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angeline invested $23,850 in two separate investment accounts. one of the accounts earned 8% annual interest while the other account earned 9% annual interest. if the combined interest earned from both accounts over one year was $2,029.00, how much money was invested in each account?

User Alexyz
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Final answer:

Angeline invested $11,750 in an account with 8% interest and $12,100 in an account with 9% interest, based on setting up and solving a system of linear equations with the given total investment and combined interest.

Step-by-step explanation:

To solve the problem posed by Angeline's investment, we need to set up a system of linear equations based on the given information. Angeline invested a total of $23,850 across two accounts with different interest rates: one at 8% and the other at 9%. The total interest earned from both accounts is $2,029.00 after one year.

Let's define two variables: x for the amount invested at 8%, and y for the amount invested at 9%. We know two things:

  • The sum of the investments is $23,850: x + y = 23,850
  • The total interest earned from both investments is $2,029: 0.08x + 0.09y = 2,029

Now we have a system of equations that we can solve simultaneously. Firstly, we can manipulate the first equation to express y in terms of x: y = 23,850 - x. We can then substitute this expression for y into the second equation:

0.08x + 0.09(23,850 - x) = 2,029

Upon simplifying, we get:

0.08x + 2,146.5 - 0.09x = 2,029

We can solve for x:

x(0.08 - 0.09) = 2,029 - 2,146.5
-0.01x = -117.5
x = 11,750

Now that we have the value for x, we can find y:

y = 23,850 - 11,750
y = 12,100

Therefore, Angeline invested $11,750 in the account with 8% annual interest and $12,100 in the account with 9% annual interest.

User Jibeex
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