Question 239034
Problem:
{{{((3a-5)/(a^2+4a+3))+((2a+2)/(a+3))=(a-3)/(a+1)}}}
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It took me several steps to solve this problem.  Look carefully as you move through the steps.
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First make eqn equal to zero:
{{{((3a-5)/(a^2+4a+3))+((2a+2)/(a+3))-((a-3)/(a+1))=0}}}
Now factor first denominator:
{{{((3a-5)/((a+3)(a+1)))+((2a+2)/(a+3))-((a-3)/(a+1))=0}}}
I would also factor the second numerator.  You'll know why in a moment... it'll make our lives easier.
{{{((3a-5)/((a+3)(a+1)))+((2(a+1))/(a+3))-((a-3)/(a+1))=0}}}
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* In order to add fractions, we need common denominators, right?
* We also know that any number multiplied by 1 is equal to itself, in other words, it doesn't change the original value.
* Furthermore, we know that any number divided by itself is equal to 1, right?
* So, in order to have equal denominators, we need to multiply the second fraction by {{{(a+1)/(a+1)}}} and the third denominator by {{{(a+3)/(a+3)}}}.
* This is how the new equation would look like:
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{{{((3a-5)/((a+3)(a+1)))+((2(a+1))/(a+3))((a+1)/(a+1))-((a-3)/(a+1))((a+3)/(a+3))=0}}}
Now in a more cohesive way:
{{{(3a-5)/((a+3)(a+1))+(2(a+1)(a+1))/((a+3)(a+1))-((a-3)(a+3))/((a+1)(a+3))=0}}}
Since we now have the same denominator, we can group all numerators above one common denominator.
{{{((3a-5)+2(a+1)(a+1)-((a-3)(a+3)))/((a+1)(a+3))=0}}}
Now perform multiplications, leave denominator as is.
{{{((3a-5)+(2(a^2+2a+1))-(a^2-9))/((a+1)(a+3))=0}}}
Distribute the negative sign in {{{-(a^2-9)}}}
{{{((3a-5)+(2(a^2+2a+1))+(-a^2+9))/((a+1)(a+3))=0}}}
Multiply the 2 in {{{2(a^2+2a+1)}}}
{{{(3a-5+2a^2+4a+2-a^2+9)/((a+1)(a+3))=0}}}
Combine like terms:
{{{(3a+4a-5+2+9+2a^2-a^2)/((a+1)(a+3))=0}}}
{{{(7a+6+a^2)/((a+1)(a+3))=0}}} 
Now in standard form to make it easier to factor: 
{{{(a^2+7a+6)/((a+1)(a+3))=0}}}
{{{((a+6)(a+3))/((a+1)(a+3))=0}}}
Now simplify and you're done.
{{{(a+6)/(a+1)}}}
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It's a long problem.  I hope you understood it and were able to follow it. :)