# How do you simplify (8a^2-6a-9)/( 6a^2-5a-6) ÷ (4a^2+11a+6)/( 9a^2+12a+4)?

Mar 29, 2018

$\frac{3 a + 2}{a + 2}$

#### Explanation:

$\left\{\frac{8 {a}^{2} - 6 a - 9}{6 {a}^{2} - 5 a - 6}\right\} \div \left\{\frac{4 {a}^{2} + 11 a + 6}{9 {a}^{2} + 12 a + 4}\right\}$

$= \left\{\frac{8 {a}^{2} - 12 a + 6 a - 9}{6 {a}^{2} - 9 a + 4 a - 6}\right\} \div \left\{\frac{4 {a}^{2} + 8 a + 3 a + 6}{9 {a}^{2} + 6 a + 6 a + 4}\right\}$

$= \left\{\frac{4 a \left(2 a - 3\right) + 3 \left(2 a - 3\right)}{3 a \left(2 a - 3\right) + 2 \left(2 a - 3\right)}\right\} \div \left\{\frac{4 a \left(a + 2\right) + 3 \left(a + 2\right)}{3 a \left(3 a + 2\right) + 2 \left(3 a + 2\right)}\right\}$

$= \left\{\frac{\cancel{\left(2 a - 3\right)} \left(4 a + 3\right)}{\cancel{\left(2 a - 3\right)} \left(3 a + 2\right)}\right\} \div \left\{\frac{\left(a + 2\right) \left(4 a + 3\right)}{\left(3 a + 2\right) \left(3 a + 2\right)}\right\}$

$= \frac{\cancel{\left(4 a + 3\right)}}{\cancel{\left(3 a + 2\right)}} \cdot \frac{\cancel{\left(3 a + 2\right)} \left(3 a + 2\right)}{\left(a + 2\right) \cancel{\left(4 a + 3\right)}}$

$\frac{3 a + 2}{a + 2}$ [Ans]