#### Answer

$(\sqrt 5, 2\sqrt 5)$ and $(-\sqrt 5, -2\sqrt 5)$

#### Work Step by Step

We can use the substitution method to solve this system of equations. The second equation has already isolated the $y$ term. We will use this expression to substitute for $y$ in the 1st equation:
$x^{2} + (2x)^{2} = 25$
Simplify the equation:
$x^{2} + 4x^{2} = 25$
Add like terms:
$5x^{2} = 25$
Divide both sides by 5:
$x^{2} = 5$
Take the square root of both sides:
$x = \sqrt 5$ or $x = -\sqrt 5$
Now that we have the values for $x$, we can substitute these values into the second equation:
$y = 2(\sqrt 5)$ or $y = 2(-\sqrt 5)$
$y = 2\sqrt 5$ or $y = -2\sqrt 5$
The points where both equations intersect are $(\sqrt 5, 2\sqrt 5)$ and
$(-\sqrt 5, -2\sqrt 5)$.