How do you solve the system $3 x^{2} - y^{2} = - 6$ $3x^2-y^2=-6$ and $y = 2 x + 1$ $y=2x+1$ ?

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Answer 1 · 2016-08-29T17:34:59

$(x, y) = (1, 3), or, (- 5, - 9)$ $(x,y)=(1,3), or, (-5,-9)$ .

We submit $y = 2 x + 1$ $y=2x+1$ in the first eqn. and get,

$3 x^{2} - {(2 x + 1)}^{2} + 6 = 0$ $3x^2-(2x+1)^2+6=0$

$\Rightarrow - x^{2} - 4 x + 5 = 0, or, x^{2} + 4 x - 5 = 0$ $rArr -x^2-4x+5=0, or, x^2+4x-5=0$ .

$\Rightarrow (x - 1) (x + 5) = 0$ $rArr (x-1)(x+5)=0$

$\Rightarrow x = 1, or, - 5$ $rArr x=1, or, -5$ .

$\Rightarrow y = 2 x + 1 = 3, or, - 9$ $rArr y=2x+1=3, or, -9$

Therefore, the solns. are $(x, y) = (1, 3), or, (- 5, - 9)$ $(x,y)=(1,3), or, (-5,-9)$ , which satisfy the set

of eqns.

How do you solve the system 3x2−y2=−63x^2-y^2=-6 and y=2x+1y=2x+1?