Got a question for you guys

The FIRST equation simpifies to:  4xy + 4y^2=0

solve the second for x= 6-y/2    and substitute into the FIRST

4y(6-y/2) + 4 y^2 =0

24y -2^y^2 +4y^2=0

2y^2 +24y = 0

y (2y +24)= 0      so y = 0   or  -12      substitute these values into the second equation to get x = 6   or  12    so x,y  =  (6,0)    or (12,-12)

Simultaneous equations

$x^2+3xy+(2y)^2=x^2-xy\qquad (1a)\\ 3xy+4y^2=-xy \qquad \qquad (1b)\\ 4xy+4y^2=0 \qquad \qquad (1c)\\ 4y(x+y)=0 \qquad \qquad (1d)\\ \mbox{True if y=0 and if y=-x}\\ \mbox{solve simultaneously with}\\ 2x+y=12 \qquad \qquad (2a)\\ If\; \;\;y=0 \;\;then\\ 2x=12\\ x=6\\ \mbox{But if y=0 in the first equation then y=0 so this is a nonsense solution}\\ If \;\;y=-x\;\;then\\ -x=-2x+12\\ \mbox{The only solution is (12,-12)}$

Here is the graph:

https://www.desmos.com/calculator/r6fjzb1hbr

x^2 + 3xy + (2y)^2 =  x^2  - xy   

2x + y = 12    (1)

The first equation  simplifies to

4xy +  4y^2  = 0     divide both sides by 4

xy  + y^2   = 0     (2)

(1) can be transformed into     y = 12 - 2x

Subbing this into (2), we have

x (12 - 2x) + (12 - 2x)^2  = 0   simplify

12x - 2x^2  +  4x^2 - 48x + 144  = 0

2x^2 - 36x + 144 = 0      divide through by 2

x^2  - 18x + 72  = 0     factor

(x - 6) (x - 12)  = 0

Setting each factor to 0,  x = 6    or  x = 12

When x = 6,  y = 12 -2(6)  = 0

When x = 12, y = 12 - 2(12)  = -12

So.....the solutions are   (6, 0)    and  (12, -12)

These two solutions are verified by WolframAlpha, here :  http://www.wolframalpha.com/input/?i=xy+%2B+y^2+%3D+0+,+2x+%2B+y+%3D+12

LOL.

Actually, (8, -4) also works