[tex]\frac{3}{x+2}+\frac{4}{x-3}=\frac{3(x-3)}{(x+2)(x-3)}+\frac{4(x+2)}{(x+2)(x-3)}=\frac{3x-9+4x+8}{(x+2)(x-3)}=\frac{7x-1}{(x+2)(x-3)}[/tex]
