Question 8(Multiple Choice Worth 1 points) \( \begin{array}{l}(05.02 \mathrm{MC}) \\ \text { Sam and Harry are family. Sam is currently six times Harry's age. Sam's age is also } 12 \text { more than four times Harry's age. The following system of equations models this scenario: } \\ x=6 y \\ x=12+4 y \\ \text { What are their current ages? }\end{array} \)

To solve the system of equations where \( x = 6y \) (Sam's age) and \( x = 12 + 4y \) (another relation involving their ages), we can substitute \( 6y \) for \( x \) in the second equation. This gives us: \[ 6y = 12 + 4y \] Now, subtract \( 4y \) from both sides: \[ 6y - 4y = 12 \] This simplifies to: \[ 2y = 12 \] Dividing both sides by 2 yields: \[ y = 6 \] Now that we know Harry’s age (6 years), we can find Sam’s age using \( x = 6y \): \[ x = 6 \times 6 = 36 \] So, Sam is 36 years old and Harry is 6 years old. Here's a little insight: Sam being six times Harry's age is a classic example of how age relationships can be mathematically represented and solved using equations. These types of problems are not only found in academic settings but often come up in real-life situations, such as family age comparisons during birthday celebrations or planning surprise parties! While solving systems of equations, it's common to make algebraic mistakes like forgetting to distribute correctly or misplacing signs. A tip is to always rewrite each equation before substituting values, double-checking that no steps are skipped. It’s like checking your math recipe before baking, to ensure all ingredients are accurately measured!