document.write( "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
\n" ); document.write( "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
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Algebra.Com's Answer #11971 by rapaljer(4671)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "Does that look like your problem after factoring the first denominator???\r
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\n" ); document.write( "\n" ); document.write( "Then your next step looked like this, right? \r
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\n" ); document.write( "\n" ); document.write( "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:\r
\n" ); document.write( "\n" ); document.write( "\"%285-3a%29+-+%282a%2B2%29%2A%28a%2B1%29+=+%283-a%29%2A%28a%2B3%29+\"\r
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\n" ); document.write( "\n" ); document.write( "Multiply the two sets of binomials, but save the negative for the next step!
\n" ); document.write( "\"%285-3a%29+-+%282a%2B2%29%2A%28a%2B1%29+=+%283-a%29%2A%28a%2B3%29+\"
\n" ); document.write( "\"5-3a+-+%282a%5E2%2B4a%2B2%29+=+%283a%2B9+-a%5E2+-+3a%29+\"\r
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\n" ); document.write( "\n" ); document.write( "Now, distribute the -1:
\n" ); document.write( "\"5-3a+-+2a%5E2+-+4a+-2+=+9+-+a%5E2+\"
\n" ); document.write( "\"3+-+7a+-+2a%5E2+=+9+-+a%5E2\"\r
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\n" ); document.write( "\n" ); document.write( "Set the quadratic equation equal to zero, by moving everything to the right side (to get a positive a^2 term!)
\n" ); document.write( "\"3+-+7a+-+2a%5E2+-3+%2B7a+%2B2a%5E2+=+9+-+a%5E2-3+%2B7a+%2B+2a%5E2+\"
\n" ); document.write( "\"0+=+a%5E2+%2B7a+%2B6\"\r
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\n" ); document.write( "\n" ); document.write( "This just happens to factor!! (Isn't life good when it works???)
\n" ); document.write( "\"0=+%28x%2B6%29%28x%2B1%29\"
\n" ); document.write( "x = -6 and x = -1\r
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\n" ); document.write( "\n" ); document.write( "Recheck to see if any denominators were accidentally made zero. They were not, so it looks like these answers are acceptable.\r
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\n" ); document.write( "\n" ); document.write( "R^2 at SCC
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