Solve the following equations for real \(x\) and \(y\):

1) \(x^2 + y^2 = 25\)
2) \(x + y = 7\)

To solve this system of equations, we can use substitution or elimination method.

From equation 2, we can express \(y\) in terms of \(x\):
\(y = 7 - x\)

Substitute this into equation 1:
\(x^2 + (7 - x)^2 = 25\)
\(x^2 + 49 - 14x + x^2 = 25\)
\(2x^2 - 14x + 24 = 0\)
\(x^2 - 7x + 12 = 0\)
\((x - 3)(x - 4) = 0\)

So, \(x = 3\) or \(x = 4\)

If \(x = 3\), then \(y = 7 - 3 = 4\)
If \(x = 4\), then \(y = 7 - 4 = 3\)

Therefore, the solutions are \(x = 3, y = 4\) and \(x = 4, y = 3\).

$$\begin{align}\begin{cases}x+2y &=3 \\\ -x+y &=5 \end{cases}\end{align}$$

Adding the two equations, we have
$$3y = 8 \quad \Rightarrow \quad y = \frac{8}{3}.$$

Substituting this into the first equation of (1), we get
$$x + 2\left(\frac{8}{3}\right) = 3 \quad \Rightarrow \quad x = 3 - \frac{16}{3} = -\frac{1}{3}.$$

Therefore, the solution to the system of equations is
$$x = -\frac{1}{3}, \quad y = \frac{8}{3}.$$