b) \( \left\{\begin{array}{l}2 x+3 y=12 \\ x=2 y-4\end{array}\right. \)

To solve the system of equations, let's first substitute \(x\) from the second equation into the first. From \(x = 2y - 4\), we can replace \(x\) in the first equation: \[ 2(2y - 4) + 3y = 12 \] This simplifies to: \[ 4y - 8 + 3y = 12 \] Combining like terms gives: \[ 7y - 8 = 12 \] Adding 8 to both sides results in: \[ 7y = 20 \] Dividing by 7, we find: \[ y = \frac{20}{7} \] Now, substituting \(y\) back into \(x = 2y - 4\): \[ x = 2\left(\frac{20}{7}\right) - 4 = \frac{40}{7} - 4 = \frac{40}{7} - \frac{28}{7} = \frac{12}{7} \] Thus, the solution to the system of equations is: \[ (x, y) = \left(\frac{12}{7}, \frac{20}{7}\right) \] Now, let's celebrate this math moment with a bit of context! Did you know that systems of equations have real-world applications? Think of scenarios like planning train schedules where different routes intersect or determining the optimal mix of ingredients in recipes! Solving these equations can help find the most efficient solutions in these cases—like figuring out the perfect timing for that delicious cake you're baking while coordinating your favorite TV show schedule! Keep in mind that when solving such systems, a common pitfall is forgetting to substitute carefully. Double-checking your substitutions makes all the difference, ensuring you catch any minor mistakes before they snowball into big problems. Happy solving!