Solve the system of equations by any method. \[ \begin{aligned} -3 x+12 y & =9 \\ x-4 y & =-3 \end{aligned} \] Enter the exact answer as an ordered pair, \( (x, y) \). If there is no solution, enter NS. If there is an infinite number of solutions, enter the general solution as an ordered pair in terms of \( x \). Include a multiplication sign between symbols. For example, \( a^{*} x \). \( \square \) \[ (x, 3+x / 4) \] Submit Assignment Quit \& Save Back Question Ment

To solve the system of equations, let's use the substitution method. First, from the second equation, we can express \( x \) in terms of \( y \): \[ x = 4y - 3 \] Next, we substitute this expression for \( x \) into the first equation: \[ -3(4y - 3) + 12y = 9 \] Expanding this gives: \[ -12y + 9 + 12y = 9 \] Which simplifies down to: \[ 9 = 9 \] This tells us that the first equation holds true for all \( y \). This indicates that there are infinitely many solutions. Substituting \( y = y \) back into the expression for \( x \), we get the general solution: \[ (x, y) = (4y - 3, y) \] Therefore, we can express the solution as: \[ (x, y) = (4 * (3+x/4) - 3, 3+x/4) \] So, the final answer as an ordered pair in terms of \( x \) is: \[ \boxed{(4y - 3, y)} \] where \( y \) can be any real number.