jmc

We notice that the denominator on the left factors, giving us $$\frac{5x+1}{(2x-1)(x+3)}=\frac{2x}{2x-1}.$$As long as $x\ne \frac{1}{2}$ we are allowed to cancel $2x-1$ from the denominators, giving $$\frac{5x+1}{x+3}=2x.$$Now we can cross-multiply to find $$5x+1=2x(x+3)=2x^2+6x.$$We simplify this to $$2x^2+x-1=0$$and then factor to $$(x+1)(2x-1)=0.$$Notice that since $2x-1$ is in the denominator of the original equation, $x=\frac{1}{2}$ is an extraneous solution. However $x=\boxed{-1}$ does solve the original equation.