Simple Algebraic Equation

Simple.

$x^2 - x - 6 = (x - 3)(x + 2)$

This can make the question easier to solve.

$\dfrac {2x - 9}{x^2 - x - 6} = \dfrac {A}{x - 3} + \dfrac {B}{x + 2}$

$\dfrac {2x - 9}{x^2 - x - 6} = \dfrac {A(x + 2)}{x^2 - x - 6} + \dfrac {B(x - 3)}{x^2 - x - 6}$

$2x - 9 = Ax + 2A + Bx - 3A$

This equation can be transferred into 2 equations:

$2x = Ax + Bx ... (1)$

$-9 = 2A - 3B ... (2)$

From the first equation, it is clear that the answer can be derived from it. Dividing both sides by $x$ yields

$2 = A + B$

Hence, our answer is 2.

$\qquad \frac { 2x-9 }{ { x }^{ 2 }-x-6 } =\frac { A }{ (x-3) } +\frac { B }{ (x+2) } \\ =>\quad 2x-9\quad =\quad A(x+2)+B(x-3)\quad ..................\quad (i)\\ Putt\quad \boxed { x=3 } \quad in\quad (i)\\ \qquad 2(3)-9\quad =\quad A[(3)+2]+B[(3)-3]\\ or\quad \quad \quad 6-9\quad =\quad 5A\quad +0\\ or\quad \quad \quad \quad -3\quad =\quad 5A\quad =>\quad \boxed { A=-\frac { 3 }{ 5 } } \\ Now\quad Putt\quad \boxed { x=-2 } \quad in\quad (i)\\ \qquad 2(-2)-9\quad =\quad A[(-2)+2]+B[(-2)-3]\\ or\quad \quad \quad -4-9\quad =\quad 0\quad -5B\\ or\quad \quad \quad \quad -13\quad =\quad -5B\quad =>\quad \boxed { B=\frac { 13 }{ 5 } } \\ Now\quad "A+B"\\ \qquad A+B=\quad -\frac { 3 }{ 5 } +\frac { 13 }{ 5 } \\ \qquad \qquad \quad =\quad \frac { -3+13 }{ 5 } \quad =\quad \frac { 10 }{ 5 } \\ or\quad \boxed { A+B=2 }$