x+y=-1 x2+y2=1

To solve the system of equations

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

we can begin by expressing one variable in terms of the other using the first equation.

From equation (1), we can express \( y \) in terms of \( x \):

\[ y = -1 - x \]

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

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

Now, expand the square:

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

Combining like terms gives:

\[ 2x^2 + 2x + 1 = 1 \]

Subtract 1 from both sides:

\[ 2x^2 + 2x = 0 \]

Now factor out the common terms:

\[ 2x(x + 1) = 0 \]

This gives us two solutions:

1. \( 2x = 0 \) ⟹ \( x = 0 \)
2. \( x + 1 = 0 \) ⟹ \( x = -1 \)

Now we can find the corresponding values of \( y \):

1. If \( x = 0 \): \[ y = -1 - 0 = -1 \] So one solution is \( (0, -1) \).

2. If \( x = -1 \): \[ y = -1 - (-1) = 0 \] So another solution is \( (-1, 0) \).

Thus, the solutions to the system of equations are:

\[ (0, -1) \quad \text{and} \quad (-1, 0) \]