In this problem: -2/a+2 + 3a/2a-1
First I made the denominators alike:
-2(2a-1)/(a+2)(2a-1) + 3a(a+2)/(a+2)(2a-1)
= -4a+2/(a+2)(2a+1) + 3a^2+6a/(a+2)(2a-1)
What I want to know is if I can eliminate the a+2 for the first fraction, on both the numerator and the denominator?
2 answers
No. You have to combine numberators first, over a common denominator.
thanks