A total of $6200 is deposited into two simple interest accounts. On one account the annual simple interest rate is 2%, and on the second account the annual simple interest rate is 4%. The amount of interest earned for 1 year was $148. How much was invested in each account?

Let x be the amount invested at 2% interest and y be the amount invested at 4% interest

x + y = 6200

0.02x + 0.04y = 148

Solve that pair of equations.

0.02x + 0.04(6200 -x) = 148

-0.02x + 248 = 148
0.02x = 100
x = 5000.
y = 1200

The simple interest paid on a loan after 6 months was $306. The annual interest rate was 8%. Find the principal.

can't figure it out yet let me know how it goes

To solve this problem, we need to set up a system of equations based on the given information.

Let's assume that x dollars were invested in the account with a 2% interest rate. Since the total amount deposited into both accounts is $6200, the amount invested in the second account will be $6200 - x.

Now we can use the formula for simple interest to calculate the interest earned in each account.

The interest earned in the first account is given by the formula: (principal) * (interest rate) * (time) = x * 0.02 * 1 = 0.02x.
The interest earned in the second account is given by the formula: (principal) * (interest rate) * (time) = (6200 - x) * 0.04 * 1 = 0.04(6200 - x).

According to the problem, the total interest earned in one year is $148. Therefore, we can write the equation:

0.02x + 0.04(6200 - x) = 148.

Now, let's solve this equation to find the value of x.

0.02x + 0.04(6200 - x) = 148
0.02x + 248 - 0.04x = 148 (distributed 0.04 to 6200-x)
-0.02x = -100 (subtract 248 from both sides)
x = 5000 (divide both sides by -0.02)

Therefore, $5000 was invested in the account with a 2% interest rate, and $6200 - $5000 = $1200 was invested in the account with a 4% interest rate.