2) \( \left\{\begin{array}{l}x-y=4 \\ x^{2}+y^{2}=16\end{array}\right. \)

To solve the system of equations \[ \begin{cases} x - y = 4 \\ x^2 + y^2 = 16 \end{cases} \] we can start by solving the first equation for \(y\): \[ y = x - 4 \] Next, we substitute this expression for \(y\) into the second equation: \[ x^2 + (x - 4)^2 = 16 \] Now, we expand \((x - 4)^2\): \[ (x - 4)^2 = x^2 - 8x + 16 \] Substituting this back into the equation gives: \[ x^2 + (x^2 - 8x + 16) = 16 \] Combining like terms, we have: \[ 2x^2 - 8x + 16 = 16 \] Subtracting 16 from both sides results in: \[ 2x^2 - 8x = 0 \] Factoring out \(2x\): \[ 2x(x - 4) = 0 \] Setting each factor to zero gives us: \[ 2x = 0 \quad \Rightarrow \quad x = 0 \] \[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \] Now we will find the corresponding \(y\) values for each \(x\): 1. For \(x = 0\): \[ y = 0 - 4 = -4 \] So one solution is \((0, -4)\). 2. For \(x = 4\): \[ y = 4 - 4 = 0 \] So another solution is \((4, 0)\). Thus, the solutions to the system of equations are: \[ \boxed{(0, -4) \text{ and } (4, 0)} \]