Let x = amount invested in the 1st account
y = amount invested in the 2nd account
z = amount invested in the 3rd account
Because the total investment is $100,000, therefore
x + y + z = 100,000 (1)
Interest earned in one year from the accounts is $3,650, therefore
0.04x + 0.03y + 0.02z = 3,650
or
4x + 3y + 2z = 365,000 (2)
Because x = 5y, therefore obtain these 2 equations:
5y +y +z = 100,000
or
6y + z = 100,000 (3)
4*(5y) +3y + 2z = 365,000
or
23y + 2z = 365,000 (4)
Substitute z=1000,000 - 6y from (3) into (4).
23y + 2(100,000 - 6y) = 365,000
23y + 200,000 - 12y = 365,000
11y = 165,000
y = $15,000
Therefore
x = 5y = $75,000
z = 100,000 - 6y = $10,000
Answer:
The amounts invested are
1st account: $75,000
2nd account: $15,000
3rd account: $10,000