Answer
$$x=-2$$
Work Step by Step
$$\frac{2x+1}{x-2}+\frac{3}{x}=-\frac{6}{x(x+2)}$$
$$(x(x-2))(\frac{2x+1}{x-2}+\frac{3}{x})=(x(x-2))(-\frac{6}{x(x+2)})$$
$$x(2x+1)+3(x-2)=-6$$
$$2x^2+x+3x-6=-6$$
$$2x^2+4x=0$$
$$x^2+2x=0$$
$$x(x+2)=0$$
$$x=0,-2$$
0 is an extraneous solution.
$$x=-2$$
