Simplify the following fully: 3.1.1 \( \frac{3}{a-4}+\frac{2}{a+3}-\frac{21}{a^{2}-a-12} \)

To simplify the expression \( \frac{3}{a-4}+\frac{2}{a+3}-\frac{21}{a^{2}-a-12} \), we first factor the quadratic in the denominator. The expression \( a^2 - a - 12 \) can be factored as \( (a - 4)(a + 3) \). Now we can rewrite the problem: \[ \frac{3}{a-4} + \frac{2}{a+3} - \frac{21}{(a-4)(a+3)} \] Next, we find the common denominator, which is \( (a-4)(a+3) \). This leads us to rewrite each term: \[ \frac{3(a+3)}{(a-4)(a+3)} + \frac{2(a-4)}{(a-4)(a+3)} - \frac{21}{(a-4)(a+3)} \] Combining the numerators gives: \[ \frac{3(a+3) + 2(a-4) - 21}{(a-4)(a+3)} \] Now, simplifying the numerator: 1. Expand \( 3(a+3) = 3a + 9 \) 2. Expand \( 2(a-4) = 2a - 8 \) 3. Combine those with \(-21\): \[ 3a + 9 + 2a - 8 - 21 = 5a - 20 \] Now we have: \[ \frac{5a - 20}{(a-4)(a+3)} \] Finally, we can factor the numerator \( 5(a - 4) \), leading to: \[ \frac{5(a - 4)}{(a - 4)(a + 3)} \] Cancelling \( (a-4) \) from the numerator and denominator (assuming \( a \neq 4 \)), we get: \[ \frac{5}{a + 3} \] So, the fully simplified expression is: \[ \frac{5}{a + 3} \quad (a \neq 4) \]