Using simple interest, it is found that:
- The amount invested at 7% is $2,000.
- The amount invested at 8% is $8,000.
Simple Interest
Simple interest is used when there is a single compounding per time period.
The interest after t years in is modeled by:
[tex]I(t) = Prt[/tex]
In which:
- r is the interest rate, as a decimal.
In this problem:
- A total of $10,000 was invested, hence [tex]P_1 + P_2 = 10000[/tex].
- Two accounts paying 7% and 8% annual interest, respectively, hence [tex]r_1 = 0.07, r_2 = 0.08[/tex].
- The total interest earned for the year was $780, hence [tex]t = 1, I_1 + I_2 = 780[/tex]
Then:
[tex]I_1 = 0.07P_1[/tex]
[tex]I_2 = 0.08P_2[/tex]
Considering [tex]P_1 + P_2 = 10000 \rightarrow P_2 = 10000 - P_1[/tex]:
[tex]I_2 = 0.08(10000 - P_1)[/tex]
Considering [tex]I_1 + I_2 = 780 \rightarrow I_2 = 780 - I_1[/tex]:
[tex]780 - I_1 = 0.08(10000 - P_1)[/tex]
Considering [tex]I_1 = 0.07P_1[/tex]:
[tex]780 - I_1 = 0.08(10000 - P_1)[/tex]
[tex]780 - 0.07P_1 = 0.08(10000 - P_1)[/tex]
[tex]0.01P_1 = 20[/tex]
[tex]P_1 = \frac{20}{0.01}[/tex]
[tex]P_1 = 2000[/tex]
And:
[tex]P_1 + P_2 = 10000[/tex]
[tex]2000 + P_2 = 10000[/tex]
[tex]P_2 = 8000[/tex]
Then:
- The amount invested at 7% is $2,000.
- The amount invested at 8% is $8,000.
You can learn more about simple interest at https://brainly.com/question/25296782
