## College Algebra 7th Edition

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