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

To solve the system of equations:

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

We can solve this system using substitution or elimination. Let's use substitution by first solving Equation 1 for \( x \):

From Equation 1:

\[ x = 1 - 2y \]

Now, substitute this expression for \( x \) into Equation 2:

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

Now simplify this:

\[ 2 - 4y + y^2 = -1 \]

Rearranging gives:

\[ y^2 - 4y + 2 + 1 = 0 \] \[ y^2 - 4y + 3 = 0 \]

Now we can factor or use the quadratic formula:

The quadratic can be factored as:

\[ (y - 3)(y - 1) = 0 \]

Setting each factor to zero gives:

\[ y - 3 = 0 \quad \text{or} \quad y - 1 = 0 \]

So,

\[ y = 3 \quad \text{or} \quad y = 1 \]

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

1. For \( y = 3 \): \[ x = 1 - 2(3) = 1 - 6 = -5 \] So one solution is \( (x, y) = (-5, 3) \).

2. For \( y = 1 \): \[ x = 1 - 2(1) = 1 - 2 = -1 \] So the other solution is \( (x, y) = (-1, 1) \).

Thus, the solutions to the system of equations are:

\[ (x, y) = (-5, 3) \quad \text{and} \quad (x, y) = (-1, 1) \]