# How do you subtract \frac{2x}{x^2+10x+25}-\frac{3x}{2x^2+7x-15}?

Dec 20, 2014

First, we're going to simplify the expressions a bit. We can do this by factoring both the denominators:
${x}^{2} + 10 x + 25 = {\left(x + 5\right)}^{2}$

$2 {x}^{2} + 7 x - 15 = \left(x + 5\right) \left(2 x - 3\right)$

To substract two fractions with a different denominator, we will always have to find the LCM of the denominators. In this case it is ${\left(x + 5\right)}^{2} \left(2 x - 3\right)$ since this can evenly be divided by both the denominators.

Let's combine what we've got already:
$\frac{2 x}{x + 5} ^ 2 - \frac{3 x}{\left(x + 5\right) \left(2 x - 3\right)}$

Now for changing to the LCM:
$\frac{2 x - 3}{2 x - 3} \cdot \frac{2 x}{x + 5} ^ 2 - \frac{x + 5}{x + 5} \frac{3 x}{\left(x + 5\right) \left(2 x - 3\right)}$

$= \frac{2 x \cdot \left(2 x - 3\right) - 3 x \cdot \left(x + 5\right)}{{\left(x + 5\right)}^{2} \left(2 x - 3\right)}$

Distribute the terms:

$\frac{4 {x}^{2} - 6 x - 3 {x}^{2} - 15 x}{{\left(x + 5\right)}^{2} \left(2 x - 3\right)}$

$= \frac{{x}^{2} - 21 x}{{\left(x + 5\right)}^{2} \left(2 x - 3\right)}$

You could've also done this without factoring, but it would've been way harder. Factoring always makes your life easier!

I really hope this helped.