We can solve the system of equations \(m^2 + n^2 = 29\) and \(m + n = 7\) as follows:

From the second equation, we can solve for one of the variables in terms of the other. Let's solve for \(n\) in terms of \(m\):

\(n = 7 - m\)

Now, we can substitute this expression for \(n\) into the first equation to get:

\(m^2 + (7 - m)^2 = 29\)

\(m^2 + 49 - 14m + m^2 = 29\)

\(2m^2 - 14m + 20 = 0\)

This is a quadratic equation in \(m\), which we can solve using the quadratic formula:

\(m = \frac{-(-14) \pm \sqrt{(-14)^2 - 4(2)(20)}}{2(2)}\)

\(m = \frac{14 \pm \sqrt{196 - 160}}{4}\)

\(m = \frac{14 \pm \sqrt{36}}{4}\)

This gives us two possible values for m:

m = \(\frac{14 + 6}{4} = 5\) and

m = \(\frac{14 - 6}{4} = 2\).

Substituting these values of m into the expression for n in terms of m, we get two corresponding values for n:

n = \(7 - m = 7 - 5 = 2\) and

n = \(7 - m = 7 - 2 = 5\).

So, the system of equations has two solutions: (m, n) = (5, 2) and (m, n) = (2, 5).