Jolene invests her savings in two bank accounts, one paying 5 percent and the other paying 9 percent simple interest per year. She puts twice as much in the lower-yielding account because it is less risky. Her annual interest is 5434 dollars. How much did she invest at each rate?
Amount invested at 5 percent interest is $\$$
Amount invested at 9 percent interest is $\$$

Step 1 ：Let's denote the amount of money Jolene invested at 9 percent as x. Then, the amount of money she invested at 5 percent is 2x.

Step 2 ：The total interest she earns in a year is the sum of the interest from both accounts, which is 0.09x (from the 9 percent account) and 0.05*2x (from the 5 percent account). This total interest equals 5434 dollars.

Step 3 ：We can set up the equation and solve for x: \(0.19x = 5434\).

Step 4 ：The solution to the equation is x = 28600. This means that Jolene invested $28600 at 9 percent interest.

Step 5 ：Since she invested twice as much at 5 percent interest, she invested 2 * $28600 = $57200 at 5 percent interest.

Step 6 ：Final Answer: Amount invested at 5 percent interest is \(\boxed{57200}\) dollars. Amount invested at 9 percent interest is \(\boxed{28600}\) dollars.