Partial Fraction Using the Long Method I've been trying to express $$\frac{x^2-13}{x^3-7x+6}$$ as a partial fraction, and, so far I have arrived at $$x^2-13=A(x-2)(x+3)+B(x-1)(x+3)+C(x-1)(x-2)$$. 
From this onwards, I substituted different values of x to remove unknowns and find the values of the unknowns individually. Yet, in my textbook, 
$$1= B+C \tag 1$$ (for the coefficients of the squared x term)
$$0=A+2B-3C \tag 2$$ (for the coefficients of the non-squared terms)
$$-13=-6A-3B+2C\tag 3$$ (for the constants) 
What I have trouble with is understanding how 1) is derived, for, from my perspective, it should be $$1=A+B+C$$. Also, I was wondering if these three equations can be solved simultaneously by using Gaussian elimination, or simply, simultaneously solve two of the equations and then solve the remaining equation form he remaining unknowns. 

As you have noticed, since $x^3-7x+6= (x-1)(x^2+x-6)=(x-1)(x-2)(x+3) $ we can write $$\frac{x^2-13}{x^3-7x+6}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x+3}$$ to get $$x^2-13=A(x-2)(x+3)+B(x-1)(x+3)+C(x-1)(x-2).$$ Then, to solve this equation for $A, B, C$, it is easier to plug in $x=1,2,-3$, i.e., the roots of denominator, to obtain:
$$x=1 \Rightarrow -12=-4A$$
$$x=2 \Rightarrow -9=5B$$
$$x=-3 \Rightarrow -4=20C$$
