What fraction of the form \(\frac{A}{x + 3}\) can be added to \(\frac{6x}{x^2 + 2x - 3}\) so that the result reduces to a fraction of the form \(\frac{B}{x - 1}\)? Here A and B are real numbers. Give the value of A as your answer.

\(\dfrac{A}{x+3}+\dfrac{6x}{x^2+2x-3}=\dfrac{B}{x-1}\\ \dfrac{A}{x+3}+\dfrac{6x}{(x+3)(x-1)}=\dfrac{B}{x-1}\\ \dfrac{A(x-1)+6x}{(x+3)(x-1)}=\dfrac{B(x+3)}{(x+3)(x-1)}\)

\((A+6)x-A=Bx + 3B\)

\(A+6 = B\\ -A = 3B\\ \text{Solve these two equations for }A, B\)