Alg3.2.7

Solve $\frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{(x-3)(2x+3)}=0\,$

Multiplying through out by $(x-3)(2x+3)\,$ ,we get

$2x(2x+3)+1(x-3)+3x+9=0\,$

$4x^2+10x+6=0\,$

$2x^2+5x+3=0\,$

$(2x+3)(x+1)=0\,$

But

$(2x+3)\,$ not equal to zero,so we get $x+1=0,x=-1\,$

Hence $x=-1\,$ is the only solution of the given equation.