How do you solve simultaneous equations?

$\begin{cases} \sqrt x \left(1 - \dfrac{12}{3x+y} \right) = 2 & \implies x \left(1 - \dfrac{12}{3x+y} \right)^2 = 4 & ...(1) \\ \sqrt y \left(1 + \dfrac{12}{3x+y} \right) = 6 & \implies y \left(1 + \dfrac{12}{3x+y} \right)^2 = 36 & ...(2) \end{cases}$

$\begin{cases} (1): & \left(1 - \dfrac{12}{3x+y} \right)^2 = \dfrac 4x & \implies 1 - \dfrac {24}{3x+y} + \left(\dfrac{12}{3x+y} \right)^2 = \dfrac 4x & ...(1a) \\ (2): & \left(1 + \dfrac{12}{3x+y} \right)^2 = \dfrac {36}y & \implies 1 + \dfrac {24}{3x+y} + \left(\dfrac{12}{3x+y} \right)^2 = \dfrac {36}y & ...(2a) \end{cases}$

$\begin{aligned} (2a)-(1a): \quad \frac {48}{3x+y} & = \frac {36}y - \frac 4x \\ & = \frac {36x-4y}{xy} \\ 12xy & = (9x-y)(3x+y) \\ & = 27x^2 + 6xy - y^2 \\ 27x^2 - 6xy - y^2 & = 0 \\ (9x+y)(3x-y) & = 0 \end{aligned}$

$\implies \begin{cases} y = -9x & \color{#D61F06}{\text{rejected, } x \text{ or } y <0 \because \sqrt x, \sqrt y \text{ in the equations.}} \\ y = 3x & \color{#3D99F6}{\text{accepted}} \end{cases}$

Now, we have:

$\begin{aligned} (1): \quad x \left(1 - \dfrac{12}{3x+y} \right)^2 & = 4 \\ x \left(1 - \dfrac{12}{3x+3x} \right)^2 & = 4 \\ (x-2)^2 & = 4x \\ x^2 - 8x + 4 & = 0 \\ \implies x & = 4 + 2\sqrt 3 & \small \color{#D61F06}{x = 4-2\sqrt 3 \text{ is not a solution upon substituting in the equations.}} \\ y & = 12 + 6\sqrt 3 \end{aligned}$

$\implies a+b+c+d = 4+2+12+6 = \boxed{24}$

First we cube both sides and we will get,

x(1- $\frac{12}{3x+y})^2$ =4

and

y(1+ $\frac{12}{3x+y})^2$ =36

Then we rearrange it

(1- $\frac{12}{3x+y})^2$ = $\frac{4}{x}$

(1+ $\frac{12}{3x+y})^2$ = $\frac{36}{y}$

then we expand both equations

1- $\frac{24}{3x+y}$ + $\frac{144}{9x^2+6xy+y^2}$ = $\frac{4}{x}$

1+ $\frac{24}{3x+y}$ + $\frac{144}{9x^2+6xy+y^2}$ = $\frac{36}{y}$

Then we plus these equations, and we will get

2+ $\frac{288}{9x^2+6xy+y^2}$ = $\frac{4}{x}$ + $\frac{36}{y}$

But we can divide this equation by 2 and we will get

1+ $\frac{144}{9x^2+6xy+y^2}$ = $\frac{2}{x}$ + $\frac{18}{y}$

Then we can completing the square on the left hand side by

(1+ $\frac{12}{3x+y})^2$ - $\frac{24}{3x+y}$ = $\frac{2}{x}$ + $\frac{18}{y}$

But

(1+ $\frac{12}{3x+y})^2$ = $\frac{36}{y}$

Therefore our equation equal to

$\frac{36}{y}$ - $\frac{24}{3x+y}$ = $\frac{2}{x}$ + $\frac{18}{y}$

and we move $\frac{24}{3x+y}$ to the right side

$\frac{36}{y}=\frac{2}{x}$ + $\frac{18}{y}$ + $\frac{24}{3x+y}$

and we can simplified the right side

$\frac{2y+18x}{xy}$ + $\frac{24}{3x+y}$ or equl to $\frac{54x^2+48xy+2y^2}{3x^2y+xy^2}$

that mean

$\frac{36}{y}=\frac{54x^2+48xy+2y^2}{3x^2y+xy^2}$

and we cross multiplying and we will get

$108x^2$ $y$ + $36x$ $y^2$ = $54x^2$ $y$ + $48x$ $y^2$ + $2y^3$

and then we will get

$54x^2$ $y$ - $12x$ $y^2$ - $2y^3$ =0

and then we divided it by 2y

$27x^2$ - $6x$ $y$ - $y^2$ =0

then we can factories this equation

(9x+y)\(3x-y)=0

Therefore y=3x or y=-9x

Then we substitute y=-9x in this equation 1+\(\frac{144}{9x^2+6xy+y^2}\)= $\frac{2}{x}$ + $\frac{18}{y}$

we will get $x^2$ =-2 (which is not possible) so y=-9x is rejected

But if we substitute y=3x in this equation 1+ $\frac{144}{9x^2+6xy+y^2}$ = $\frac{2}{x}$ + $\frac{18}{y}$ we will get that,

$x^2$ - $8x$ + $4$ =0

and if we used quadratic formula we will get that

x=4+2 $\sqrt{3}$ and y=12+6 $\sqrt{3}$

or

x=4-2 $\sqrt{3}$ and y=12-6 $\sqrt{3}$

But if we substitute x=4-2 $\sqrt{3}$ and y=12-6 $\sqrt{3}$ into the original equation we will get -2 there fore x can not be 4-2 $\sqrt{3}$

Therefore the answer is

x=4+2 $\sqrt{3}$ and y=12+6 $\sqrt{3}$

Comparing coefficient a=4 b=2 c=12 d=6

a+b+c+d=24