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?

$$\text{Given: }\frac{(Ax + B)(x - 3) + (Cx + D)(x + 5)}{(x + 5)(x - 3)} = \frac{x^2 - 4}{(x + 5)(x - 3)}$$
$$(Ax^2-3Ax+BX-3B)+(Cx^2+5Cx+Dx+5D)-(x^2-4)=0$$
$$(A+C-1)x^2+(-3A+B+5C+D)x+(-3B+5D+4)=0$$
$$(\mathbf{A+C}-1)x^2+(\mathbf{-3A-3C}+B+\mathbf{5C+3C} -D)x-(3B-5D-4)=0$$
Noting the coefficient of $x^2$: $A+C=1$, we can  eliminate $x^2$
$$(B+8C+D-3)x=3B-5D-4$$
Noting the coefficients of $x: (B+8C+D-3)=0\implies3B-5D-4=0$
$$\implies 3B=5D+4\implies B=\frac{5D+4}{3}\implies 5D+4+24C+3D-9=8D+24C-5=0\\
\implies 3C+D=\frac{5}{8}$$
We now  have the values for $A+C$ and $3C+D$. If you can set up $4$ simultaneous equations using these and other combinations from the original equations, you can find a unique solution, if the determinant of a matrix of those equations' coefficients is non-zero. Otherwise, you can experiment with one or two values as you did in your question and some solution(s) should pop out. I do not know how you would go about proving there are infinite solutions if that is the case.
