Question

PLEASE HELP

Answer 1

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:

$I(t) = Prt$

In which:

• P is the initial amount.

• r is the interest rate, as a decimal.

In this problem:

• A total of $10,000 was invested, hence $P_1 + P_2 = 10000$.

• Two accounts paying 7% and 8% annual interest, respectively, hence $r_1 = 0.07, r_2 = 0.08$.

• The total interest earned for the year was $780, hence $t = 1, I_1 + I_2 = 780$

Then:

$I_1 = 0.07P_1$

$I_2 = 0.08P_2$

Considering $P_1 + P_2 = 10000 \rightarrow P_2 = 10000 - P_1$:

$I_2 = 0.08(10000 - P_1)$

Considering $I_1 + I_2 = 780 \rightarrow I_2 = 780 - I_1$:

$780 - I_1 = 0.08(10000 - P_1)$

Considering $I_1 = 0.07P_1$:

$780 - I_1 = 0.08(10000 - P_1)$

$780 - 0.07P_1 = 0.08(10000 - P_1)$

$0.01P_1 = 20$

$P_1 = \frac{20}{0.01}$

$P_1 = 2000$

And:

$P_1 + P_2 = 10000$

$2000 + P_2 = 10000$

$P_2 = 8000$

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 brainly.com/question/25296782