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

How do you find the exact solutions to the system $\frac{{x}^{2}}{30}+\frac{{y}^{2}}{6}=1$ and x=y?
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Jaylene Tyler
$\frac{{y}^{2}}{30}+\frac{{y}^{2}}{6}=1$
$\frac{{y}^{2}}{30}×30+\frac{{y}^{2}}{6}×30=1×30$
${y}^{2}+5{y}^{2}=30$
$6{y}^{2}=30$
${y}^{2}=\frac{30}{6}=5$
${y}^{2}-5=0$
$\left(y+\sqrt{5}\right)\left(y-\sqrt{5}\right)=0$
Hence, either $y+\sqrt{5}=0$ i.e. $y=-\sqrt{5}$
or $y-\sqrt{5}=0$ i.e. $y=\sqrt{5}$
Hence, $x=y=-\sqrt{5}$ or $x=y=\sqrt{5}$