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Question 250822: solve the system
y^2-x^2+2x-5=0
x^2+y^2-2x-3=0
Answer by drk(1908)   (Show Source):
You can put this solution on YOUR website! Here are the original equations:
(i) y^2 - x^2 + 2x - 5 = 0
(ii) x^2 + y^2 - 2x - 3 = 0.
Let's rewrite them so that each term is below its corresponding term:
solve the system
(iii) y^2 - x^2 + 2x - 5 = 0
((iv) y^2 + x^2 - 2x - 3 = 0.
You notice if you add down the X^2 and 2x parts drop out.
2Y^2 = 8.
Y^2 = 4
Y = +-2.
If Y = +2, then (ii) becomes
x^2 + 2^2 - 2x - 3 = 0
x^2 - 2x +1 = 0
(x-1)^2 = 0
X = 1.
If Y = -2, then (ii) becomes
x^2 + (-2)^2 - 2x -3 = 0
x^2 - 2x +1 = 0
(x-1)^2 = 0
X = 1.
So, our answers are (1,2) ; (1,-2).
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