Can You Solve This Real Number System of Equations?

[anemone]

Here is this week's POTW:

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Let $a,\,b,\,c$ and $d$ be the real numbers which satisfy the system of equations below:

$a+b+2ab=3$
$b+c+2bc=4$
$c+d+2cd=5$

Evaluate $d+a+2da$.

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[anemone]

Congratulations to the following members for their correct solution!(Cool)

1. castor28
2. Opalg
3. Olinguito
4. kaliprasad

Solution from castor28:

Spoiler

We write the equations as:
\begin{align*}
a &= \frac{3-b}{2b+1}&[1]\\
b &= \frac{4-c}{2c+1}&[2]\\
c &= \frac{5-d}{2d+1}&[3]
\end{align*}
Substituting $b$ from $[2]$ into $[1]$ gives:
$$
a = \frac{7c-1}{9}
$$
Substituting $c$ from $[3]$ in this equation gives:
$$
a = \frac{34-9d}{9(2d+1)}
$$
Finally, substituting $a$ from this equation into $d+a+2da$ gives:
\begin{align*}
d+a+2da &= d + \frac{34-9d}{9(2d+1)} + 2d\frac{34-9d}{9(2d+1)}\\
&= d + \frac{(34-9d)(2d+1)}{9(2d+1)}\\
&= \bf\frac{34}{9}
\end{align*}

Alternate solution from Olinguito:

Spoiler

We write the equations as:
\begin{align*}
a &= \frac{3-b}{2b+1}&[1]\\
b &= \frac{4-c}{2c+1}&[2]\\
c &= \frac{5-d}{2d+1}&[3]
\end{align*}
Substituting $b$ from $[2]$ into $[1]$ gives:
$$
a = \frac{7c-1}{9}
$$
Substituting $c$ from $[3]$ in this equation gives:
$$
a = \frac{34-9d}{9(2d+1)}
$$
Finally, substituting $a$ from this equation into $d+a+2da$ gives:
\begin{align*}
d+a+2da &= d + \frac{34-9d}{9(2d+1)} + 2d\frac{34-9d}{9(2d+1)}\\
&= d + \frac{(34-9d)(2d+1)}{9(2d+1)}\\
&= \bf\frac{34}{9}
\end{align*}

Second alternate solution from Olinguito:

Spoiler

Let
$$A\ =\ \frac1{\sqrt2}+a\sqrt2 \\\\ B\ =\ \frac1{\sqrt2}+b\sqrt2 \\\\ C\ =\ \frac1{\sqrt2}+c\sqrt2 \\\\ D\ =\ \frac1{\sqrt2}+d\sqrt2.$$
Then
$$AB\ =\ \frac12+a+b+2ab\ =\ \frac72 \\\\ BC\ =\ \frac12+b+c+2bc\ =\ \frac92 \\\\ CD\ =\ \frac12+c+d+2cd\ =\ \frac{11}2.$$
$\implies\ AD\ =\ \dfrac{(AB)(CD)}{(BC)}\ =\ \dfrac{\left(\dfrac72\right)\left(\dfrac{11}2\right)}{\left(\dfrac92\right)}\ =\ \dfrac{77}{18}.$

Hence

$d+a+2da\ =\ DA-\dfrac12\ =\ \dfrac{77}{18}-\dfrac12\ =\ \boxed{\dfrac{34}9}.$