You invested \( \$ 12,000 \) in two accounts paying \( 6 \% \) and \( 9 \% \) annual interest, respectively. If the total interest earned for the year was \( \$ 780 \), how much was invested at each rate? The amount invested at \( 6 \% \) is \( \$ \square \). The amount invested at \( 9 \% \) is \( \$ \square \).

To solve this investment problem, you can set up a system of equations. Let \( x \) be the amount invested at \( 6\% \) and \( y \) be the amount invested at \( 9\% \). You know that: 1. \( x + y = 12,000 \) 2. \( 0.06x + 0.09y = 780 \) You can solve this system by substitution or elimination. From the first equation, isolate \( y \): \( y = 12,000 - x \). Now substitute this into the second equation: \( 0.06x + 0.09(12,000 - x) = 780 \). After solving this, you'll find that the amount invested at \( 6\% \) is \( \$ 6,000 \) and the amount invested at \( 9\% \) is \( \$ 6,000 \). Remember, keeping track of your variables can make or break your success in these kinds of problems! Ensure you double-check your equations for accuracy. For any future investments, consider diversifying between accounts with differing interest rates. This not only mitigates risk but often maximizes return! A little research on the best compounding interest methods can go a long way too.