How do you subtract (-4)/(9-x^2)-(2x+1)/(x^2-3x)?

Aug 4, 2016

Here's what I got.

Explanation:

The first thing to do here is make sure that the two fractions have the same denominator.

Notice that the first denominator can be written as

$9 - {x}^{2} = {3}^{2} - {x}^{2} = \left(3 - x\right) \cdot \left(3 + x\right)$

The second denominator can be written as

${x}^{2} - 3 x = x \cdot \left(x - 3\right)$

Now, you can rewrite this as

$x \cdot \left(x - 3\right) = - x \cdot \left(3 - x\right)$

since

$- x \cdot \left(3 - x\right) = - 3 x - x \cdot \left(- x\right) = {x}^{2} - 3 x$

This means that the second fraction can be rewritten s

$\frac{2 x + 1}{{x}^{2} - 3 x} = \frac{2 x + 1}{- \left[x \cdot \left(3 - x\right)\right]} = - \frac{2 x + 1}{x \left(3 - x\right)}$

The original expression becomes

$- \frac{4}{\left(3 - x\right) \cdot \left(3 + x\right)} - \left[- \frac{2 x + 1}{x \left(3 - x\right)}\right]$

$- \frac{4}{\left(3 - x\right) \left(3 + x\right)} + \frac{2 x + 1}{x \left(3 - x\right)}$

Now, in order to find the common denominator, you must multiply the first fraction by $1 = \frac{x}{x}$ and the second fraction by $1 = \frac{3 + x}{3 + x}$.

This will get you

$- \frac{4}{\left(3 - x\right) \left(3 + x\right)} \cdot \frac{x}{x} + \frac{2 x + 1}{x \left(3 - x\right)} \cdot \frac{3 + x}{3 + x}$

$- \frac{4 x}{x \left(3 - x\right) \left(3 + x\right)} + \frac{\left(2 x + 1\right) \left(3 + x\right)}{x \left(3 - x\right) \left(3 + x\right)}$

Now you're ready to focus on the numerators. You have

$- 4 x + \left(2 x + 1\right) \left(3 + x\right)$

$- 4 x + \left(6 x + 2 {x}^{2} + 3 + x\right)$

$2 {x}^{2} + 3 x + 3$

Put this back into the resulting expression to get

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

Keep in mind that you need to have $x \ne \left\{0 , \pm 3\right\}$.