$\large \begin{align} \color{black}{\normalsize \text{Solve }\hspace{.33em}\~\
3x^2+2y^2=5xy~~and~~ 12x+12y=5xy \hspace{.33em}\~\
x,y\in \mathbb{R} }\end{align}$

Question

$\large \begin{align} \color{black}{\normalsize \text{Solve }\hspace{.33em}\~\
3x^2+2y^2=5xy~~and~~ 12x+12y=5xy   \hspace{.33em}\~\
x,y\in \mathbb{R} }\end{align}$

freckles · Answer

12x+12y=5xy
12x=5xy-12y
12x=y(5x-12)
y=12x/(5x-12)
So I would try to replace the y's in the first equation with 12x/(5x-12)
and solve for x

mathmath333 · Answer

painful

freckles · Answer

you think we can find a pretty trick?

freckles · Answer

like it is one of those questions?

mathmath333 · Answer

yes i hope

freckles · Answer

well like I don't know if this helps but we have an ellispe if we write 3x^2+2y^2=12x+12y

mathmath333 · Answer

if it were only integers not real numbers then how could we do it

mathmath333 · Answer

wolfram does gives two pairs of integers http://www.wolframalpha.com/input/?i=solve+3x%5E2%2B2y%5E2%3D5xy%2C12x%2B12y%3D5xy+over+reals

ganeshie8 · Answer

$3x^2+2y^2=5xy$ factors like below 
$$(x-y)(3x-2y)=0$$

we get two cases :
when $x =y$, secod equation becomes $$12x+12x=5x\cdot x \implies   x = 0, ~\frac{24}{5}=y $$

you an work the other case similarly..

ganeshie8 · Answer

fun fact : 
$ax^2+by^2+cxy=0$ represents a pair of straight lines passing through origin, so we can always factor it as below product :
$$ax^2+by^2+cxy =0= (y-m_1x)(y-m_2x)$$

mathmath333 · Answer

i will rote this fact