5) \( \begin{array}{l}(-6 x+y=-8) \\ (12 x-9 y=30)\end{array} \)

To solve the system of linear equations: \[ \begin{cases} -6x + y = -8 \quad \text{(Equation 1)} \\ 12x - 9y = 30 \quad \text{(Equation 2)} \end{cases} \] we can use the **substitution method**. Here's a step-by-step solution: ### Step 1: Solve Equation 1 for \( y \) Starting with Equation 1: \[ -6x + y = -8 \] Add \( 6x \) to both sides to isolate \( y \): \[ y = 6x - 8 \] ### Step 2: Substitute \( y \) into Equation 2 Now, substitute \( y = 6x - 8 \) into Equation 2: \[ 12x - 9y = 30 \] Plugging in \( y \): \[ 12x - 9(6x - 8) = 30 \] ### Step 3: Simplify and Solve for \( x \) Expand the equation: \[ 12x - 54x + 72 = 30 \] Combine like terms: \[ -42x + 72 = 30 \] Subtract 72 from both sides: \[ -42x = 30 - 72 \] \[ -42x = -42 \] Divide both sides by -42: \[ x = \frac{-42}{-42} = 1 \] ### Step 4: Find \( y \) Using the Value of \( x \) Now that we have \( x = 1 \), substitute it back into the expression for \( y \): \[ y = 6x - 8 \] \[ y = 6(1) - 8 \] \[ y = 6 - 8 = -2 \] ### Solution The solution to the system of equations is: \[ x = 1 \quad \text{and} \quad y = -2 \] ### Verification Let's verify the solution by plugging \( x = 1 \) and \( y = -2 \) back into both equations. **Equation 1:** \[ -6(1) + (-2) = -6 - 2 = -8 \quad \checkmark \] **Equation 2:** \[ 12(1) - 9(-2) = 12 + 18 = 30 \quad \checkmark \] Both equations are satisfied, confirming that \( x = 1 \) and \( y = -2 \) is the correct solution.