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x2 - 4y2 = 16
∴ (x - 2y)(x + 2y) = 16

2y - x = 2
∴ x = 2y - 2

∴ ([2y - 2] - 2y)([2y - 2] + 2y) = 16
∴ (y - 1 - y)(y - 1 + y) = 16
∴ -1(2y - 1) = 16
∴ 1 - 2y = 16
∴ -2y = 15
∴ y = -7.5

2y - x = -2
∴ -15 - x = -2
∴ x = -13

So the point of intersection is (-13, -7.5)

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Q: How do you find the solution to x2-4y2 equals 16 and 2y-x equals 2?
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