Larry Mitchell invested part of his \( \$ 13,000 \) advance at \( 5 \% \) annual simple interest and the rest at \( 3 \% \) annual simple interest. If his total yearly interest from both accounts was \( \$ 510 \), find the amount invested at each rate. The amount invested at \( 5 \% \) is \( \$ \square \). The amount invested at \( 3 \% \) is \( \$ \square \).

Let's denote the amount invested at \( 5\% \) as \( x \). Therefore, the amount invested at \( 3\% \) would be \( 13,000 - x \). The total interest can be expressed as: \[ 0.05x + 0.03(13000 - x) = 510 \] Now let's distribute and simplify the equation: \[ 0.05x + 390 - 0.03x = 510 \] \[ 0.02x + 390 = 510 \] Subtract \( 390 \) from both sides: \[ 0.02x = 120 \] Now, divide by \( 0.02 \): \[ x = 120 / 0.02 = 6000 \] So the amount invested at \( 5\% \) is \( \$ 6000 \). Now, we can calculate the amount invested at \( 3\% \): \[ 13000 - 6000 = 7000 \] Thus, the amounts invested are: The amount invested at \( 5\% \) is \( \$ 6000 \). The amount invested at \( 3\% \) is \( \$ 7000 \).