5A) Solve the system of equations using a method of your choice. \[ \begin{array}{l}8 x-3 y=-1 \\ x+2 y=-12\end{array} \] 5B) Solve the system of equations using a different method.

Certainly! Let's solve the given system of equations using two different methods for parts **5A** and **5B**. \[ \begin{cases} 8x - 3y = -1 \quad \text{(Equation 1)} \\ x + 2y = -12 \quad \text{(Equation 2)} \end{cases} \] --- ### **5A) Solving by Substitution Method** **Step 1: Solve one equation for one variable** Let's solve **Equation 2** for \( x \): \[ x + 2y = -12 \\ \Rightarrow x = -12 - 2y \] **Step 2: Substitute this expression into the other equation** Substitute \( x = -12 - 2y \) into **Equation 1**: \[ 8(-12 - 2y) - 3y = -1 \\ -96 - 16y - 3y = -1 \\ -96 - 19y = -1 \] **Step 3: Solve for \( y \)** \[ -19y = -1 + 96 \\ -19y = 95 \\ y = \frac{95}{-19} \\ y = -5 \] **Step 4: Substitute \( y \) back to find \( x \)** Using \( x = -12 - 2y \): \[ x = -12 - 2(-5) \\ x = -12 + 10 \\ x = -2 \] **Solution:** \[ x = -2, \quad y = -5 \] --- ### **5B) Solving by Elimination Method** **Step 1: Align the equations** \[ \begin{cases} 8x - 3y = -1 \quad \text{(Equation 1)} \\ x + 2y = -12 \quad \text{(Equation 2)} \end{cases} \] **Step 2: Make the coefficients of one variable the same** Let's eliminate \( x \). Multiply **Equation 2** by 8: \[ 8(x + 2y) = 8(-12) \\ 8x + 16y = -96 \quad \text{(Equation 3)} \] **Step 3: Subtract **Equation 1** from **Equation 3** to eliminate \( x \)** \[ (8x + 16y) - (8x - 3y) = -96 - (-1) \\ 8x + 16y - 8x + 3y = -95 \\ 19y = -95 \] **Step 4: Solve for \( y \)** \[ y = \frac{-95}{19} \\ y = -5 \] **Step 5: Substitute \( y \) back to find \( x \)** Using **Equation 2**: \[ x + 2(-5) = -12 \\ x - 10 = -12 \\ x = -12 + 10 \\ x = -2 \] **Solution:** \[ x = -2, \quad y = -5 \] --- Both methods lead to the same solution: \[ \boxed{(x, y) = (-2,\ -5)} \]