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

To solve the system of equations, let’s first rewrite them for clarity: 1) \(3x + y = 20\) 2) \(x + y = 12\) Next, let’s eliminate \(y\) by rearranging the second equation: \(y = 12 - x\). Now, substitute \(y\) in the first equation: \(3x + (12 - x) = 20\). This simplifies to: \(3x + 12 - x = 20\) \(2x + 12 = 20\) \(2x = 8\) \(x = 4\). Now substitute \(x\) back into \(y = 12 - x\): \(y = 12 - 4 = 8\). The solution to the system is \(x = 4\) and \(y = 8\)! Now that you've cracked those equations, let’s see how this can help in real life. Think about budgeting—if your monthly expenses can be represented by equations like these, you'll easily figure out how much you can save or invest each month! And hey, when diving into systems of equations, a common mistake is not double-checking your substitutions. Always plug your values back into the original equations to ensure everything balances out! Happy calculating!