An investor deposited some money at $1.7 \%$ annual interest, and two equal but larger amounts at $2.3 \%$ and $2.4 \%$. The total amount invested was $\$ 26,000$, and the total annual interest earned was $\$ 598$. How much was invested at each rate?
Answer
So, the investor invested \(\boxed{2000}\) dollars at 1.7%, and \(\boxed{12000}\) dollars each at 2.3% and 2.4%.
Step 1 ：Let's denote the amount invested at 1.7% as x, the amount invested at 2.3% as y, and the amount invested at 2.4% as z.
Step 2 ：We know that the total amount invested was $26,000, which gives us the equation \(x + y + z = 26000\).
Step 3 ：We also know that the amounts invested at 2.3% and 2.4% were equal, so we have \(y = z\).
Step 4 ：The total annual interest earned was $598, which gives us another equation: \(0.017x + 0.023y + 0.024z = 598\).
Step 5 ：We can solve these equations simultaneously to find the values of x, y, and z.
Step 6 ：The solution to these equations is \(x = 2000\), \(y = 12000\), and \(z = 12000\).
Step 7 ：So, the investor invested \(\boxed{2000}\) dollars at 1.7%, and \(\boxed{12000}\) dollars each at 2.3% and 2.4%.
