Question 207635
{{{(3a-5)/(a^2+4a+3)+(2a+2)/(a+3)=(a-3)/(a+1)}}}
-----
{{{(a^2+4a+3)}}} factors out to be {{{(a+3)*(a+1)}}} so your equation becomes:
{{{(3a-5)/((a+3)*(a+1))+(2a+2)/(a+3)=(a-3)/(a+1)}}}
-----
if you multiply both sides of the equation by {{{(a+3)*(a+1)}}} you will get:
{{{(3a-5)+(2a+2)*(a+1)=(a-3)(a+3)}}}
-----
if you simplify this you get:
{{{(3a-5)+2a^2 + 4a + 2=a^2 - 9}}}
if you subtract {{{a^2 - 9}}} from both sides of the equation you get:
{{{3a-5+2a^2 + 4a + 2 -a^2 + 9=0}}}
if you combine like terms  you get:
{{{a^2+7a+6=0}}}
this factors out to be {{{(a+1)*(a+6) = 0}}}
{{{a = -1}}}
or
{{{a = -6}}}
-----
when we substitute a = -1 in the original equation we get a division by 0 so a = -1 is no good.
-----
when we substitute a = -6 in the original equation we get (27) = (27) so a = -6 is good.
-----
computations for a = -6 are as follows:
-----
{{{(3a-5)/(a^2+4a+3)+(2a+2)/(a+3)=(a-3)/(a+1)}}}
becomes:
{{{(-18-5)/(36-24+3)+(-12+2)/(-6+3)=(-6-3)/(-6+1)}}}
which becomes:
{{{(-23)/(15)+(-10)/(-3)=(-9)/(-5)}}}
if we multiply both sides of the equation by 15 we get:
{{{(-23)+(50)=(27)}}}
{{{(27)=(27)}}}