The problem is to add two rational expressions: $\frac{t}{t+6} + \frac{t-3}{t-6}$.

1. Problem Description

The problem is to add two rational expressions: $\frac{t}{t+6} + \frac{t-3}{t-6}$.

2. Solution Steps

To add the two rational expressions, we need to find a common denominator. The common denominator is $(t+6)(t-6)$. We then rewrite each fraction with the common denominator and add the numerators.

$\frac{t}{t+6} + \frac{t-3}{t-6} = \frac{t(t-6)}{(t+6)(t-6)} + \frac{(t-3)(t+6)}{(t-6)(t+6)}$

Combining the fractions, we have:

$\frac{t(t-6) + (t-3)(t+6)}{(t+6)(t-6)}$

Expand the numerator:

$\frac{t^2 - 6t + (t^2 + 6t - 3t - 18)}{(t+6)(t-6)}$

$\frac{t^2 - 6t + t^2 + 3t - 18}{(t+6)(t-6)}$

Combine like terms in the numerator:

$\frac{2t^2 - 3t - 18}{(t+6)(t-6)}$

Expand the denominator:

$(t+6)(t-6) = t^2 - 36$

So, the expression becomes:

$\frac{2t^2 - 3t - 18}{t^2 - 36}$

We attempt to factor the numerator:

$2t^2 - 3t - 18 = (2t+a)(t+b) = 2t^2 + (a+2b)t + ab$

We need $ab=-18$ and $a+2b = -3$. If $a = 6$ and $b = -3$, then $a+2b = 6-6 = 0$.

If $a=-6$ and $b=3$, then $a+2b = -6+6 = 0$.

If $a = 3$ and $b = -6$, then $a+2b = 3-12 = -9$.

If $a = -3$ and $b=6$, then $a+2b = -3+12 = 9$.

If $a=9$ and $b=-2$, then $a+2b = 9-4 = 5$.

If $a=-9$ and $b=2$, then $a+2b = -9+4 = -5$.

If $a=12$ and $b=-\frac{3}{2}$, then $a+2b = 12-3 = 9$.

Consider $(2t+6)(t-3) = 2t^2-6t+6t-18 = 2t^2-18$. This doesn't work.

Consider $(2t-9)(t+2) = 2t^2 + 4t - 9t - 18 = 2t^2 - 5t - 18$. This doesn't work.

The numerator does not appear to factor nicely. Thus, we leave the expression as:

$\frac{2t^2 - 3t - 18}{t^2 - 36}$

3. Final Answer

$\frac{2t^2 - 3t - 18}{t^2 - 36}$