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

2. Solution Steps

We can solve this system of equations by adding the two equations to eliminate $y$.

Adding the two equations gives:

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

$2x^2 = 50$

Dividing both sides by 2:

$x^2 = 25$

Taking the square root of both sides:

$x = \pm 5$

Now we can substitute the values of $x$ back into either of the original equations to solve for $y$. Let's use the first equation $x^2 + y^2 = 40$.

If $x = 5$, then

$(5)^2 + y^2 = 40$

$25 + y^2 = 40$

$y^2 = 40 - 25$

$y^2 = 15$

$y = \pm \sqrt{15}$

If $x = -5$, then

$(-5)^2 + y^2 = 40$

$25 + y^2 = 40$

$y^2 = 40 - 25$

$y^2 = 15$

$y = \pm \sqrt{15}$

Therefore, we have four solutions: $(5, \sqrt{15})$, $(5, -\sqrt{15})$, $(-5, \sqrt{15})$, and $(-5, -\sqrt{15})$.

3. Final Answer

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