We have
[tex]\frac{x-2}{x+3}+\frac{10x}{x{}^2-9}[/tex]first, we need to factorize the next term
[tex]x^2-9=(x+3)(x-3)[/tex]so we have
[tex]\frac{x-2}{x+3}+\frac{10x}{(x+3)(x-3)}[/tex]Remember in order to sum a fraction the denominator must be the same
[tex]\frac{(x-2)(x-3)+10x}{(x+3)(x-3)}[/tex]then we solve the multiplications (x-2)(x-3)
[tex]\frac{x^2-3x-2x+6+10x}{(x+3)(x-3)}=\frac{x^2+5x+6}{(x+3)(x-3)}[/tex]then we can factorize the numerator
[tex]x^2+5x+6=(x+3)(x+2)[/tex]so the simplification will be
[tex]\frac{x^2+5x+6}{(x+3)(x-3)}=\frac{(x+3)(x+2)}{(x+3)(x-3)}=\frac{(x+2)}{(x-3)}[/tex]the final result is
[tex]\frac{(x+2)}{(x-3)}[/tex]