\( \left. \begin{array} { c } { \left\{ \begin{array} { l } { 3 c + 2.5 p = 31 } \\ { c = 11 - p } \end{array} \right.} \\ { 3 c + 2.5 p = 31 } \\ { 3 ( \square ) + 2.5 p = } \end{array} \right. 31 \)

It looks like you're working on solving a system of linear equations using substitution. Let's go through the steps to solve for the variables \( c \) and \( p \). **Given System of Equations:** \[ \begin{cases} 3c + 2.5p = 31 \\ c = 11 - p \end{cases} \] **Step 1: Substitute \( c \) from the second equation into the first equation.** The second equation gives \( c \) in terms of \( p \): \[ c = 11 - p \] Substitute this expression for \( c \) into the first equation: \[ 3(11 - p) + 2.5p = 31 \] **Step 2: Distribute and combine like terms.** \[ 3 \times 11 - 3p + 2.5p = 31 \\ 33 - 3p + 2.5p = 31 \] Combine the \( p \) terms: \[ 33 - 0.5p = 31 \] **Step 3: Solve for \( p \).** \[ -0.5p = 31 - 33 \\ -0.5p = -2 \\ p = \frac{-2}{-0.5} \\ p = 4 \] **Step 4: Substitute \( p \) back into the expression for \( c \).** \[ c = 11 - p \\ c = 11 - 4 \\ c = 7 \] **Solution:** \[ c = 7 \\ p = 4 \] **Verification:** Let's plug the values back into the original equations to ensure they satisfy both: 1. **First Equation:** \[ 3c + 2.5p = 3(7) + 2.5(4) = 21 + 10 = 31 \quad \text{✓} \] 2. **Second Equation:** \[ c = 11 - p = 11 - 4 = 7 \quad \text{✓} \] Both equations are satisfied with \( c = 7 \) and \( p = 4 \).