\( \begin{array}{ll}\text { 3. } & 4 a+1=-11 \\ \text { 4. } & 6 b-3=9 \\ \text { 5. } & 5 c+8=3\end{array} \)

Certainly! Let's solve each of the equations step by step. --- ### **3. \( 4a + 1 = -11 \)** **Step 1:** Subtract 1 from both sides to isolate the term with \( a \). \[ 4a + 1 - 1 = -11 - 1 \] \[ 4a = -12 \] **Step 2:** Divide both sides by 4 to solve for \( a \). \[ \frac{4a}{4} = \frac{-12}{4} \] \[ a = -3 \] --- ### **4. \( 6b - 3 = 9 \)** **Step 1:** Add 3 to both sides to isolate the term with \( b \). \[ 6b - 3 + 3 = 9 + 3 \] \[ 6b = 12 \] **Step 2:** Divide both sides by 6 to solve for \( b \). \[ \frac{6b}{6} = \frac{12}{6} \] \[ b = 2 \] --- ### **5. \( 5c + 8 = 3 \)** **Step 1:** Subtract 8 from both sides to isolate the term with \( c \). \[ 5c + 8 - 8 = 3 - 8 \] \[ 5c = -5 \] **Step 2:** Divide both sides by 5 to solve for \( c \). \[ \frac{5c}{5} = \frac{-5}{5} \] \[ c = -1 \] --- ### **Summary of Solutions** - \( a = -3 \) - \( b = 2 \) - \( c = -1 \) If you have any further questions or need additional assistance, feel free to ask!