Two investments totaling $63,500$63,500 produce an annual income of $3250$3250. One investment yields 4%4% per year, while the other yields 6%6% per year. How much is invested at each rate?
Question
Two investments totaling 63,500 produce an annual income of 3250. One investment yields 4%4% per year, while the other yields 6%6% per year. How much is invested at each rate?
Solution
Let's denote the amount invested at 4% as x and the amount invested at 6% as y.
From the problem, we know two things:
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The total amount invested is $63,500. So, we can write this as an equation: x + y = 63,500.
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The total annual income from both investments is $3,250. The income from the first investment is 4% of x and the income from the second investment is 6% of y. We can write this as another equation: 0.04x + 0.06y = 3,250.
Now we have a system of two equations, and we can solve it step by step:
Step 1: Multiply the first equation by 0.04: 0.04x + 0.04y = 2,540.
Step 2: Subtract the first equation from the second: 0.02y = 710.
Step 3: Solve for y: y = 710 / 0.02 = 35,500.
Step 4: Substitute y = 35,500 into the first equation: x + 35,500 = 63,500.
Step 5: Solve for x: x = 63,500 - 35,500 = 28,000.
So, 35,500 is invested at 6%.
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