## Algebra 2 (1st Edition)

$$\frac{x+7}{2\left(x-1\right)}$$
Simplifying the expression by cancelling terms in the numerator and the denominator, we find: $$\frac{3\left(x^2-3\right)\left(x+7\right)}{\left(x-1\right)\left(6x^2-18\right)} \\ \frac{3\left(x^2-3\right)\left(x+7\right)}{\left(x-1\right)\cdot \:6\left(x^2-3\right)} \\ \frac{x+7}{2\left(x-1\right)}$$