\( 1 \leftarrow \quad \) You invested \( \$ 20,000 \) in two accounts paying \( 5 \% \) and \( 9 \% \) annual interest, respectively. If the total interest earned for the year 1 was \( \$ 1040 \), how much was invested at each rate? The amount invested at \( 5 \% \) is \( \$ \square \).

To solve this problem, let's define the amount invested at \( 5\% \) as \( x \). Then, the amount invested at \( 9\% \) would be \( 20,000 - x \). The total interest earned from both accounts can be set up as an equation based on the following interest calculations: \[ 0.05x + 0.09(20,000 - x) = 1040 \] Now, let's simplify this equation step by step. 1. Distributing the \( 9\% \): \[ 0.05x + 1800 - 0.09x = 1040 \] 2. Combine like terms: \[ -0.04x + 1800 = 1040 \] 3. Now, isolate \( x \): \[ -0.04x = 1040 - 1800 \] \[ -0.04x = -760 \] \[ x = \frac{-760}{-0.04} = 19000 \] So, the amount invested at \( 5\% \) is \( \$ 19,000 \) and at \( 9\% \) is \( 20,000 - 19,000 = \$ 1,000 \). Now, let’s confirm this! The interest from \( 5\% \): \[ 0.05 \times 19000 = 950 \] And the interest from \( 9\% \): \[ 0.09 \times 1000 = 90 \] Adding them together: \[ 950 + 90 = 1040 \] All checks out! The amount invested at \( 5\% \) is \( \$ 19,000 \).