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