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

Marisol Rivers
2022-07-22
Answered

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

Answered 2022-07-23
Author has **10** answers

$\frac{{y}^{2}}{30}+\frac{{y}^{2}}{6}=1$

$\frac{{y}^{2}}{30}\times 30+\frac{{y}^{2}}{6}\times 30=1\times 30$

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

$6{y}^{2}=30$

${y}^{2}=\frac{30}{6}=5$

${y}^{2}-5=0$

$(y+\sqrt{5})(y-\sqrt{5})=0$

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 $x=y=\sqrt{5}$

$\frac{{y}^{2}}{30}\times 30+\frac{{y}^{2}}{6}\times 30=1\times 30$

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

$6{y}^{2}=30$

${y}^{2}=\frac{30}{6}=5$

${y}^{2}-5=0$

$(y+\sqrt{5})(y-\sqrt{5})=0$

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 $x=y=\sqrt{5}$

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