4. Helene invested a total of $1000 in two simple-interest bank accounts. One account paid 5% annual interest; the other paid 6% annual interest. The total amount of interest she earned after 1 year was $58. Find the amount invested in each account

Let $x be the amount of money Helene invested in first account. Then $ (1000-x) is the amount of money she invested in second account.

1. First account paid 5% annual interest. 5% from $x is $0.05x.

2. Second accoubt paid 6% annual interest. 6% from $ (1000-x) is $0.06 (1000-x).

3. The total amount of interest she earned after 1 year was $58. This means that

0.05x+0.06 (1000-x) = 58.

Solve this equation:

0.05x+0.06·1000-0.06x=58,

0.05x+60-0.06x=58,

-0.01x=58-60,

-0.01x=-2,

0.01x=2,

x=200.

1000-x=1000-200=800.

Answer: Helene invested $200 in 1st account and $800 in 2nd account.