Question

Here is a question from a college algebra exercise. The answer is supposed to be "1" but I keep getting something else. Here is the problem:
(6a^2+a-1)/(6a^2+5a+1) + (3a^2+2a-1)/(3a^2+4a+1). I have factored, canceled, and ended up with (6a-2)/(3a+1). Can anyone help me see how to get "1" for the answer?

Answer 1

Factoring gives:
\[\Large \frac{(3a-1)(2a+1)}{(6a-1)(a+1)} + \frac{(3a-1)(a+1)}{(3a+1)(a+1)}\]

Answer 2

Now, to add those two things, you first need to get a common denominator. Both denominators have a factor of (a+1) already, so multiply the left by (3a+1) on top and bottom, and multiply the right by (6a-1) on the top and bottom. One you have a common denominator, you can join the two into one fraction thusly
\[\Large \frac{(3a+1)(3a-1)(2a+1)+(6a-1)(3a-1)(a+1)}{(6a-1)(3a+1)(a+1)}\]

Answer 3

Thanks for looking. The denominator of the first term is 6a^2+5a+1 which factors to (3a+a)(2a+1), not (6a-1)(a+1). Right?

Answer 4

As either a math question, you should not be posting this in the OpenStudy Feedback group.

Answer 5

Ah, good catch, Sterna. Except that it doesn't factor to (3a+a)(2a+1). It factors to (3a+1)(2a+1)

Answer 6

Resulting in
\[\frac{(3a-1)(2a+1)}{(3a+1)(2a+1)}+\frac{(3a-1)(a+1)}{(3a+1)(a+1)}\]