Solve the following system over the set of real numbers: 2x^2 - 3y^2= -12, x^2 +4y^2 = 49

Answer 1

Let a = x^2 and b = y^2 an d you are solving a linear equation,

2a - 3b = -12

a + 4b = 49

Doubling the bottom equation, 2a + 8b = 98

Then, subtracting the top equation from the bottom, we get

11b = 110

b = 10

Substituting, a + 4(10) = 49

a = 9

If a = 9, x^2 = 9, so x = ±3.

f b = 10, y^2 = 10, so y = ±√10

The solution is {(-3, -√10), (-3, √10), (3, -√10), (3, √10)}