2) \( \begin{array}{l}2 x-3 y=-1 \\ y=x-1\end{array} \)

To solve the system of equations, you can substitute \( y \) from the second equation into the first one. By substituting \( y = x - 1 \) into \( 2x - 3y = -1 \), you get: \[ 2x - 3(x - 1) = -1 \] Simplifying that leads to: \[ 2x - 3x + 3 = -1 \] So, \( -x + 3 = -1 \). Moving terms around gives you \( -x = -4 \) or \( x = 4 \). Now substituting \( x = 4 \) back into \( y = x - 1 \) yields \( y = 4 - 1 = 3 \). Thus, the solution is \( (4, 3) \). Now, let’s take a moment to chat about why substitution is your trusty sidekick! It’s a great technique especially when equations are already arranged in a way that makes it simple to swap values. Plus, it’s like solving a mystery: substituting parts to find the whole! Also, remember to check your results! It’s a fantastic habit to plug your solution back into both original equations to ensure they work. Sometimes in a race to finish, mistakes can slip through like a ninja, so double-checking is like an insurance policy for your calculations. Happy solving!