# How do you simplify ((2x-5)/(x^2-9))/((3x-1)/(x+3))?

Feb 18, 2017

$\frac{2 x - 5}{\left(x + 3\right) \left(3 x - 1\right)}$

#### Explanation:

Rewrite the problem by moving the bottom fraction in line with the top fraction and adding a division symbol. Rewrite ${x}^{2} - 9$ into $\left(x - 3\right) \left(x + 3\right)$:

$\frac{2 x - 5}{\left(x - 3\right) \left(x + 3\right)} \div \frac{3 x - 1}{x + 3}$

Flip the fraction on the left and rewrite the problem as a multiplication problem:

$\frac{2 x - 5}{\left(x - 3\right) \left(x + 3\right)} \times \left(\frac{x + 3}{3 x - 1}\right)$

Cancel out the x+3

$\frac{2 x - 5}{x + 3} \times \left(\frac{1}{3 x - 1}\right)$

Consolidate the denominator:

$\frac{2 x - 5}{\left(x + 3\right) \left(3 x - 1\right)}$

Feb 18, 2017

$\frac{2 x - 5}{3 {x}^{2} - 10 x + 3}$

#### Explanation:

Using an example:

Consider $\text{ } 6 \div 2 = 3$
Although not normally done it is both correct and permissible to write this as: $\frac{6}{1} \div \frac{2}{1}$

The shortcut method is: Turn $\frac{2}{1}$ upside down and multiply

$\frac{6}{1} \times \frac{1}{2} = \frac{6 \times 1}{1 \times 2} = \frac{6}{2} = 3$

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Given$\text{ } \frac{2 x - 5}{{x}^{2} - 9} \div \frac{3 x - 1}{x + 3}$

Write the given expression as: $\text{ } \frac{2 x - 5}{{x}^{2} - 9} \times \frac{x + 3}{3 x - 1}$

But ${x}^{2} - 9$ is the same as ${x}^{2} - {3}^{2} = \left(x - 3\right) \left(x + 3\right)$ giving:

$\frac{2 x - 5}{\left(x + 3\right) \left(x - 3\right)} \times \frac{x + 3}{3 x - 1}$

$\frac{2 x - 5}{{\cancel{\left(x + 3\right)}}^{1} \left(x - 3\right)} \times \frac{{\cancel{\left(x + 3\right)}}^{1}}{3 x - 1}$

$\frac{2 x - 5}{\left(x - 3\right) \left(3 x - 1\right)}$

$\frac{2 x - 5}{3 {x}^{2} - 10 x + 3}$