Solve the following system of equations: $x^2 + y^2 = 41$ $x^2 - y = 41$

1. Problem Description

Solve the following system of equations:

$x^2 + y^2 = 41$

$x^2 - y = 41$

2. Solution Steps

We are given two equations:

$x^2 + y^2 = 41$ (1)

$x^2 - y = 41$ (2)

From equation (2), we can express $x^2$ in terms of $y$:

$x^2 = y + 41$ (3)

Substitute (3) into equation (1):

$(y + 41) + y^2 = 41$

$y^2 + y + 41 - 41 = 0$

$y^2 + y = 0$

$y(y+1) = 0$

So we have two possible values for $y$:

$y = 0$ or $y = -1$

Case 1: $y = 0$

Substitute $y = 0$ into equation (3):

$x^2 = 0 + 41$

$x^2 = 41$

$x = \pm \sqrt{41}$

Case 2: $y = -1$

Substitute $y = -1$ into equation (3):

$x^2 = -1 + 41$

$x^2 = 40$

$x = \pm \sqrt{40} = \pm 2\sqrt{10}$

Therefore, the solutions are $(\sqrt{41}, 0)$, $(-\sqrt{41}, 0)$, $(2\sqrt{10}, -1)$, and $(-2\sqrt{10}, -1)$.

3. Final Answer

The solutions are $x = \sqrt{41}, y = 0$; $x = -\sqrt{41}, y = 0$; $x = 2\sqrt{10}, y = -1$; $x = -2\sqrt{10}, y = -1$.

Final Answer: The solutions are $(\sqrt{41}, 0)$, $(-\sqrt{41}, 0)$, $(2\sqrt{10}, -1)$, and $(-2\sqrt{10}, -1)$.