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