Penelope invested $6650 into theee investment accounts, earning 8%, 12.5%, and 7.5% simple interest per year. If the annual interest earned totaled $671.25 and the amount invested at 12.5% was $300 more than twice the amount invested at 8%, find the amount invested into each account.

$1500 was invested in the account earning 8%.

$3300 was invested in the account earning 12.5%.

$1850 was was invested in the account earning 7.5%.

Step-by-step explanation:

Let x represent the amount invested in the account earning 8%.

Let y represent the amount invested in the account earning 12.5%.

Let z represent the amount invested in the account earning 7.5%.

Penelope invested $6650 into theee investment accounts, earning 8%, 12.5%, and 7.5% simple interest per year. It means that

x + y + z = 6650- - - - - - - - - - -1

Interest earned from the $x investment is

8/100 × x = 0.08x

Interest earned from the $y investment is

12.5/100 × x = 0.125y

Interest earned from the $z investment is

7.5/100 × x = 0.075z

If the annual interest earned totaled $671.25, it means that

0.08x + 0.125y + 0.075z = 671.25- - - - - - - 2

The amount invested at 12.5% was $300 more than twice the amount invested at 8%. It means that

y = 2x + 300

Substituting y = 2x + 300 into equation 1 and equation 2, it becomes

x + 2x + 300 + z = 6650

3x + z = 6650 - 300

3x + z = 6350- - - - - - - - - 4

0.08x + 0.125(2x + 300) + 0.075z = 671.25

0.08x + 0.25x + 37.5 + 0.075z = 671.25

0.33x + 0.075z = 671.25 - 37.5

0.33x + 0.075z = 633.75- - - - - - - 5

From equation 4, z = 6350 - 3x

Substituting z = 6350 - 3x into equation 5, it becomes

0.33x + 0.075(6350 - 3x) = 633.75

0.33x + 476.25 - 0.225x = 633.75

0.33x - 0.225x = 633.75 - 476.25

0.105x = 157.5

x = 157.5/0.105

x = 1500

y = 2x + 300

y = 2(1500) + 300

y = 3300

Substituting x = 1500 and y = 3300 into equation 1, it becomes

1500 + 3300 + z = 6650

z = 6650 - 4800

z = 1850