How do you solve the system $10y=x^2$ and $x^2-6=-2y$ ?

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Answer 1 · 2016-07-15T07:26:18

We have two solutions $(-sqrt5,1/2)$ and $(sqrt5,1/2)$

As $10y=x^2$ and $x^2-6=-2y$ , putting value of $x^2$ from first equation into second we get

$10y-6=-2y$ or

$10y+2y+6-6=-2y+2y+6$ or

$12y=6$ or

$y=6/12=1/2$ .

Hence, $x^2=10×1/2=5$ or

$x^2-5=0$ or

$x^2-(sqrt5)^2=0$ or

$(x-sqrt5)(x+sqrt5)=0$

Hence, either $x-sqrt5=0$ i.e. $x=sqrt5$

or $x+sqrt5=0$ i.e. $x=-sqrt5$

Hence we have two solutions $(-sqrt5,1/2)$ and $(sqrt5,1/2)$

How do you solve the system 10y=x^210y=x^2 and x^2-6=-2yx^2-6=-2y?