How do you solve the system $3x^2 - y^2 =1 1$ , $x^2 + 2y = 2$ ?

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Answer 1 · 2018-03-07T19:36:01

Pairs of the solution are $(-2sqrt3, -5)$ , $(-2, -1)$ , $(2, -1)$ and $(2sqrt3, -5)$

$3*(x^2+2y)-(3x^2-y^2)=2*3-11$

$3x^2+6y-3x^2+y^2=-5$

$y^2+6y=-5$

$y^2+6y+5=0$

$(y+1)*(y+5)=0$

$y_1=-5$ and $y_2=-1$

For $y=-5$ , $x^2+2*(-5)=2$ , so $x^2=12$

Hence $x_1=-2sqrt3$ and $x_2=2sqrt3$

For $y=-1$ , $x^2+2*(-1)=2$ , so $x^2=4$

Hence $x_3=-2$ and $x_4=2$

Thus, pairs of the solution are $(-2sqrt3, -5)$ , $(-2, -1)$ , $(2, -1)$ and $(2sqrt3, -5)$

How do you solve the system 3x^2 - y^2 =1 1 3x^2 - y^2 =1 1 , x^2 + 2y = 2x^2 + 2y = 2?