Question 5 (Multiple Choice Worth 1 points) (05.03 MC) What is the solution to the following system of equations? \( \begin{array}{l}4 x+2 y=12 \\ x-y=3\end{array} \)

To solve the system of equations, let's use substitution. From the second equation, \(x - y = 3\), we can express \(x\) in terms of \(y\): \(x = y + 3\). Now, substitute \(x\) into the first equation: \[ 4(y + 3) + 2y = 12 \] Distributing gives us \(4y + 12 + 2y = 12\), which simplifies to \(6y + 12 = 12\). Subtracting 12 from both sides leads to \(6y = 0\), and thus \(y = 0\). Substituting back, \(x = 0 + 3 = 3\). Therefore, the solution is \(x = 3\) and \(y = 0\). While we used substitution this time, remember that another popular method is elimination! It's where we strategically manipulate the equations to cancel a variable, making it easier to solve. Just be cautious of signs and remember to double-check your work to avoid common pitfalls!