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If: x -y = 2 then x = 2 +y

If: x^2 -4y^2 = 5 then (2 +y)^2 -4y^2 -5 = 0

Removing brackets: 4 +4y +y^2 -4y^2 -5 = 0

Collecting like terms: 4y -3y^2 -1 = 0

Divide all terms by -3: y^2 -4/3y +1/3 = 0

Completing the square: (y -2/3)^2 -4/9 +1/3 = 0 => (y -2/3)^2 = 1/9

Square root both sides: y -2/3 = -1/3 or +1/3

Add 2/3 to both sides: y = 1/3 or 1

By substitution into original equation intersection are at: (7/3, 1/3) and (3, 1)

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x - y = 2 => x = y + 2Then x2 - 4y2 = 5 => (y + 2)2 - 4y2 = 5 => y2 + 4y + 4 - 4y2 = 5 => 3y2 - 4y + 1 = 0

Dividing by 3 gives y2 - (4/3)y + 1/3 = 0

y2 - (4/3)y = - 1/3

(y - 2/3)2 = (2/3)2 - 1/3 = 4/9 - 1/3 = 1/9

Taking square roots: y - 2/3 = +/- 1/3

and so y = 2/3 +/- 1/3 => y = 1 or y = 1/3

Then x = y + 2 => x = 3 or x = 21/3.


So the points are (1, 3) and (21/3 , 1/3)


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Q: Where are the points of intersection of the line x -y equals 2 with the curve x2 - 4y2 equals 5 by completing the square?
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