Answer: $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
