If possible, solve the system of equations. Use any method. If there is not a unique solution to the system, state a reason. \[ \begin{array}{ll}4 x-y=22 & \text { (1) } \\ 7 x+3 y=48 & \text { (2) }\end{array} \]

To solve the system of equations: \[ \begin{cases} 4x - y = 22 \quad \text{(1)} \\ 7x + 3y = 48 \quad \text{(2)} \end{cases} \] we can use the **substitution method**. Here's a step-by-step solution: ### Step 1: Solve Equation (1) for \( y \) Starting with equation (1): \[ 4x - y = 22 \] Subtract \( 4x \) from both sides: \[ -y = -4x + 22 \] Multiply both sides by \( -1 \) to solve for \( y \): \[ y = 4x - 22 \quad \text{(3)} \] ### Step 2: Substitute \( y \) into Equation (2) Now, substitute equation (3) into equation (2): \[ 7x + 3y = 48 \] Replace \( y \) with \( 4x - 22 \): \[ 7x + 3(4x - 22) = 48 \] ### Step 3: Simplify and Solve for \( x \) Expand the equation: \[ 7x + 12x - 66 = 48 \] Combine like terms: \[ 19x - 66 = 48 \] Add \( 66 \) to both sides: \[ 19x = 114 \] Divide both sides by \( 19 \): \[ x = \frac{114}{19} = 6 \] ### Step 4: Find \( y \) Using the Value of \( x \) Now that we have \( x = 6 \), substitute this back into equation (3): \[ y = 4(6) - 22 = 24 - 22 = 2 \] ### Conclusion The system has a **unique solution**: \[ x = 6, \quad y = 2 \] --- **Verification:** To ensure the solution is correct, plug \( x = 6 \) and \( y = 2 \) back into both original equations. 1. **Equation (1):** \[ 4(6) - 2 = 24 - 2 = 22 \quad \text{✓} \] 2. **Equation (2):** \[ 7(6) + 3(2) = 42 + 6 = 48 \quad \text{✓} \] Both equations are satisfied, confirming that \( x = 6 \) and \( y = 2 \) is indeed the correct solution.