Solve this system.?

Question

Solve the system:
2x2 - 3y2 = 16
x2 + 2y2 = 15.

Answer 1

2x² - 3y² = 16
x² + 2y² = 15

2x² - 3y² = 16
2x² + 4y² = 30
--------------------
.......7y² = 14
y² = 2
y = ±√2

x² +2(2) = 15
x² = 11
x = ±√11

4 symmetric points of intersection:
(±√11, ±√2)

Answer 2

2x2 - 3y2 = 16
x2 + 2y2 = 15

multiply the second equation by 2, then subtract:
2x^2 - 3y^2 = 16
2x^2 + 4y^2 = 30

-7y^2 = =-14
y^2 = 2
y= +/-sqrt2
now substitute back into equation one:
2x^2 - 3y^2 = 16
2x^2-3(2) =16
2x^2 = 16+6
x^2 = 11
x= +/- sqrt(11)

Answer 3

2x^2 - 3y^2 = 16
-
x^2 + 2y^2 = 15
----------------------
x^2 - 5y^2 = 1

use the resulting equation to solve for x^2, then, substitute it to any of the first two equations...

x^2 = 1 + 5y^2

x^2 + 2y^2 = 15
1 + 5y^2 + 2y^2 = 15
1 + 7y^2 = 15
y = 2^(1/2)

substitute the y value

2x^2 - 3y^2 = 16
2x^2 - 3(2^(1/2))^2 = 16
2x^2 - 6 = 16
x = 11^(1/2)