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

##### 1 Answer

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 = (3-x) * (3+x)#

The second denominator can be written as

#x^2 - 3x = x * (x - 3)#

Now, you can rewrite this as

#x * (x-3) = - x * (3 - x)#

since

#-x * (3 -x ) = -3x - x * (-x) = x^2 - 3x#

This means that the second fraction can be rewritten s

#(2x+1)/(x^2 - 3x) = (2x + 1)/(-[x * (3-x)]) = - (2x+1)/(x (3-x))#

The original expression becomes

#-4/((3-x) * (3 + x)) - [ - (2x+1)/(x (3-x))]#

#-4/((3-x)(3 + x)) + (2x + 1)/(x(3-x))#

Now, in order to find the **common denominator**, you must multiply the first fraction by

This will get you

#-4/((3-x)(3+x)) * x/x + (2x+1)/(x(3-x)) * (3+x)/(3+x)#

#-(4x)/(x(3-x)(3+x)) + ((2x+1)(3+x))/(x(3-x)(3+x))#

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

#-4x + (2x+1)(3+x)#

#-4x + (6x + 2x^2 + 3 + x)#

#2x^2 + 3x + 3#

Put this back into the resulting expression to get

#(2x^2 + 3x + 3)/(x(3-x)(3+x))#

Keep in mind that you need to have