An investor has 7000 to invest in two accounts. The first account earns 8% annual simple intrest, and the second account earns 13% annual simple intrest .How much money Should be invested in each account so that the simple intrest earned is 500?
.08 x + .13(7000-x) = 500
To determine the amount of money to be invested in each account, we can use a system of equations.
Let's say the amount invested in the first account is x dollars, and the amount invested in the second account is 7000 - x dollars.
The simple interest earned from the first account, at 8% annual interest rate, can be calculated as 0.08x. Similarly, the simple interest earned from the second account, at 13% annual interest rate, can be calculated as 0.13(7000 - x).
We are given that the total simple interest earned from both accounts is 500. Therefore, we can set up the following equation:
0.08x + 0.13(7000 - x) = 500
Now, let's solve the equation:
0.08x + 0.13(7000 - x) = 500
0.08x + 910 - 0.13x = 500
-0.05x = -410
x = -410 / -0.05
x = 8200
So, the investor should invest $8200 in the first account and $7000 - $8200 = $-1200 in the second account.
However, investing a negative amount does not make sense, so it seems like there is no solution to the problem as stated.
Please verify the given information to ensure accuracy in solving the problem.