Joe invested $12000, part at 6% interest and part at 8% interest. In the first year he earned $860 in interest. How much was invested at each rate?
Let x = amount invested.
x(0.06) + (12000-x)(0.08) = 860.
Solve for x.
To find the amounts invested at each interest rate, let's assume Joe invested x dollars at 6% interest and (12000 - x) dollars at 8% interest.
The interest earned on the first investment is x * 6%. The interest earned on the second investment is (12000 - x) * 8%.
According to the problem, in the first year, Joe earned $860 in interest. Therefore, we can write the following equation to represent the total interest earned:
x * 6% + (12000 - x) * 8% = $860
Now, let's solve this equation to find the value of x.
To eliminate percentages, we divide the equation by 100:
0.06x + 0.08(12000 - x) = $860
Simplifying, we have:
0.06x + 960 - 0.08x = $860
Combining like terms, we get:
-0.02x = $860 - $960
-0.02x = -$100
Dividing both sides by -0.02 gives us:
x = $100 / -0.02
x = -$5000
Since the number of dollars cannot be negative, this result doesn't make sense in the context of the problem. Therefore, there might be an error in the given information or the question itself. Please check the problem again to ensure the accuracy of the information provided.