\( \begin{array}{cc}\text { System 1 } & \text { System 2 } \\ 2 x+3 y=6 & 2 x+3 y=6 \\ -2 x-y=-6 & x+y=4\end{array} \)

Sure, let's solve both **System 1** and **System 2** step by step. --- ### **System 1:** \[ \begin{cases} 2x + 3y = 6 \quad \text{(Equation 1)} \\ -2x - y = -6 \quad \text{(Equation 2)} \end{cases} \] **Step 1: Add Equation 1 and Equation 2** \[ \begin{align*} (2x + 3y) + (-2x - y) &= 6 + (-6) \\ 2x - 2x + 3y - y &= 0 \\ 2y &= 0 \\ y &= 0 \end{align*} \] **Step 2: Substitute \( y = 0 \) into Equation 1** \[ 2x + 3(0) = 6 \\ 2x = 6 \\ x = 3 \] **Solution for System 1:** \[ x = 3, \quad y = 0 \] --- ### **System 2:** \[ \begin{cases} 2x + 3y = 6 \quad \text{(Equation 1)} \\ x + y = 4 \quad \text{(Equation 2)} \end{cases} \] **Step 1: Solve Equation 2 for \( x \)** \[ x = 4 - y \] **Step 2: Substitute \( x = 4 - y \) into Equation 1** \[ 2(4 - y) + 3y = 6 \\ 8 - 2y + 3y = 6 \\ 8 + y = 6 \\ y = 6 - 8 \\ y = -2 \] **Step 3: Substitute \( y = -2 \) back into Equation 2 to find \( x \)** \[ x = 4 - (-2) \\ x = 4 + 2 \\ x = 6 \] **Solution for System 2:** \[ x = 6, \quad y = -2 \] --- ### **Summary of Solutions:** - **System 1:** \( x = 3 \), \( y = 0 \) - **System 2:** \( x = 6 \), \( y = -2 \) If you have any further questions or need additional explanations, feel free to ask!