# How do you combine (t+1)/(t+3)-(t-2)/(t-3)+6/(t^2-9)?

May 17, 2018

See a solution process below:

#### Explanation:

First, we need to understand:

${t}^{2} - 9 = \left(t + 3\right) \left(t - 3\right)$

Therefore, we can multiply the two fractions on the left by the appropriate form of $1$ to put them over a common denominator with the fraction on the right:

$\left(\frac{t - 3}{t - 3} \cdot \frac{t + 1}{t + 3}\right) - \left(\frac{t + 3}{t + 3} \cdot \frac{t - 2}{t - 3}\right) + \frac{6}{{t}^{2} - 9} \implies$

$\frac{\left(t - 3\right) \left(t + 1\right)}{{t}^{2} - 9} - \frac{\left(t + 3\right) \left(t - 2\right)}{{t}^{2} - 9} + \frac{6}{{t}^{2} - 9} \implies$

$\frac{{t}^{2} - 3 t + 1 t - 3}{{t}^{2} - 9} - \frac{{t}^{2} + 3 t - 2 t - 6}{{t}^{2} - 9} + \frac{6}{{t}^{2} - 9} \implies$

$\frac{{t}^{2} + \left(- 3 + 1\right) t - 3}{{t}^{2} - 9} - \frac{{t}^{2} + \left(3 - 2\right) t - 6}{{t}^{2} - 9} + \frac{6}{{t}^{2} - 9} \implies$

$\frac{{t}^{2} + \left(- 3\right) t - 3}{{t}^{2} - 9} - \frac{{t}^{2} + 1 t - 6}{{t}^{2} - 9} + \frac{6}{{t}^{2} - 9} \implies$

$\frac{{t}^{2} - 3 t - 3}{{t}^{2} - 9} - \frac{{t}^{2} + 1 t - 6}{{t}^{2} - 9} + \frac{6}{{t}^{2} - 9}$

Now, we can combine like terms in the numerator over the common denominator:

$\frac{{t}^{2} - 3 t - 3 - {t}^{2} - 1 t + 6 + 6}{{t}^{2} - 9} \implies$

$\frac{{t}^{2} - {t}^{2} - 3 t - 1 t - 3 + 6 + 6}{{t}^{2} - 9} \implies$

$\frac{\left({t}^{2} - {t}^{2}\right) + \left(- 3 - 1\right) t + \left(- 3 + 6 + 6\right)}{{t}^{2} - 9} \implies$

$\frac{0 + \left(- 4\right) t + 9}{{t}^{2} - 9} \implies$

$\frac{- 4 t + 9}{{t}^{2} - 9}$

Or

$\frac{9 - 4 t}{{t}^{2} - 9}$