1. ## Stuck on simple rational function addition problem

Hey guys, I'm stuck on some simple rational function problems. I know this is probably really easy to do but I'm a bit rusty on my math skills.

Any help would be appreciated.

1.

((3)/(y^2-3y+2)) + ((7)/(y^2-1))

2.

((6)/(x-8)) - ((x)/(x-1)) + ((x^2+6)/(x^2-9x+8))

Thanks guys.

2. (1) What do you think your first move should be? Show us some of your work on this problem so we can see what you're thinking, and to improve upon it.

(2) The simple approach to getting a common denominator is the "brute force" method:

$\frac{A}{B} + \frac{C}{D} = \frac{AD}{BD} + \frac{BC}{BD} = \frac{AD + BC}{BD}$

3. Hmmm, I've never heard of the "brute force" method. However, I though the first step would be to find the LCD of the denominators.

Please excuse my lack of knowledge on this. I'm trying to help my daughter with her math homework and can't figure it out for the life of me.

4. It's brute force in the sense that you're just mashing together both of the denominators. It would be wiser, if possible, to find something in common, but I wasn't going to do all that work. Using the formula given above, you can get a single-rational term that has a lot of stuff that can be simplified. E.g., using the brute force approach outlined above (i.e., it isn't always clean, but it will always get the job done), we get,

$\frac{(3y^2 - 3) + (7y^2 - 21y + 14)}{(y^2 -3y + 2)(y^2 - 1)} = \frac{10y^2 - 21y + 11}{(y^2 - 3y+2)(y^2 - 1)}$

From here you can check if these quadratic expressions can be simplified to reduce the term even further (i.e., can we present the numerator in terms of one of those factors in the denominator?).

5. Okay I think I get it. The top would factor out to (y-1)(10y-11) and the bottom would factor out to (Y+1)(y-2)(y^2-1). So then I can eliminate a (y-1) from both the top and bottom. Is that correct?

How about the second problem, how would I apply the brute force method to that one?

6. That looks good, save for your typo on the numerator. The answer is then,

$\frac{10y - 11}{(y^2 - 1)(y-2)}$

The bottom is done in exactly the same way. In this case you have something of the form,

$\frac{A}{B} + \frac{C}{D} + \frac{E}{F}$

The solution is just the same: multipy each term by "1" as the denominator of the off-terms. For instance, multiply the first term by 1 = D/D and 1 = F/F, since any number multiplied by 1 is still that number. Thus, we get

$\frac{ADF}{BDF} + \frac{BCF}{BDF} + \frac{BDE}{BDF} = \frac{ADF + BCF + BDE}{BDF}$

Problem two is identical to the above formulation with the proper substitutions. After some algebra, something is sure to clean up.