We need to find the solution of the system of equations: $x + \sqrt{y} = 1$ $x^2 + y = 25$

1. Problem Description

We need to find the solution of the system of equations:

$x + \sqrt{y} = 1$

$x^2 + y = 25$

2. Solution Steps

From the first equation, we can express $x$ in terms of $y$:

$x = 1 - \sqrt{y}$

Substitute this into the second equation:

$(1 - \sqrt{y})^2 + y = 25$

$1 - 2\sqrt{y} + y + y = 25$

$2y - 2\sqrt{y} + 1 = 25$

$2y - 2\sqrt{y} - 24 = 0$

$y - \sqrt{y} - 12 = 0$

Let $z = \sqrt{y}$. Then $z^2 = y$. Substituting, we get:

$z^2 - z - 12 = 0$

$(z - 4)(z + 3) = 0$

$z = 4$ or $z = -3$

Since $z = \sqrt{y}$, $z$ must be non-negative. Thus, $z = 4$.

Then, $\sqrt{y} = 4$, so $y = 4^2 = 16$.

Now, we can find $x$:

$x = 1 - \sqrt{y} = 1 - \sqrt{16} = 1 - 4 = -3$.

Therefore, the solution is $(x, y) = (-3, 16)$.

3. Final Answer

(-3, 16)