We are given the following system of equations: $x^2 + y^2 = 29$ $3x^2 + 25y^2 = 100$ We need to solve for $x$ and $y$.

1. Problem Description

We are given the following system of equations:

$x^2 + y^2 = 29$

$3x^2 + 25y^2 = 100$

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

2. Solution Steps

Let the given equations be:

$x^2 + y^2 = 29$ (1)

$3x^2 + 25y^2 = 100$ (2)

Multiply equation (1) by 3:

$3(x^2 + y^2) = 3(29)$

$3x^2 + 3y^2 = 87$ (3)

Subtract equation (3) from equation (2):

$(3x^2 + 25y^2) - (3x^2 + 3y^2) = 100 - 87$

$22y^2 = 13$

$y^2 = \frac{13}{22}$

Substitute $y^2 = \frac{13}{22}$ into equation (1):

$x^2 + \frac{13}{22} = 29$

$x^2 = 29 - \frac{13}{22}$

$x^2 = \frac{29*22 - 13}{22} = \frac{638 - 13}{22} = \frac{625}{22}$

Now, solve for x and y:

$x = \pm \sqrt{\frac{625}{22}} = \pm \frac{25}{\sqrt{22}} = \pm \frac{25\sqrt{22}}{22}$

$y = \pm \sqrt{\frac{13}{22}} = \pm \frac{\sqrt{13}}{\sqrt{22}} = \pm \frac{\sqrt{13*22}}{22} = \pm \frac{\sqrt{286}}{22}$

The solutions are:

$x = \pm \frac{25\sqrt{22}}{22}, y = \pm \frac{\sqrt{286}}{22}$

3. Final Answer

$x = \pm \frac{25\sqrt{22}}{22}, y = \pm \frac{\sqrt{286}}{22}$