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Two investments totaling $38,500$⁢38,500 produce an annual income of $2805$⁢2805.  One investment yields 3%3% per year, while the other yields 9%9% per year. How much is invested at each rate?

Question

Two investments totaling 38,50038,500⁢38,500 produce an annual income of 28052805⁢2805.  One investment yields 3%3% per year, while the other yields 9%9% per year. How much is invested at each rate?

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Solution

Let's denote the amount invested at 3% as x and the amount invested at 9% as y.

From the problem, we have two equations:

  1. x + y = 38,500(Thisisbecausethetotalamountinvestedis38,500 (This is because the total amount invested is 38,500)

  2. 0.03x + 0.09y = 2805(Thisisbecausethetotalincomefrombothinvestmentsis2805 (This is because the total income from both investments is 2805, and each investment contributes its amount times its interest rate to the total income)

We can solve this system of equations step by step:

Step 1: Multiply the first equation by 0.03 to make the coefficients of x in both equations the same:

0.03x + 0.03y = $1155

Step 2: Subtract the new equation from the second equation:

0.06y = $1650

Step 3: Solve for y:

y = 1650/0.06=1650 / 0.06 = 27,500

Step 4: Substitute y = $27,500 into the first equation to solve for x:

x = 38,50038,500 - 27,500 = $11,000

So, 11,000isinvestedat311,000 is invested at 3% and 27,500 is invested at 9%.

This problem has been solved

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