Question 84322
Assuming that you mean {{{3/(x+3) + 1/x}}}, the first step is to find the LCD, which must contain both factors of x and x+3.  This LCD= x(x+3).


The second step is to build each fraction so it's denominator is x(x+3):
{{{(3/(x+3))*(x/x) + (1/x)*((x+3)/(x+3)) }}}
{{{(3x+ x+3)/(x(x+3)) }}}
{{{(4x+3)/(x(x+3)) }}}


The fraction does not reduce.


R^2 at SCC