SOLUTION: I am in elementary algebra in college. The title of the section I am in is "solving rational equations". The problem I am having trouble with is 5-3a over a^2+4a+3 minus 2a+2 ove

Algebra ->  Systems-of-equations -> SOLUTION: I am in elementary algebra in college. The title of the section I am in is "solving rational equations". The problem I am having trouble with is 5-3a over a^2+4a+3 minus 2a+2 ove      Log On


   

Question 21945: I am in elementary algebra in college. The title of the section I am in is "solving rational equations". The problem I am having trouble with is
5-3a over a^2+4a+3 minus 2a+2 over a+3 equals 3-a over a+1. I factored the a^2+4a+3 to (a+3)(a+1). Then multiplied 2a+2 by (a+1) and 3-a by (a+3). Now I am stuck, could you please show me how to do this problem! :( Thank You
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!


Does that look like your problem after factoring the first denominator???

Then your next step looked like this, right?


After this, just multiply both sides of the equation by the LCD, which is (x+3)(x+1). This eliminates the denominators, giving you an equation with NO fractions:
%285-3a%29+-+%282a%2B2%29%2A%28a%2B1%29+=+%283-a%29%2A%28a%2B3%29+

Multiply the two sets of binomials, but save the negative for the next step!
%285-3a%29+-+%282a%2B2%29%2A%28a%2B1%29+=+%283-a%29%2A%28a%2B3%29+
5-3a+-+%282a%5E2%2B4a%2B2%29+=+%283a%2B9+-a%5E2+-+3a%29+

Now, distribute the -1:
5-3a+-+2a%5E2+-+4a+-2+=+9+-+a%5E2+
3+-+7a+-+2a%5E2+=+9+-+a%5E2

Set the quadratic equation equal to zero, by moving everything to the right side (to get a positive a^2 term!)
3+-+7a+-+2a%5E2+-3+%2B7a+%2B2a%5E2+=+9+-+a%5E2-3+%2B7a+%2B+2a%5E2+
0+=+a%5E2+%2B7a+%2B6

This just happens to factor!! (Isn't life good when it works???)
0=+%28x%2B6%29%28x%2B1%29
x = -6 and x = -1


Recheck to see if any denominators were accidentally made zero. They were not, so it looks like these answers are acceptable.

R^2 at SCC