2·x + 3·y = 2
x = 1 - 1.5·y
Nun in die Kreisgleichung einsetzen
x^2 + y^2 - 4·x + 10·y - 36 = 0
(1 - 1.5·y)^2 + y^2 - 4·(1 - 1.5·y) + 10·y - 36 = 0
2.25·y^2 - 3·y + 1 + y^2 + 6·y - 4 + 10·y - 36 = 0
3.25·y^2 + 13·y - 39 = 0
Lösen mit abc-Formel
y = -6 --> x = 1 - 1.5·(-6) = 10
y = 2 --> x = 1 - 1.5·(2) = -2