Fill in each blank with an integer (signed number), \begin{tabular}{|l|l|}\hline (a) \( -4-\square=-4+(-1) \) & ( \( \square) \) \\ (b) \( -3-(-6)=-3+\square \) & \( \times \) \\ \hline\end{tabular}
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Let's solve the equations step by step. (a) For the equation \( -4 - \square = -4 + (-1) \): \[ -4 + (-1) = -4 - 1 = -5 \] So, the equation becomes: \[ -4 - \square = -5 \] To isolate \( \square \), we add 4 to both sides: \[ -\square = -5 + 4 \] \[ -\square = -1 \] Now, multiplying both sides by -1 gives: \[ \square = 1 \] (b) For the equation \( -3 - (-6) = -3 + \square \): First, calculate the left side: \[ -3 - (-6) = -3 + 6 = 3 \] So, the equation now is: \[ 3 = -3 + \square \] To isolate \( \square \), we add 3 to both sides: \[ 3 + 3 = \square \] \[ \square = 6 \] Thus, the filled-in table is: \[ \begin{tabular}{|l|l|}\hline (a) \( -4-\square=-4+(-1) \) & \( \square = 1 \) \\ (b) \( -3-(-6)=-3+\square \) & \( \square = 6 \) \\ \hline \end{tabular} \]