College Algebra: Solving with Partial Fractions

Had no idea partial fractions could do that!

07/20/2009

~ brittanie

Great lesson to prep for the math portion of the GRE!

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Now, this is going to be a little application I just want to show you really fast just to give you another sense that, in fact, you can use these simultaneous systems of many variables to solve things. This one, I have to admit, is a little bit lame, unless you plan to go on to calculus, in which case it actually can be really useful at some point.
Anyway, this will actually come up again if you see calculus a little bit. Here is the basic question. Now the basic question is going to sound a little bit weird. Suppose I give you this rational function? That’s just a polynomial divided by a polynomial. Suppose I said to you, “Hey, you know what? That actually is the answer to a question. And the question was, I had two fractions.” One fraction had some mystery thing on top, but 3x + 5 on the bottom, and some other mystery thing on this top, and x - 2. Then I added them together, and I got that as the answer. Now, of course, I had to get a common denominator. So the common denominator was the product of these things.
Suppose I wanted to know what the fractions were individually before I added? That is to say, suppose I’m given this rational function, and I want to split it apart into two pieces, this piece plus that piece. What’s the right tops to make this thing actually work out? I want to ? instead of taking two fractions and adding them like we used to do, I now want to take this and break it apart into the sum of two fractions. This actually is called the technique of partial fractions, and believe it or not, people in calculus actually use this stuff.
But let’s not worry about why it’s useful. Let’s just see how we could do it, and how it actually leads to a problem about two equations and two unknowns. The trick is this, since we don’t know what the tops are, let’s just give them names. I’ll call this A and I’ll call this B. In fact, that will equal A, some unknown thing. Over 3x + 5 + and then B, some unknown thing over x - 2. Now to figure out what A and B should be, let’s just add these up. I have to get a common bottom, so here I multiply top and bottom by x - 2. Here I multiply top and bottom by 3x + 5. Now I add, the bottoms are the same. In fact, this equals. Let’s see, what does it equal? Well, if I distribute here, I’d see Ax - 2A + 3Bx + 5B, all over the common bottom. I’m not going to multiply that out. I’m just going to write it like this. Actually, I can combine these x terms here, and what I’d see is (A + 3B)x + and these constant terms are 5B - 2A over the bottom, (3x + 5)(x - 2).
That’s what we get when we take these two fractions and add them. But I know what the answer should be, it should be x - 5. And this is what I have. So what that means is that the thing multiplying the x, that value right there, must be a 1, because that’s what’s multiplying this. So A + 3B = 1, and this thing here, the number part, 5B - 2A, that should equal this number part, which is -5. In fact, to have this thing equal the actual answer, I need to have these match up. The thing in front of the x needs to be the same thing. The thing with the constant needs to be the same thing.
That gives me two equations and two unknowns. In fact, what I see here is that A + 3B = 1, and I also see that -2A + 5B = -5. Those two equations came from the fact that 1 has to equal the term in front of the x, and this has to equal the term here, because I am equating this with this fraction. Then I add it up and saw how it came out.
So, in fact, I now have two equations and two unknowns. Then I solve this using a substitution method or any other method that you want. Let’s take this and solve it for A. So A = 1 - 3B. If I now insert that here for A, what I would see is -2(1 - 3B) + 5B = -5. So if I distribute, I see -2 + 6B + 5B = -5. So 6B + 5B = 11B. If I take that -2 and bring it over here, it pops over as +2, but I’ve got -5, so that’s -3. I see that B = -3/11.
What would A be? Well, I can go back to here and solve for A. A = 1 - 3 times this. Or a minus times a minus becomes a plus, 3 x 3 = 9/11. 1 = 11/11 + 9/11 = 20/11. So A = 20/11.
So, what’s the answer to the original question? The answer to the original question is A = 20/11, and B = -3/11. If I bring this back here, what I see is, if I put in 20/11 as the number on top here, and -3/11 as the number on top here, when you take those two fractions and combine them, what you get is that answer. So, I just took this one fraction and split it into two fractions, the top here is 20/11. The top here is -3/11, but if you combined getting a common denominator and adding, you’ll end up with exactly this.
This is the technique of partial fractions. And this is actually really important when you study calculus. It’s a nice application of taking a fact and setting up two equations and two unknowns.