SOLUTION: An investment of ​$35,000 was made by a business club. The investment was split into three parts and lasted for one year. The first part of the investment earned​ 8%

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Question 1081051: An investment of ​$35,000 was made by a business club. The investment was split into three parts and lasted for one year. The first part of the investment earned​ 8% interest, the second​ 6%, and the third​ 9%. Total interest from the investments was $ 2610. The interest from the first investment was 22 times the interest from the second. Find the amounts of the three parts of the investment.
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
as far as i can tell, there is no solution to this problem as stated.

i let the first investment = A and the second investment = B and the third investment = C

the first two rows of the matrix solution were derived from:

A + B + C = 35000
.08A + .06B + .09C = 2610

i used a matrix calculator to solve it mechanically after repeated tries to solve it manually.

the matrix only used the coefficients of each variable and he constant

the problem appears to be with the fact that the interest from the first investment is 22 times the interest from the second investment.

the calculator solves it but the value of C is negative which can't be.

i went back and solved it stating that the interest from the first investment is equal to the interest from the second investment and got a valid answer.

the matrix calculator i used can be found here.

http://www.gregthatcher.com/Mathematics/GaussJordan.aspx

it did solve it, but the value of the third investment was negative, therefore invalid.

as a test, i made the interest on the first investment equal to the interest on the second investment and was able to get a valid answer both manually and through the use of the calculator.

following are the inputs and outputs when the interest on the first investment is 22 times the interest on the second investment.

note that inputs have to be in fractional form.
therefore .08 is equal to 8/100 and .06 is equal to 6/100, etc.

$$$

$$$

following are the inputs and outputs when the interest on the first investment is equal to the interest on the second investment.

$$$

$$$

the third row in the matrix is derived as follows:

when interest on the first investment is equal to 22 times interest on the second investment, you get 8/100 * A = 22 * 6/100 * B.

this results in 8/100 * A = 132/100 * B

subtract 132/100 * b from both sides of the equation to get:

8/100 * A - 132/100 * B = 0

since C is equal to 0 in this equation, you put it in matrix row form by entering it as:

8/100 * A - 132/100 * B + 0 * C = 0

since the matrix only uses the coefficients, then the third row is:

8/100 -132/100 0 0, as shown.

similarly, this results in a third input row of 8/100 -6/100 0 0 when the interest on the first investment is equal to the interest on the second investment.

start with 8/100 * A = 6/100 * B

subtract 6/100 * B from both sides of the equation to get:

8/100 * A - 6/100 * B = 0

final form for the third row is 8/100 -6/000 0 0, as shown in the second set inputs.

bottom line is there is no valid solution with the problem as given.





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