# If a/(x-2)+b/(x+3)=(2x+11)/(x^2+x-6), then (a,b) = ?

Dec 7, 2017

$\frac{3}{x - 2} - \frac{1}{x + 3}$

#### Explanation:

This can be solvec by using partial fractions.

We know that $\frac{2 x + 11}{{x}^{2} + x - 6} = \frac{a}{x - 2} + \frac{b}{x + 3}$

We can add these two fractions to get:
$\frac{a}{x - 2} + \frac{b}{x + 3} = \frac{a \left(x + 3\right)}{\left(x - 2\right) \left(x + 3\right)} + \frac{b \left(x - 2\right)}{\left(x - 2\right) \left(x + 3\right)}$
$= \frac{a \left(x + 3\right) + b \left(x - 2\right)}{\left(x + 3\right) \left(x - 2\right)}$

By multiplying by $\left(x - 3\right) \left(x + 2\right)$, we get:

$2 x + 11 = a \left(x + 3\right) + b \left(x - 2\right)$

First, we will use $x = 2$ as this will cancel out $b$ and allow us to work out $a$.

$2 \left(2\right) + 11 = a \left(2 + 3\right) + b \left(2 - 2\right)$

$4 + 11 = a \left(5\right) + b \left(0\right)$

$15 = 5 a$

$a = \frac{15}{5} = 3$

Now, we do the same for $x = - 3$

$2 \left(- 3\right) + 11 = a \left(- 3 + 3\right) + b \left(- 3 - 2\right)$

$- 6 + 11 = a \left(0\right) + b \left(- 5\right)$

$5 = - 5 b$

$b = - \frac{5}{5} = - 1$

$a = 3 , b = - 1$

$\frac{2 x + 11}{{x}^{2} + x - 6} = \frac{3}{x - 2} + \frac{- 1}{x + 3}$

$= \frac{3}{x - 2} - \frac{1}{x + 3}$