How do you solve the system $x + 2 y^{2} = - 6$ $x+2y^2=-6$ and $x + 8 y = 0$ $x+8y=0$ ?

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Answer 1 · 2016-07-17T08:59:21

Solution set is $(- 8, 1)$ $(-8,1)$ or $(- 24, 3)$ $(-24,3)$ .

The set of equations can be solved using substitution method. The second equation gives us $x = - 8 y$ $x=-8y$ and putting this in first we get

$(- 8 y) + 2 y^{2} = - 6$ $(-8y)+2y^2=-6$ or

$2 y^{2} - 8 y + 6 = 0$ $2y^2-8y+6=0$ and diving by $2$ $2$ we get

$y^{2} - 4 y + 3 = 0$ $y^2-4y+3=0$ or

$y^{2} - 3 y - y + 3 = 0$ $y^2-3y-y+3=0$ or

$y (y - 3) - 1 (y - 3) = 0$ $y(y-3)-1(y-3)=0$ or

$(y - 1) (y - 3) = 0$ $(y-1)(y-3)=0$ i.e. either

$y - 1 = 0$ $y-1=0$ or $y = 1$ $y=1$ and $x = - 8$ $x=-8$ or

$y - 3 = 0$ $y-3=0$ or $y = 3$ $y=3$ and $x = - 24$ $x=-24$

Hence solution set is $(- 8, 1)$ $(-8,1)$ or $(- 24, 3)$ $(-24,3)$ .

How do you solve the system x+2y^2=-6x+2y2=−6x+2y^2=-6 and x+8y=0x+8y=0x+8y=0?