. 8x + 7 times negative 6x^3. Notice 8x + 7 divided by 8x + 7 is 1. If you were to do that, you would get back to your original expression right over here, the negative 6x^3 over 3x + 1.

And now we're ready to add. This is all going to be equal to... I'll write the denominator in white so we have our common denominator, 8x + 7 times 3x + 1. Now, in the magenta, I would want to distribute the negative 5x. So negative 5x times positive 3x is negative 15x^2, and then negative 5x times 1 is minus 5x.

Then in the green, I would have... let's see, I'll distribute the negative 6x^3. So negative 6x^3 times positive 8x is going to be negative 48x^4, and then negative 6x^3 times positive 7 is going to be negative 42x^3.

I think I'm done because there's no more... I only have one fourth degree term, one third degree term, one second degree term, one first degree term, and that's it. There's no more simplification here. Some of you might want to just write it in descending degree order. So you could write it as negative 48x^4 minus 42x^3 minus 15x^2 minus 5x, all of that over 8x + 7 times 3x + 1.

But either way, we are all done, and it looks like up here... yeah, there's nothing to factor out. These two are divisible by 5, these are divisible by 6. But even if I were to factor that out, nothing over here down here, no five or six to factor out. Yeah, so it looks like we are all done.

More Articles

View All
Introduction to electron configurations | AP Chemistry | Khan Academy
In a previous video, we’ve introduced ourselves to the idea of an orbital. Electrons don’t just orbit a nucleus the way that a planet might orbit a star, but really, in order to describe where an electron is at any given point in time, we’re really thinki…
Division with partial quotients example
Let’s say we want to figure out what 473 divided by 5 is, and like always, why don’t you pause this video and try to work through it? If you’re familiar with the idea of division with partial quotients, I encourage you to try it out that way. All right, …
2015 AP Calculus AB 6a | AP Calculus AB solved exams | AP Calculus AB | Khan Academy
Consider the curve given by the equation (y^3 - xy = 2). It can be shown that the derivative of (y) with respect to (x) is equal to (\frac{y}{3y^2 - x}). All right, write an equation for the line tangent to the curve at the point ((-1, 1)). So, we could…
Humans Are Exceptional
Three out of these four theories have an interesting pattern to them, with good explanations in epistemology. We’re saying conjectures and their refutations, and error correction is how we improve knowledge. With genetic evolution, genetic mutations, vari…
Partial sums: formula for nth term from partial sum | Series | AP Calculus BC | Khan Academy
Partial sum of the series we’re going from one to infinity summing it up of a sub n is given by, and they tell us the formula for the sum of the first n terms. They say write a rule for what the actual nth term is going to be. Now to help us with this, l…
How Warren Buffett Made His First $1 Million
So, in this video, we’re going to talk about how Warren Buffett made his first million dollars and what you can learn from it to make yours. Warren Buffett is currently worth $100 billion and built a company that is worth $650 billion. If you’re watching …