(a) Solve for \( x \) and \( y \) simultaneously: (1) \( y=x+2 \) \( x+x y=4 \)

First, let's substitute the expression for \( y \) from the first equation into the second equation. We know that \( y = x + 2 \). Substituting this into the second equation gives us: \[ x + x(x + 2) = 4 \] This simplifies to: \[ x + x^2 + 2x = 4 \] Combine like terms: \[ x^2 + 3x - 4 = 0 \] Now, we can factor the quadratic equation: \[ (x + 4)(x - 1) = 0 \] Setting each factor to zero gives: \[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \] \[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \] Next, we substitute these values back into \( y = x + 2 \) to find the corresponding \( y \) values: 1. For \( x = -4 \): \[ y = -4 + 2 = -2 \] So, one solution is \( (x, y) = (-4, -2) \). 2. For \( x = 1 \): \[ y = 1 + 2 = 3 \] So, another solution is \( (x, y) = (1, 3) \). The solutions are \( (x, y) = (-4, -2) \) and \( (x, y) = (1, 3) \).