Solving a rational equation with multiple and nested fractions 
This is the equation to solve: $\dfrac{\dfrac{x+\dfrac{1}{2}} {\dfrac{1}{2}+\dfrac{x}{3}}}{\dfrac{1}{4}+\dfrac{x}{5}}=3$

What I did:
$x+\dfrac{1}{2}=\dfrac{2x+1}{2}$
$\dfrac{x}{3}+\dfrac{1}{2}=\dfrac{2x+3}{6}$
$\dfrac{x}{5}+\dfrac{1}{4}=\dfrac{4x+5}{20}$
$\dfrac{2x+1}{2}\div \dfrac{2x+3}{6} = \dfrac{2x+1}{2}\times \dfrac{6}{2x+3}=   \dfrac{6x+3}{2x+3}$ 
$\dfrac{6x+3}{2x+3}\div\dfrac{4x+5}{20}=\dfrac{6x+3}{2x+3}\times\dfrac{20}{4x+5}=3$
$\implies\dfrac{120x+60}{(2x+3)(4x+5)}=3$
I know how to solve the equation. But right now I tried several times and I got wrong answers.
So I appreciate your help. Thank you.

$$\frac { \frac { 2x+1 }{ \frac { 2 }{ \frac { 3+2x }{ 6 }  }  }  }{ \frac { 5+4x }{ 20 }  } =3\quad \Rightarrow \frac { \frac { 2x+1 }{ 2 } \cdot \frac { 6 }{ 2x+3 }  }{ \frac { 5+4x }{ 20 }  } =3\quad \Rightarrow \frac { \frac { 2x+1 }{ 2x+3 }  }{ \frac { 5+4x }{ 20 }  } =1\Rightarrow \frac { 2x+1 }{ 2x+3 } =\frac { 5+4x }{ 20 } \Rightarrow \\ \Rightarrow 8{ x }^{ 2 }-18x-5=0\Rightarrow x=\frac { 9\pm 11 }{ 8 } $$
