Two investments totaling $18,000$18,000 produce an annual income of $1370$1370. One investment yields 7%7% per year, while the other yields 8%8% per year. How much is invested at each rate?
Question
Two investments totaling 18,000 produce an annual income of 1370. One investment yields 7%7% per year, while the other yields 8%8% per year. How much is invested at each rate?
Solution
Let's denote the amount invested at 7% as x and the amount invested at 8% as y.
From the problem, we know two things:
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The total amount invested is $18,000. So, we can write this as an equation: x + y = 18,000.
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The total income from both investments is $1370. The income from the first investment is 7% of x (or 0.07x) and the income from the second investment is 8% of y (or 0.08y). So, we can write this as another equation: 0.07x + 0.08y = 1370.
Now we have a system of two equations, and we can solve it step by step:
Step 1: Multiply the first equation by 0.07 to make the coefficients of x in both equations the same: 0.07x + 0.07y = 1260.
Step 2: Subtract the new first equation from the second equation: 0.01y = 110.
Step 3: Solve for y by dividing both sides by 0.01: y = 11000.
Step 4: Substitute y = 11000 into the first equation: x + 11000 = 18000.
Step 5: Solve for x: x = 7000.
So, 11000 is invested at 8%.
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