How to determine algebraically whether an equation has an infinite solutions or not?

I was learning for the first time about partial fraction decomposition. Whoever explains it, emphasises that the fraction should be proper in order to be able to decompose the fraction. I was curious about knowing what happens if I try to decompose an improper fraction, So I tried to do one:

$\frac{x^2 - 4}{(x + 5)(x - 3)}$

I got the equation: $\frac{(Ax + B)(x - 3) + (Cx + D)(x + 5)}{(x + 5)(x - 3)} = \frac{x^2 - 4}{(x + 5)(x - 3)}$. I have 4 unknowns: A, B, C and D.

$\therefore (Ax + B)(x - 3) + (Cx + D)(x + 5) = x^2 - 4$

After expanding and regrouping the coefficients:

$(A + C) x^2 + (-3A + B + 5C + D)x + (-3B + 5D) = x^2 - 4$

Here the coefficient of the term $x^2$ is 1 therefore:

$(A+C) = 1$

similarly:

$(-3A + B + 5C + D) = 0$

$(-3B + 5D) = -4$

I still have to get one more equation to be able to solve this system so I substituted 1 for x and I got this equation:

$-2A - 2B + 6C + 6D = -3$

After getting four equations I used this site to solve the system of equations. Unfortinetly I got no soultion. Tried another site and also the same result.

I've tried to use different values for x and got another equaitons like:

for x = 2 : $-2A - B + 24C + 7D$

for x = -1 : $4A - 4B - 4C + 4D$

for x = -2 : $10A - 5B - 6C + 3D$

But also that didn't work. Always the system of equations have an infinite solutions.

After tring to figure out why this is happening, I've managed to prove logically that this equation:

$(Ax + B)(x - 3) + (Cx + D)(x + 5) = x^2 - 4$

has an infinite solutions and my approach was as follows:

After doing polynomial long division and decomposing the fraction using the traditional way, the result should be:

$\frac{5}{8(x - 3)} - \frac{21}{8(x - 5)} + 1$

Now I can add the last term (the one) to the first term and get the follows:

$\frac{8x-19}{8(x-3)} - \frac{21}{8(x+5)}$

From that solution I can see that $A = 0, B = \frac{-21}{8}, C = 1, D = \frac{-19}{8}$. After all these are just the coefficients of the terms. and this solution worked fine.

Alternatively I can add the one to the second term instead and get:

$\frac{5}{8(x-3)} + \frac{8x + 19}{8(x+5)}$

Now $A = 1, B = \frac{19}{8}, C = 0, D = \frac{5}{8}$

Generally, after adding the one to any of the terms, I can add any number to one of the terms and add its negative to the other term and the equation will remain the same, But the value of the 4 constants (A, B, C, and D) will change. And from that I got convinced that there are an infinite solutions to this equation.

But Algebraically? I'm not able to prove that it has an infinite solutions algebraically. And my questions is how to prove algebraically that this equation has an infinite solutions? Or generally how to know whether the equation has just one solution or an infinite?

Solution 1:

Steven Stadnicki has given an easy algebraic approach not very different from yours to see why there are infinitely many solutions. But here are some comments on what you did:

1. You correctly wrote down equations for the coefficients of $x^0,x^1,x^2$. Notice that if you find $A,B,C,D$ satisfying all those three equations, then you do have $\frac{(Ax + B)(x - 3) + (Cx + D)(x + 5)}{(x + 5)(x - 3)}$ $= \frac{x^2 - 4}{(x + 5)(x - 3)}$ for every real/complex $x∉\{3,-5\}$. Thus if you show that that system of three equations has infinitely many solutions, then the original problem of finding a partial fraction decomposition also has infinitely many solutions. There is also the not-so-trivial fact that if $\frac{(Ax + B)(x - 3) + (Cx + D)(x + 5)}{(x + 5)(x - 3)}$ $= \frac{x^2 - 4}{(x + 5)(x - 3)}$ for infinitely many real/complex numbers, then the coefficients must match. This is why you want to match coefficients in the first place.

2. You said that you were told that you must have a proper fraction to be able to perform partial fraction decomposition. Depending on what you mean by partial fractions, that is either wrong or misleading. As you observed, you have multiple ways to decompose in a certain fashion. But it is also true that there is a unique decomposition of $\frac{x^2 - 4}{(x + 5)(x - 3)}$ into the sum $\frac{Ax+B}{x+5}+\frac{C}{x-3}$ for constants $A,B,C$. If that does not count as a partial fraction decomposition, yours would not either, and matching of coefficients as you did shows that $\frac{x^2 - 4}{(x + 5)(x - 3)}$ cannot be decomposed into the sum $\frac{A}{x+5}+\frac{B}{x-3}$ for constants $A,B$.

More generally, if we want to express something in a certain form with some constant parameters, we can often write down a system of equations that the parameters must satisfy, and transform it into other equivalent systems, and hope that we reach a system that we can easily find a solution to. This is applicable not only to linear equations but other kinds of equations as well. For example, consider the following system:

$a^2+b = 1$.

$b^2+a = 2$.

We can transform it into an equivalent system by using the first equation to eliminate one variable, as follows:

$a^2+b = 1$ and $b^2+a = 2$

  $⇔$ $b = 1-a^2$ and $b^2+a = 2$

  $⇔$ $b = 1-a^2$ and $(1-a^2)^2+a = 2$

Observe that if you find a solution to second equation $(1-a^2)^2+a = 2$, you get a solution to the last line because "$b$" occurs only in the first equation and isolated on one side.

If you do this elimination procedure systematically on a system of linear equations, you will essentially do what is called Gaussian elimination, and you will essentially arrive at a row echelon form. In that form it is easy to immediately determine whether the system has zero, one or infinitely many solutions, and also how many degrees of freedom there are.

Solution 2:

Yes there is a way. After expanding and regrouping the coefficients, you have a system of equations to solve for $A,B,C,D$.

Consider the example system of equations $x+y=2$, and $2x+2y = 4$. If I solve the first equation for $x$, I get $$x=2-y$$ Then if I plug this into the second equation: $$2(2-y) + 2y=4$$ $$4 - 2y + 2y = 4$$ $$4 = 4$$ Notice that the $y$ disappeared. I could solve for $x$, but then not for $y$. Similarly, if I solved for $y$ first, the $x$ would disappear. This means that the only constraint on my solution is that $x=2-y$ (there is no requirement for $x$ and $y$ to be any particular numbers).

So possible solutions to my system are $x=2, y=0$, or also $x=20, y=-18$, or any pair of numbers of the form $(x,y)=(2-t,t), t\in \mathbb{R}$.

In my example, I can describe the solution space as $\{(-t,t)+(2,0):t\in \mathbb{R}\}$ which contains infinitely many points.

For your problem, you can do the same thing. By attempting to solve your system of equations, you should be able demonstrate that the number of solutions is infinite.