# How do you express (x-9)/[(x+5)(x-2)]  in partial fractions?

$\frac{x - 9}{\left(x + 5\right) \left(x - 2\right)} = \frac{2}{x + 5} - \frac{1}{x - 2}$

#### Explanation:

solution: set up the working equations first by using variables A and B.

$\frac{x - 9}{\left(x + 5\right) \left(x - 2\right)} = \frac{A}{x + 5} + \frac{B}{x - 2}$

using LCD-least common denominator $= \left(x + 5\right) \left(x - 2\right)$ simplify

$\frac{x - 9}{\left(x + 5\right) \left(x - 2\right)} = \frac{A \left(x - 2\right) + B \left(x + 5\right)}{\left(x + 5\right) \left(x - 2\right)}$

$\frac{x - 9}{\left(x + 5\right) \left(x - 2\right)} = \frac{A x - 2 A + B x + 5 B}{\left(x + 5\right) \left(x - 2\right)}$

rearrange the numerators

$\frac{1 \cdot x - 9 \cdot {x}^{0}}{\left(x + 5\right) \left(x - 2\right)} = \frac{\left(A + B\right) \cdot x + \left(- 2 A + 5 B\right) \cdot {x}^{0}}{\left(x + 5\right) \left(x - 2\right)}$

We can have the equations now to solve for A and B

$A + B = 1$
$- 2 A + 5 B = - 9$

Solve for A and B simultaneosly

$7 B = - 7$

$B = - 1$ and $A = 2$

God bless....I hope the explanation is useful..