# How do you simplify (28n^2-175) / (2n^2-7n+5) div(14n^2+31n-10 )/ (n^2-6n+5)?

Nov 12, 2017

$\frac{7 n - 35}{7 n - 2}$

#### Explanation:

Once, I simplified fractions separately,

$\frac{28 {n}^{2} - 175}{2 {n}^{2} - 7 n + 5}$

=$\frac{7 \cdot \left(4 {n}^{2} - 25\right)}{2 {n}^{2} - 7 n + 5}$

=$\frac{7 \left(2 n + 5\right) \left(2 n - 5\right)}{\left(2 n - 5\right) \left(n - 1\right)}$

=$\frac{7 \cdot \left(2 n + 5\right)}{n - 1}$

$\frac{14 {n}^{2} + 31 n - 10}{{n}^{2} - 6 n + 5}$

$\frac{14 {n}^{2} + 35 n - 4 n - 10}{\left(n - 1\right) \cdot \left(n - 5\right)}$

=[7n*(2n+5)-2*(2n+5)]/[(n-1)*(n-5)

=$\frac{\left(7 n - 2\right) \left(2 n + 5\right)}{\left(n - 1\right) \left(n - 5\right)}$

Thus,

$\frac{\frac{28 {n}^{2} - 175}{2 {n}^{2} - 7 n + 5}}{\frac{14 {n}^{2} + 31 n - 10}{{n}^{2} - 6 n + 5}}$

=$\frac{28 {n}^{2} - 175}{2 {n}^{2} - 7 n + 5} \cdot \frac{{n}^{2} - 6 n + 5}{14 {n}^{2} + 31 n - 10}$

=$\frac{7 \cdot \left(2 n + 5\right)}{n - 1}$*$\frac{\left(n - 1\right) \left(n - 5\right)}{\left(7 n - 2\right) \left(2 n + 5\right)}$

=$\frac{7 \cdot \left(n - 5\right)}{7 n - 2}$

=$\frac{7 n - 35}{7 n - 2}$