\( \left. \begin{array} { | l l | l l } \hline 11 & { - 8 m - 9 = 15 } & { 12 } & { - 3 = \frac { 4 + p } { 5 } } \end{array} \right. \)

Sure, let's solve the given equations step by step. You provided the following system of equations: \[ \begin{array}{|l l|l l|} \hline 11 & -8m - 9 = 15 & 12 & -3 = \frac{4 + p}{5} \\ \hline \end{array} \] It appears that **Equation 11** and **Equation 12** are to be solved for variables \( m \) and \( p \) respectively. Let's tackle each one individually. --- ### **Equation 11:** \[ -8m - 9 = 15 \] **Step 1:** Add 9 to both sides to isolate the term with \( m \). \[ -8m = 15 + 9 \] \[ -8m = 24 \] **Step 2:** Divide both sides by -8 to solve for \( m \). \[ m = \frac{24}{-8} \] \[ m = -3 \] **Solution:** \( m = -3 \) --- ### **Equation 12:** \[ -3 = \frac{4 + p}{5} \] **Step 1:** Multiply both sides by 5 to eliminate the denominator. \[ -3 \times 5 = 4 + p \] \[ -15 = 4 + p \] **Step 2:** Subtract 4 from both sides to solve for \( p \). \[ -15 - 4 = p \] \[ -19 = p \] **Solution:** \( p = -19 \) --- ### **Final Answers:** - **\( m = -3 \)** - **\( p = -19 \)**