You invested $17,000 in two accounts paying 6% and 9% annual interest, respectively.

If the total interest earned for the year was $1200, how much was invested at each rate?

The amount invested at 6% is how much?

The amount invested at 6% is $11,000

The amount invested at 9% is $6,000

Step-by-step explanation:

* Lets explain how to solve the problem

- The invested amount is $17,000

- There are two accounts one paying 6% annual interest and the

other paying 9% annual interest

- The total interest earned for a year is $1200

- We need to know how much was invested at each rate

* Lets change these information to system of equations

- Assume that the amount invested in the account with 6% rat is x

and the amount invested in the account with 9% rat is y

- The rule of interest is I = Prt, where I is the interest amount,

P is the money invested, r is the rate in decimal, t is the time

- For The first account

∵ r = 6% = 6/100 = 0.06

- For The second account

∵ r = 9% = 9/100 = 0.09

∵ t = 1 ⇒ for both accounts

- The total interest earned for the year was $1200

∴ I = x (0.06) (1) + y (0.09) (1)

∴ I = 0.06 x + 0.09 y

∵ I = $1200

∴ 0.06 x + 0.09 y = 1200

- Multiply both sides by 100

∴ 6 x + 9 y = 120,000 ⇒ (1)

- The amount of money invested is $17,000 in both accounts

∴ x + y = 17,000 ⇒ (2)

* Lets solve the two equations to find x and y

- Multiply equation (2) by - 6 to eliminate x

∴ - 6 x + - 6 y = - 102,000

∴ - 6 x - 6y = - 102,000 ⇒ (3)

- Add equations (1) and (3)

∴ 3 y = 18,000

- Divide both sides by 3

∴ y = 6,000

- Substitute the value of y in equation (2) to find the value of x

∴ x + 6,000 = 17,000

- Subtract 6,000 from both sides

∴ x = 11,000

∵ x represents the amount invested in the account with 6% rat

∴ The amount invested at 6% is $11,000

∵ y represents the amount invested in the account with 9% rat

∴ The amount invested at 9% is $6,000