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