2. \( \left\{\begin{array}{r}3 x+y=20 \\ x+y=12\end{array}\right. \)

To solve the system of equations, we can use substitution or elimination. Start with the second equation \( x + y = 12 \). Rearranging it gives \( y = 12 - x \). Now substitute this expression for \( y \) into the first equation: \[ 3x + (12 - x) = 20 \] This simplifies to: \[ 3x + 12 - x = 20 \] Combining like terms, we have: \[ 2x + 12 = 20 \] Next, subtract 12 from both sides: \[ 2x = 8 \] Dividing both sides by 2 gives: \[ x = 4 \] Now substitute \( x = 4 \) back into the equation \( y = 12 - x \): \[ y = 12 - 4 = 8 \] Thus, the solution to the system of equations is \( x = 4 \) and \( y = 8 \). However, did you know that systems of equations were used as far back as ancient civilizations? The Babylonians already had methods to solve these types of problems—now that’s old school math! In real life, you might encounter systems of equations in planning budgets or allocating resources, such as when determining how to maximize profits while staying within constraints like expenditures or materials. It’s like solving a balancing act for your money!