How do you find the exact solutions to the system $x^2/30+y^2/6=1$ and $x=y$ ?

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Answer 1 · 2016-11-23T05:53:24

$x=y=-sqrt5$ or $x=y=sqrt5$

$y^2/30+y^2/6=1$

$y^2/30xx30+y^2/6xx30=1xx30$

$y^2+5y^2=30$

$6y^2=30$

$y^2=30/6=5$

$y^2-5=0$

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

Hence, either $y+sqrt5=0$ i.e. $y=-sqrt5$

or $y-sqrt5=0$ i.e. $y=sqrt5$

Hence, $x=y=-sqrt5$ or $x=y=sqrt5$

How do you find the exact solutions to the system x^2/30+y^2/6=1x^2/30+y^2/6=1 and x=yx=y?