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
Parallel & perpendicular lines from graph
In this video, we’re going to do a couple of examples that deal with parallel and perpendicular lines. So you have parallel, you have perpendicular, and of course, you have lines that are neither parallel nor perpendicular. Just as a bit of a review, if …
Photographing the Wild Wolves of Yellowstone | Exposure
In Rogard Kipling’s The Jungle Book, he has a quote that says, “For the strength of the pack is the wolf, and the strength of the wolf is the pack.” Yellowstone lives and breathes wolves. In the last 20 years, I wanted to photograph them and bring that to…
How We Can Keep Plastics Out of Our Ocean | National Geographic
8 million metric tons of plastic trash enters the sea from land every year; the equivalent of five plastic bags filled with trash for every foot of coastline in the world. Across our ocean, plastic trash blows into circulation, dispersed almost everywhere…
Geoff Ralston's Intro - Startup Investor School Day 1
Welcome everyone to my competitors’ startup investor school. If you think you’re at a different class, you should leave now. So it’s great to see you all here. I’m Jeff Ralston, and I’m going to act kind of as the master of ceremonies. I’ll be introducin…
My Investing Plan For 2023 (How To Prepare)
What’s up, Graham? It’s guys here. So 2023 is probably going to be one of the most confusing years for investing. After all, stocks are the cheapest they’ve been in two years, but there’s a chance they could drop even further. Real estate has only starte…
Power dissipation in resistors in series versus in parallel
A student builds a circuit with a battery and two light bulbs in series. Then the student builds a second circuit with two light bulbs in parallel. Which battery runs out of power first? Assume all bulbs have equal resistance. Assume both batteries have …