system of equations. \( x \) to the corresponding \[ \begin{array}{l} x+3 y=6 \\ 2 x-3 y=12 \end{array} \] \( x+2 y=4 \) \( -3 x+y=2 \) \( \square \) \( x+3 y=12 \) \( 2 x+y=39 \) \( \square \) \( 4 x-5 y=15 \) \( 2 x-5 y=11 \) \( \square \) \( 4 x+2 y=-2 \) \[ x+3 y=12 \] \( \square \)

Sure, let's solve each of the given systems of equations step by step. We'll find the values of \( x \) and \( y \) that satisfy each system. --- ### **System 1** \[ \begin{cases} x + 3y = 6 \\ 2x - 3y = 12 \end{cases} \] **Solution:** 1. **Add the two equations to eliminate \( y \):** \[ (x + 2x) + (3y - 3y) = 6 + 12 \\ 3x = 18 \\ x = 6 \] 2. **Substitute \( x = 6 \) into the first equation:** \[ 6 + 3y = 6 \\ 3y = 0 \\ y = 0 \] **Solution:** \( \boxed{(x, y) = (6,\ 0)} \) --- ### **System 2** \[ \begin{cases} x + 2y = 4 \\ -3x + y = 2 \end{cases} \] **Solution:** 1. **Express \( x \) from the first equation:** \[ x = 4 - 2y \] 2. **Substitute \( x = 4 - 2y \) into the second equation:** \[ -3(4 - 2y) + y = 2 \\ -12 + 6y + y = 2 \\ 7y = 14 \\ y = 2 \] 3. **Find \( x \):** \[ x = 4 - 2(2) = 0 \] **Solution:** \( \boxed{(x, y) = (0,\ 2)} \) --- ### **System 3** \[ \begin{cases} x + 3y = 12 \\ 2x + y = 39 \end{cases} \] **Solution:** 1. **Express \( x \) from the first equation:** \[ x = 12 - 3y \] 2. **Substitute \( x = 12 - 3y \) into the second equation:** \[ 2(12 - 3y) + y = 39 \\ 24 - 6y + y = 39 \\ -5y = 15 \\ y = -3 \] 3. **Find \( x \):** \[ x = 12 - 3(-3) = 12 + 9 = 21 \] **Solution:** \( \boxed{(x, y) = (21,\ -3)} \) --- ### **System 4** \[ \begin{cases} 4x - 5y = 15 \\ 2x - 5y = 11 \end{cases} \] **Solution:** 1. **Subtract the second equation from the first to eliminate \( y \):** \[ (4x - 2x) + (-5y + 5y) = 15 - 11 \\ 2x = 4 \\ x = 2 \] 2. **Substitute \( x = 2 \) into the second equation:** \[ 2(2) - 5y = 11 \\ 4 - 5y = 11 \\ -5y = 7 \\ y = -\frac{7}{5} \quad \text{or} \quad y = -1.4 \] **Solution:** \( \boxed{(x, y) = \left(2,\ -\frac{7}{5}\right)} \) --- ### **System 5** \[ \begin{cases} 4x + 2y = -2 \\ x + 3y = 12 \end{cases} \] **Solution:** 1. **Express \( x \) from the second equation:** \[ x = 12 - 3y \] 2. **Substitute \( x = 12 - 3y \) into the first equation:** \[ 4(12 - 3y) + 2y = -2 \\ 48 - 12y + 2y = -2 \\ -10y = -50 \\ y = 5 \] 3. **Find \( x \):** \[ x = 12 - 3(5) = 12 - 15 = -3 \] **Solution:** \( \boxed{(x, y) = (-3,\ 5)} \) --- Feel free to reach out if you need further assistance with these systems or any other questions!