Question

Mae Jackson has a total of $5,500 invested in two simple interest accounts. The annual simple interest rate on one account is 9%. The annual simple interest rate on the second account is 6%. How much is invested in each account if both accounts earn the same amount of interest?

Answer 1

Mae Jackson has $2,200 invested at a 9% interest rate and $3,300 at 6%, with both accounts earning the same annual simple interest.

Step-by-step explanation:

To find out how much is invested in each account when both accounts earn the same amount of interest, we set up two equations. Let's denote the amount invested in the 9% account as x and the amount in the 6% account as y.

The total investment is $5,500, so:
x + y = 5,500 ... (1)

The annual simple interest for both accounts is the same, so the interest from the first account at 9% equals the interest from the second account at 6%. Thus:
0.09x = 0.06y ... (2)

Solving these equations simultaneously, we multiply equation (2) by 100 to get rid of decimals:
9x = 6y

We then substitute for y from equation (1) in the new version of equation (2), leading to:
9x = 6(5,500 - x)

Simplifying, we get:
9x = 33,000 - 6x15x = 33,000x = 2,200

Substituting x = 2,200 back into equation (1) gives:
2,200 + y = 5,500y = 3,300

So, $2,200 is invested at 9% and $3,300 at 6%.