x^2+4y=8 x+y=2

To solve the system of equations given by:

1. \( x^2 + 4y = 8 \)
2. \( x + y = 2 \)

We can first solve the second equation for \( y \):

\[ y = 2 - x \]

Now, we can substitute this expression for \( y \) in the first equation:

\[ x^2 + 4(2 - x) = 8 \]

Expanding this gives:

\[ x^2 + 8 - 4x = 8 \]

Next, we can simplify this equation:

\[ x^2 - 4x + 8 - 8 = 0 \]

This simplifies to:

\[ x^2 - 4x = 0 \]

Now, we can factor it:

\[ x(x - 4) = 0 \]

Setting each factor to zero gives us:

\[ x = 0 \quad \text{or} \quad x = 4 \]

Next, we substitute these values back into the equation \( y = 2 - x \) to find the corresponding \( y \) values.

1. If \( x = 0 \):

\[ y = 2 - 0 = 2 \]

So one solution is \( (0, 2) \).

2. If \( x = 4 \):

\[ y = 2 - 4 = -2 \]

So the other solution is \( (4, -2) \).

In conclusion, the solution set for the system of equations is:

\[ (0, 2) \quad \text{and} \quad (4, -2) \]