x+y=-1 x^2+y^2=25

To solve the system of equations:

1. \( x + y = -1 \)
2. \( x^2 + y^2 = 25 \)

We can use the first equation to express \( y \) in terms of \( x \):

From equation (1): \[ y = -1 - x \]

Next, substitute this expression for \( y \) into equation (2):

\[ x^2 + (-1 - x)^2 = 25 \]

Expanding the squared term: \[ x^2 + (1 + 2x + x^2) = 25 \] \[ 2x^2 + 2x + 1 = 25 \]

Now, simplify this equation: \[ 2x^2 + 2x + 1 - 25 = 0 \] \[ 2x^2 + 2x - 24 = 0 \]

Divide the entire equation by 2: \[ x^2 + x - 12 = 0 \]

Next, we can factor this quadratic equation: \[ (x + 4)(x - 3) = 0 \]

Setting each factor to zero gives us: \[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \] \[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]

Now we will find the corresponding values of \( y \) for each \( x \).

1. If \( x = -4 \): \[ y = -1 - (-4) = -1 + 4 = 3 \] Thus, one solution is \( (x, y) = (-4, 3) \).

2. If \( x = 3 \): \[ y = -1 - 3 = -4 \] Thus, the other solution is \( (x, y) = (3, -4) \).

The two solutions to the system of equations are: \[ (x, y) = (-4, 3) \quad \text{and} \quad (x, y) = (3, -4) \]