an investor has a total of $24,900 deposited in three different accounts which earn annual interest rate of 7%, 6%, and 4%. the amount deposited in the 7% account is twice the amount in the 6% account. if the three accounts earn a total annual interest of about $1412 how much money is deposited in each account?

Call the account A, B, C

.07A+.06B+.04C=1412
A=2B or A-2B=0
A+B+C=24900

I assume you can solve now.

To solve this problem, let's start by assigning variables to the amounts of money deposited in each account.

Let x be the amount of money deposited in the 6% account.
Since the amount deposited in the 7% account is twice the amount in the 6% account, the amount deposited in the 7% account is 2x.
The third account, which earns 4% interest, will have the remaining amount, which is (24900 - x - 2x) = 24900 - 3x.

Now, let's calculate the amounts of interest earned by each account.
The 6% account earns an amount of 0.06x in interest.
The 7% account earns an amount of 0.07(2x) = 0.14x in interest.
The 4% account earns an amount of 0.04(24900 - 3x) = 996 - 0.12x in interest.

According to the problem, the total interest earned by the three accounts is $1412. So, we can set up the equation:

0.06x + 0.14x + (996 - 0.12x) = 1412

Simplifying the equation, we have:

0.06x + 0.14x + 996 - 0.12x = 1412

Combine the x terms: 0.08x + 996 = 1412

Subtract 996 from both sides: 0.08x = 416

Divide both sides by 0.08: x = 5200

Therefore, the amount of money deposited in the 6% account is $5200.
The amount deposited in the 7% account is twice that, so it is $10400.
The remaining amount in the 4% account is 24900 - 5200 - 10400 = $9300.

So, the investor has $5200 deposited in the 6% account, $10400 deposited in the 7% account, and $9300 deposited in the 4% account.