Robert invested $10000 and earned a total of $550 interest. He invested part of the $10000 in an account paying 5% and rest in an account paying 6%. How much did he invest in each account.

(x * .05) + [(10000 - x) * .06] = 550

if x at 5% then the rest (10000-x) is at 6%. So, add up the interest:

.05x + .06(10000-x) = 550

.05x +.06(10,000-x) = 550

.05 x + 600 -.06 x = 550

50 = .01 x
x = 5000

To solve this problem, we can use a system of equations based on the information given. Let's assume Robert invested x dollars in the 5% account and ($10000 - x) dollars in the 6% account.

The interest earned from the 5% account can be calculated by multiplying the amount invested (x) by the interest rate (5%) and dividing by 100:

Interest from 5% account = (x * 5) / 100 = 0.05x

Similarly, the interest earned from the 6% account can be calculated using ($10000 - x) as the amount invested in the 6% account:

Interest from 6% account = (($10000 - x) * 6) / 100 = 0.06(10000 - x)

According to the problem, the total interest earned is $550. Therefore, we can set up the equation:

0.05x + 0.06(10000 - x) = 550

To solve this equation, we can distribute 0.06 to (10000 - x):

0.05x + 600 - 0.06x = 550

Combine like terms:

-0.01x = -50

Divide by -0.01:

x = 5000

Therefore, Robert invested $5000 in the 5% account and ($10000 - $5000) = $5000 in the 6% account.