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

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Answer 1 · 2016-07-17T08:59:21

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

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

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

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

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

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

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

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

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

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

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

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