Answer:
The amount invested at 6% is $11,000
The amount invested at 9% is $6,000
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