please help me solve this nonlinear system x^2 + 4y^2 = 116 and x^2 - 3y^2= -59?

Answer 1

x^2 + 4y^2 = 116 ---------- (1)
x^2 - 3y^2 = -59 ---------- (2)

(1) - (2):
7y^2 = 175
y^2 = 25
y = -5 or 5

Substitute y = -5 into (1)
x^2 + 4(-5)^2 = 116
x^2 + 100 = 116
x^2 = 16
x = -4 or 4

Substitute y = 5 into (1)
x^2 + 4(5)^2 = 116
x^2 + 100 = 116
x^2 = 16
x = -4 or 4

Hence,
when y = -5, x = -4 or 4
when y = 5, x = -4 or 4

Answer 2

Subtract the equations to eliminate x^2. This gives
7y^2 = 175
<=> y^2 = 25 <=> y = ± 5
Substituting this into either of the other equations gives x^2 = 16 <=> x = ± 4.

So the four solutions are
x = 4, y = 5
x = 4, y = -5
x = -4, y = 5
x = -4, y = -5.

Answer 3

x = 4 and -4, y = 5 and -5

Answer 4

x^2 + 4y^2 = 116 .............(1)
x^2 - 3y^2 = -59 ...............(2)

Substract (2) from (1),

x^2 + 4y^2 - x^2 + 3y^2 = 116 + 59
7y^2 = 175
y^2 = 25
y = 5 .................(3)

Substitute (3) into (1),
x^2 + 4(5^2) = 116
x^2 + 100 = 116
x^2 = 16
x = 4

Answer 5

Third responder is correct. What we have are an ellipse and a hyperbola; these intersect at four points. The solution can be easily visualized by letting w = x^2, v = y^2, which gives you a linear system which can be solved routinely.

Answer 6

x² = 3y² - 59

3y² - 59 + 4y² = 116
7y² = 175
y² ~ 11