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

Question

1673 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-21T13:58:57

$y = 5$ $y = 5$
$x = \pm 4$ $x=+-4$

First substitute $y = 5$ $y=5$ into $y^{2} = x^{2} + 9$ $y^2=x^2+9$

You get: $5^{2} = x^{2} + 9$ $5^2 = x^2 +9$

Square the 5 and take 9 away from both sides: $25 - 9 = x^{2}$ $25-9=x^2$

Simplify: $16 = x^{2}$ $16=x^2$
Therefore $x = \pm \sqrt{16}$ $x=+-sqrt(16)$
So $x = \pm 4$ $x=+-4$

You already know that $y = 5$ $y = 5$ and now you know that $x = \pm 4$ $x=+-4$

(Techincally $y = \pm 5$ $y = +-5$ as $5^{2} = 25$ $5^2=25$ and ${(- 5)}^{2} = 25$ $(-5)^2 = 25$ which are the same answers.)

How do you find the exact solutions to the system y=5y=5y=5 and y^2=x^2+9y2=x2+9y^2=x^2+9?