# Homework Help: Solving for Q and R in Seemingly Unsolvable Problem

1. Feb 14, 2012

### erok81

This isn't from my homework, but it came up during a test for my sister's husband. He works at an Air Force base as some type of engineer and they are required to take this general math class. So as far as I can tell, it's nothing advanced but I can't figure it out, and swear it is unsolvable - three unknowns and only one equation.

Maybe it's something I don't know. The professor gave an answer for both Q and R, nothing for x.

1. The problem statement, all variables and given/known data

Solve for Q and R.

$\frac{(x+4)}{(x+1)}(x+2) = \frac{Q}{(x+1)} + \frac{R}{(x+2)}$

2. Relevant equations

Nothing.

3. The attempt at a solution

So I got it simplified down to x+4 = Q(x+2) + R(x+1) but that's about as far as I can get. The only thing I can think of is Q and R are some engineering terms I've never heard of (physics student here). I tried solving it for x, but that didn't get me anywhere

Q and R end up being regular old whole numbers. I can post those values as well.

2. Feb 14, 2012

### Dick

It looks like it's just a partial fractions question. You know those, don't you?

3. Feb 14, 2012

### DivisionByZro

I am assuming this is the equation you have:

$$\frac{x+4}{(x+1)(x+2)} = \frac{P}{x+1} + \frac{Q}{x+2}$$

This type of decomposition is useful when integrating the quotient of two polynomial functions.
It is typically called "partial fractions decomposition".

Here's an example:

$$\frac{5x-4}{(x+1)(2x-1)} = \frac{P}{x+1} + \frac{Q}{2x-1}$$

You can multiply both sides by

$$(x+1)(2x-1)$$

to get

$$5x-4 = P(x+1) + Q(2x-1)$$

Now, you can expand the above, collect like-terms, and then compare coefficients of both sides:

$$5x-4 = (P+2Q)x + (P-Q)$$

We now have a system of 2 equations in 2 unknowns:

$$5 = P+2Q$$

$$-4 = P-Q$$

Does this help?

4. Feb 14, 2012

### erok81

Yes! Thank you both. That helps. I can't believe I didn't even think of trying partial fractions. I just saw so many variables and had a negative outlook on it from the inset. Plus I never saw the original problem, so I also assumed something was left off.