Solve quadratic system

X^2/16 + y^2/4 = 1
x^2/4 - y^2/3 = 1

(x^2)/16 + (y^2)/4 = 1
(x^2)/16 + (4y^2)/16 = 1
(x^2+4y^2)/16 = 1
16(x^2+4y^2)/16 = 1(16)
x^2+4y^2 = 16
x^2+4y^2-4y^2 = 16-4y^2
x^2 = 16-4y^2
√x^2 = √(16-4y^2)
x = √(16-4y^2)
x = √(4(4-y^2))
x = √4√(4-y^2)
x = 2√(4-y^2)

(x^2)/4 - (y^2)/3 = 1
(3x^2)/12 - (4y^2)/12 = 1
(3x^2-4y^2)/12 = 1
12(3x^2-4y^2)/12 = 1(12)
3x^2-4y^2 = 12
3x^2-4y^2+4y^2 = 12+4y^2
3x^2 = 12+4y^2
(3x^2)/3 = (12+4y^2)/3
x^2 = (12+4y^2)/3
√x^2 = √((12+4y^2)/3)
x = √(4(3+y^2)/3)
x = √4√((3+y^2)/3)
x = 2√((3+y^2)/3)

2√(4-y^2) = 2√((3+y^2)/3)
2√(4-y^2)/2 = 2√((3+y^2)/3)/2
√(4-y^2) = √((3+y^2)/3)
√(4-y^2)^2 = √((3+y^2)/3)^2
4-y^2 = (3+y^2)/3
3(4-y^2) = 3(3+y^2)/3
12-3y^2 = 3+y^2
12-3y^2+3y^2 = 3+y^2+3y^2
12 = 3+4y^2
12-12 = 3+4y^2-12
0 = 4y^2-9
0 = (2y+3)(2y-3)
Either 2y+3=0 or 2y-3=0.
Therefore y=-3/2 or y=3/2.
y = +-3/2

(x^2)/16 + (y^2)/4 = 1
(x^2)/16 + ((+-3/2)^2)/4 = 1
(x^2)/16 + (9/4)/4 = 1
(x^2)/16 + 4(9/4)/16 = 1
(x^2)/16 + 9/16 = 1
(x^2+9)/16 = 1
16(x^2+9)/16 = 1(16)
x^2+9 = 16
x^2+9-9 = 16-9
x^2 = 7
√x^2 = √7
x = +-7

Answer: x=+-7 and y=+-3/2

Answer: x=+-7 and y=+-3/2 is incorrect.

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X^2/16 + y^2/4 = 1 ......(1)
x^2/4 - y^2/3 = 1 ......(2)
4*(1) - (2): (4/3)y^2 = 3
y = +/- 3/2
x^2 = 4(1+3/4) = 7
Answer: Four solutions (+/-sqrt(7), +/-3/2)

x^2/16 + y^2/4 = 1
x^2/4 - y^2/3 = 1

x^2 + 4y^2 = 16
3x^2 - 4y^2 = 12
4x^2 = 28
x^2 = 7
x = ±√7
x = -√7 , y = ± 3/2
x = √7 , y = ± 3/2