Question 1159372: Solve the following word problem using matrices:
Part of $30,000 is invested at 6% yearly interest, another part at 7%, and the remainder at 8% yearly interest. The total yearly interest income from the three investments adds up to $2200. The sum of the amounts invested at 6% and 7% equals the amount of money invested at 8%. How much is invested at each rate?
Found 2 solutions by MowMow, MathTherapy:
Answer by MowMow(42)   (Show Source):
You can put this solution on YOUR website! Using matrices to solve this problem:
The 3x4 matrix is:
x y z Amount
1 1 1 30000
.06 .07 .08 2200
1 1 -1 0
The answer is
x = $ 5,000
y = $10,000
z = $15,000
Answer by MathTherapy(10551)  (Show Source):
You can put this solution on YOUR website! Solve the following word problem using matrices:
Part of $30,000 is invested at 6% yearly interest, another part at 7%, and the remainder at 8% yearly interest. The total yearly interest income from the three investments adds up to $2200. The sum of the amounts invested at 6% and 7% equals the amount of money invested at 8%. How much is invested at each rate?
With the amount invested at 8% being the total of the 6% and 7% investments, it goes without saying that $15,000 was invested in the 6% and 7%, and $15,000 was invested at 8%.
Let the amount invested at 6% be S
Then amount invested at 7% = 15,000 - S
We then get the following EARNINGS equation: .06S + .07(15,000 - S) + .08(15,000) = 2,200
.06S + 1,050 - .07S + 1,200 = 2,200
.06S - .07S = 2,200 - 2,250
- .01S = - 50
Amount invested at 6% = 
It then follows that $15,000 - $5,000, or $10,000 was invested at 7%.
It was already revealed how much was invested at 8%.
Just realized that this asks for a matrices-solution. The other person did it anyway!
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