You invest $20,000 in two accounts. Account A earned 4.5% annual simple interest and Account B earned 6% annual simple interest. the total amount of interest earned for 1 year was $987.50, how much was invested into each account?

.045x + .06(20000-x) = 987.50

To find out how much was invested in each account, let's call the amount invested in Account A as 'x' and the amount invested in Account B as '20000 - x' (since the total investment is $20,000).

Interest earned from Account A = Amount invested in Account A (x) * Interest rate for Account A (4.5%) = 0.045x
Interest earned from Account B = Amount invested in Account B (20000 - x) * Interest rate for Account B (6%) = 0.06(20000 - x)

We are given that the total interest earned for 1 year is $987.50, so we can set up the equation:

0.045x + 0.06(20000 - x) = 987.50

To solve this equation, we can simplify and solve for 'x':

0.045x + 1200 - 0.06x = 987.50
-0.015x + 1200 = 987.50
-0.015x = 987.50 - 1200
-0.015x = -212.50
x = (-212.50) / (-0.015)
x ≈ 14,166.67

Therefore, approximately $14,166.67 was invested in Account A (at 4.5% interest rate), and the remaining amount ($20,000 - $14,166.67 ≈ $5,833.33) was invested in Account B (at 6% interest rate).