x+y=9, y^2+x=29.

To solve the system of equations given by:

1. \( x + y = 9 \)
2. \( y^2 + x = 29 \)

we can start by expressing \( x \) in terms of \( y \) from the first equation:

\[ x = 9 - y \]

Now, we substitute this expression for \( x \) into the second equation:

\[ y^2 + (9 - y) = 29 \]

Simplifying this, we get:

\[ y^2 + 9 - y = 29 \]

Rearranging the equation gives us:

\[ y^2 - y + 9 - 29 = 0 \]

This simplifies to:

\[ y^2 - y - 20 = 0 \]

Next, we can factor the quadratic equation:

\[ (y - 5)(y + 4) = 0 \]

From this, we can find the values of \( y \):

\[ y - 5 = 0 \quad \Rightarrow \quad y = 5 \]

\[ y + 4 = 0 \quad \Rightarrow \quad y = -4 \]

Now, we will find the corresponding values of \( x \) for both \( y \) values using the equation \( x = 9 - y \).

1. For \( y = 5 \):

\[ x = 9 - 5 = 4 \]

2. For \( y = -4 \):

\[ x = 9 - (-4) = 9 + 4 = 13 \]

Thus, the solutions of the system of equations are:

1. \( (x, y) = (4, 5) \)
2. \( (x, y) = (13, -4) \)

So the final answer is:

\[ (4, 5) \quad \text{and} \quad (13, -4) \]