Fill in each blank with an integer (signed number). \begin{tabular}{|l|l|}\hline (a) \( 2-(-1)=\square+1 \) & ( \( \square) \) \\ \hline (b) \( 3-6=3+\square \) & \( \times \) \\ \hline\end{tabular}

Let's solve each equation step by step. For (a): We have the equation \( 2 - (-1) = \square + 1 \). First, simplify the left side: \( 2 - (-1) = 2 + 1 = 3 \). Now we set up the equation: \( 3 = \square + 1 \). To isolate \( \square \), subtract 1 from both sides: \( \square = 3 - 1 = 2 \). So, the answer for (a) is 2. For (b): We have the equation \( 3 - 6 = 3 + \square \). First, simplify the left side: \( 3 - 6 = -3 \). Now we set up the equation: \( -3 = 3 + \square \). To isolate \( \square \), subtract 3 from both sides: \( -3 - 3 = \square \). This simplifies to: \( \square = -6 \). So, the answer for (b) is -6. Putting it all together, we have: (a) \( \square = 2 \) (b) \( \square = -6 \) Thus, the answers are: \[ \begin{tabular}{|l|l|}\hline (a) \( 2 - (-1) = 2 + 1 \) & \( 2 \) \\ \hline (b) \( 3 - 6 = 3 + (-6) \) & \( -6 \) \\ \hline \end{tabular} \]