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?
No. You have to combine numberators first, over a common denominator.
thanks
To determine if you can eliminate the (a+2) factor in the numerator and denominator of the first fraction, you need to check if it can be canceled out completely.
In this case, since (a+2) is present in both the numerator and denominator of the first fraction, you can indeed cancel it out. This is because dividing a term by the same term results in a value of 1, effectively eliminating it.
So, you can simplify the expression further by canceling out the (a+2) factor:
(-4a + 2) / (a+2)(2a-1) = (-4a + 2) / (2a-1)
Therefore, in the simplified form, the first fraction no longer has the (a+2) factor in its numerator or denominator.