Partial fractions, disagreement with Wolfram Alpha On my math homework, I have this problem, and WolframAlpha says that $A=-\frac{1}{7}$ and $B=\frac{1}{7}$. However, while solving the problem on my own, I found a restriction that $A \neq -B$. How is it that this answer works? The problem in question is:
$$\dfrac{1}{x^2 - 3 x - 10} = \dfrac{A}{x+2} + \dfrac{B}{x - 5}$$

Here's the full derivation: $$\frac{1}{x^2-3x-10}=\frac{A}{x+2}+\frac{B}{x-5}\implies$$ $$(x+2)(x-5)\left(\frac{1}{x^2-3x-10}\right)=(x+2)(x-5)\left(\frac{A}{x+2}+\frac{B}{x-5}\right)\implies$$ $$1=A(x-5)+B(x+2)=A(x)-5A+B(x)+2B=x(A+B)+1(2B-5A)$$
Hence we need $A+B=0\space\text{and}\space2B-5A=1$. So substitute $A=-B$ into the second equation: $$2B-5(-B)=1\implies7B=1\implies B=\frac{1}{7}$$
But we have the condition that $A+B=0$ and we know $B=\frac{1}{7},$ hence $$A+\frac{1}{7}=0\implies A=-\frac{1}{7}$$ Thus $$\frac{1}{x^2-3x-10}=\frac{A}{x+2}+\frac{B}{x-5}=\frac{1}{7(x-5)}-\frac{1}{7(x+2)}$$
