Fill in each blank with an integer (signed number). \begin{tabular}{|l|}\hline \( \begin{array}{ll}\text { (a) }-4-\square=-4+(-1) & \text { ( } \square) \\ \text { (b) }-3-(-6)=-3+\square & \times\end{array} \)\end{tabular}

1. For part (a), the equation is \[ -4 - \square = -4 + (-1). \] Let the unknown integer be \( x \). Then, the equation becomes \[ -4 - x = -4 - 1. \] Simplify the right-hand side: \[ -4 - x = -5. \] To solve for \( x \), add \(-4\) to the left side: \[ -x = -5 + 4, \] \[ -x = -1. \] Multiply both sides by \(-1\): \[ x = 1. \] 2. For part (b), the equation is \[ -3 - (-6) = -3 + \square. \] Calculate the left-hand side by subtracting a negative: \[ -3 - (-6) = -3 + 6 = 3. \] Let the unknown integer be \( y \). Then, the equation is \[ -3 + y = 3. \] To solve for \( y \), add \( 3 \) to both sides: \[ y = 3 + 3, \] \[ y = 6. \] Thus, the blanks should be filled with: - (a) \(1\) - (b) \(6\)