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