We are given a system of two equations: $x^2 + y^2 = 40$ $x^2 - y^2 = 10$ We are asked to solve for $x$ and $y$.

1. Problem Description

We are given a system of two equations:

$x^2 + y^2 = 40$

$x^2 - y^2 = 10$

We are asked to solve for $x$ and $y$.

2. Solution Steps

We can solve this system of equations by adding the two equations.

$(x^2 + y^2) + (x^2 - y^2) = 40 + 10$

$2x^2 = 50$

$x^2 = \frac{50}{2}$

$x^2 = 25$

Taking the square root of both sides, we get

$x = \pm \sqrt{25}$

$x = \pm 5$

Now we can substitute the values of $x^2$ into either equation to solve for $y^2$. Let's use the first equation:

$x^2 + y^2 = 40$

$25 + y^2 = 40$

$y^2 = 40 - 25$

$y^2 = 15$

Taking the square root of both sides, we get

$y = \pm \sqrt{15}$

Therefore, the solutions are $(5, \sqrt{15})$, $(5, -\sqrt{15})$, $(-5, \sqrt{15})$, and $(-5, -\sqrt{15})$.

3. Final Answer

The solutions are $x = \pm 5$ and $y = \pm \sqrt{15}$.

The solution set is $\{(5, \sqrt{15}), (5, -\sqrt{15}), (-5, \sqrt{15}), (-5, -\sqrt{15})\}$.