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 ->  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

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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(4551) About Me  (Show Source):
You can put this solution on YOUR website!
%285-3a%29%2F%28%28a%2B3%29%28a%2B1%29%29-+%282a%2B2%29%2F%28a%2B3%29+=+%283-a%29%2F%28a%2B1%29+

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

Then your next step looked like this, right?
%285-3a%29%2F%28%28a%2B3%29%28a%2B1%29%29-+%28%282a%2B2%29%2F%28a%2B3%29%29%2A%28%28a%2B1%29%2F%28a%2B1%29%29+=+%28%283-a%29%2F%28a%2B1%29%29%2A%28%28a%2B3%29%2F%28a%2B3%29%29+

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