SOLUTION: I am stuck on this problem...I found that the LCD is (a+1)(a+3) but I am not sure where to go from there. Thanks for your help! Problem: (solve) (3a-5/a^2+4a+3)+(2a+2/a+3)=(a-3/

Question 128226This question is from textbook Intro Algebra
: I am stuck on this problem...I found that the LCD is (a+1)(a+3) but I am not sure where to go from there. Thanks for your help!
Problem: (solve)
(3a-5/a^2+4a+3)+(2a+2/a+3)=(a-3/a+1) This question is from textbook Intro Algebra

Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
((3a-5)/(a^2+4a+3))+((2a+2)/(a+3))=((a-3)/(a+1)) I assume this is the problem.
Use parentheses more liberally to avoid possible confusion.
((3a-5)+(2(a+1)(a+1))=((a-3)(a+3)) Multiply each side by LCD to eliminate fractions.
3a-5+2a^2+4a+2=a^2-9
3a-5+2a^2+4a+2-a^2+9=0
a^2+7a+6=0
(a+6)(a+1)=0
a=-6, a=-1
.
Ed

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