## College Algebra (11th Edition)

$$x=-9$$
$$\frac{4}{(x-2)(x+3)}-\frac{1}{(x+2)(x-2)}=\frac{2}{(x+3)(x+2)}$$ $$((x-2)(x+3)(x+2))(\frac{4}{(x-2)(x+3)}-\frac{1}{(x+2)(x-2)})=((x-2)(x+3)(x+2))(\frac{2}{(x+3)(x+2)})$$ $$4(x+2)-(x+3)=2(x-2)$$ $$4x+8-x-3=2x-4$$ $$3x+5=2x-4$$ $$x=-9$$ -9 is not an extraneous solution; $x=-9$.