We are given a system of two equations: $x^2 + y^2 = 5$ $x^2 + x^2 + 3y^2 = 14$, which simplifies to $2x^2 + 3y^2 = 14$. We need to solve for $x$ and $y$.

1. Problem Description

We are given a system of two equations:

$x^2 + y^2 = 5$

$x^2 + x^2 + 3y^2 = 14$, which simplifies to $2x^2 + 3y^2 = 14$.

We need to solve for $x$ and $y$.

2. Solution Steps

We have the system of equations:

$x^2 + y^2 = 5$ (1)

$2x^2 + 3y^2 = 14$ (2)

From equation (1), we can express $x^2$ in terms of $y^2$:

$x^2 = 5 - y^2$ (3)

Substitute equation (3) into equation (2):

$2(5 - y^2) + 3y^2 = 14$

$10 - 2y^2 + 3y^2 = 14$

$y^2 = 14 - 10$

$y^2 = 4$

$y = \pm 2$

Now, substitute the values of $y$ into equation (3) to find the corresponding values of $x$:

If $y = 2$:

$x^2 = 5 - (2)^2 = 5 - 4 = 1$

$x = \pm 1$

If $y = -2$:

$x^2 = 5 - (-2)^2 = 5 - 4 = 1$

$x = \pm 1$

Therefore, the solutions are $(1, 2)$, $(1, -2)$, $(-1, 2)$, and $(-1, -2)$.

3. Final Answer

The solutions are $(1, 2)$, $(1, -2)$, $(-1, 2)$, and $(-1, -2)$.