\[ \qquad(-4 x-y)^{2}+(x-4)^{2}=0 \] Solve for \( x \) and \( y \) Answer: • \( x=\square \) • \( y=\square \)

To solve the equation \[ (-4x - y)^2 + (x - 4)^2 = 0, \] we can analyze the equation step by step. 1. **Understanding the equation**: The sum of two squares is equal to zero. This means that both squares must individually be equal to zero because squares of real numbers are always non-negative. 2. **Setting each square to zero**: - From the first term: \[ (-4x - y)^2 = 0 \implies -4x - y = 0 \implies y = -4x. \] - From the second term: \[ (x - 4)^2 = 0 \implies x - 4 = 0 \implies x = 4. \] 3. **Substituting \( x \) back to find \( y \)**: - Substitute \( x = 4 \) into \( y = -4x \): \[ y = -4(4) = -16. \] Thus, the solutions are: - \( x = 4 \) - \( y = -16 \) Final answer: - \( x = 4 \) - \( y = -16 \)