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.
 A: You have, essentially,
$$
\frac{120x+60}{(2x+3)(4x+5)} = 3
$$
Multiply both sides by $(2x+3)(4x+5)$ to get
$$
120x+60 = 3(2x+3)(4x+5) = 24x^2+66x+45
$$
Subtract $120x+60$ from both sides to get
$$
24x^2-54x-15 = 0
$$
Divide both sides by $3$ to get
$$
8x^2-18x-5 = 0
$$
which factors as
$$
(4x+1)(2x-5) = 0
$$
to yield $x = -1/4$ or $5/2$.  (Or, you can use the quadratic formula.)
A: Notice, $$\frac{\frac{x+\frac{1}{2}}{\frac{1}{2}+\frac{x}{3}}}{\frac{1}{4}+\frac{x}{5}}=3$$
$$\frac{\frac{3(2x+1)}{(2x+3)}}{\frac{4x+5}{20}}=3$$
$$\frac{60(2x+1)}{(2x+3)(4x+5)}=3$$
$$8x^2-18x-5=0$$
$$x=\frac{18\pm\sqrt{(-18)^2-4(8)(-5)}}{2(8)}$$
$$x=\frac{18\pm22}{16}$$ $$x=\frac{5}{2}$$ or $$ x=-\frac{1}{4}$$
A: $$\frac { \frac { 2x+1 }{ \frac { 2 }{ \frac { 3+2x }{ 6 }  }  }  }{ \frac { 5+4x }{ 20 }  } =3\quad \Rightarrow \frac { \frac { 2x+1 }{ 2 } \cdot \frac { 6 }{ 2x+3 }  }{ \frac { 5+4x }{ 20 }  } =3\quad \Rightarrow \frac { \frac { 2x+1 }{ 2x+3 }  }{ \frac { 5+4x }{ 20 }  } =1\Rightarrow \frac { 2x+1 }{ 2x+3 } =\frac { 5+4x }{ 20 } \Rightarrow \\ \Rightarrow 8{ x }^{ 2 }-18x-5=0\Rightarrow x=\frac { 9\pm 11 }{ 8 } $$
