Find the real solution of the system 3x^2-2y^2=-5 xy=12

This is a homogenous system and the first step is to
eliminate the numbers alone.

For this purpose, we'll
multiply the 1st equation by 12 and the 2nd equation by
5:

36x^2 - 24y^2 = -60 (3)

5xy
= 60 (4)

We'll add (3) and
(4):

36x^2 + 5xy - 24y^2 =
0

We'll divide by x^2:

36 +
5y/x - 24y^2/x^2 = 0

We'll substitute y/x =
t:

-24t^2 + 5t + 36 = 0

24t^2
- 5t - 36 = 0

We'll apply quadratic
formula:

t1 =
[5+sqrt(25+3456)]/48

t1 =
(5+59)/48

t1 = 4/3

t2 =
(5-59)/48

t2 = -9/8

We'll put
y/x = 4/3

y = 4x/3

We'll
substitute y in the 2nd equation:

x*(4x/3) =
12

4x^2 = 4*9

x^2 =
9

x1 = 3 and x2 = -3

y1 = 4
and y2 = -4

y/x = -9/8

y =
-9x/8

-9x^2 = 8*12

-3x^2 =
32

x^2 = -32/3

This equation
has no real solutions.

The system will have
the following real solutions {(3 ; 4) ; (-3 ;
-4)}.