Linda has $9000 to invest. She invested x dollars in an account that earned 3% simple interest and the remainder of the money in an account that earned 2.6% simple interest. In one year, both accounts together earned $250 interest. How much was invested in each account?

Question

Linda has $9000 to invest. She invested x dollars in an account that earned 3% simple interest and the remainder of the money in an account that earned 2.6% simple interest. In one year, both accounts together earned $250 interest. How much was invested in each account?

Answer 1

If x is at 3%, the remainder (9000-x) is at 2.6%. So, add up the interest:

.03x +.026(9000-x) = 250

Answer 2

To solve this problem, we can set up an equation based on the given information.

Let's assume Linda invested x dollars in an account that earned 3% simple interest. Therefore, she invested (9000 - x) dollars in the other account that earned 2.6% simple interest.

The interest earned from the first account can be calculated as (x * 0.03), since 3% is equivalent to 0.03 as a decimal.

The interest earned from the second account can be calculated as ((9000 - x) * 0.026), since 2.6% is equivalent to 0.026 as a decimal.

According to the problem, the total interest earned from both accounts is 250 dollars. So, we can write the equation:

(x * 0.03) + ((9000 - x) * 0.026) = 250

Now, let's solve for x.

0.03x + 0.026(9000 - x) = 250
0.03x + 234 - 0.026x = 250
0.004x = 16
x = 16 / 0.004
x = 4000

Therefore, Linda invested $4000 in the account that earned 3% simple interest, and the remaining amount ($9000 - $4000 = $5000) in the account that earned 2.6% simple interest.