How do you find the exact solutions to the system $5x^2+y^2=30$ and $9x^2-y^2=-16$ ?

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Answer 1 · 2016-11-13T15:20:54

The solution set is ${-1, 5}; {-1, -5}; {1, 5}; {1, -5}$

$y^2 = 30 - 5x^2 -> 9x^2 - (30 - 5x^2) = -16$

$9x^2 - 30 + 5x^2 + 16 = 0$

$14x^2 - 14 = 0$

$14(x^2 - 1) = 0$

$x^2 - 1 = 0$

$(x + 1)(x- 1) = 0$

$x = -1 and 1$

Substitute back into one of the original equations.

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

$5(-1)^2 + y^2 = 30 and 5(1)^2 + y^2 = 30$

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

$y^2 = 25 and y^2 = 25$

$y = +-5 and y = +-5$

So, the solution sets are as follows:

${-1, 5}; {-1, -5}; {1, 5}; {1, -5}$

Hopefully this helps!

How do you find the exact solutions to the system 5x^2+y^2=305x^2+y^2=30 and 9x^2-y^2=-169x^2-y^2=-16?