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SecureThttps/www ayerfomework.aspahomeworkld 483808397questionld-7&flushed; falseld 501847 MAC 1105 10 week online summer 2018 ( 11500) Alyssa Olivares &1 /2/18 11:39 PM Homework: Section 5.2-Systems of Linear Equations in 3 Var Score: 0 of 1 pt 5.2.39 Save 9 of 9 (8 complete) HW Score: 77.78%, 7 of 9 pts Skill Builder Question Help A person invested $7300 tr 1 year, part at 4%, part at 10%, and the remainder at 12% The total annual income trom these nvestments was $736 The amount of money r vested at 12% was SS00 more than the amounts invested at 4% and 10% combined. Find the amount invested at each rate The person nvested s?at 4%, S?at 10%, and at 12% Enter your answer in the edit fields and then click Check Answer Clear All Check Answer te 5 8 9

Explanation / Answer

Let the amounts invested at 4 % , 19 % and 12 % be $ x, $ y and $ z respectively. Then x+y +z = 7300…(1), z = x+y +500 or, x+y-z = -500…(2) and x*0.04+y*0.10+z*0.12 = 736 or, 4x+10y+12z = 73600 or, 2x+5y+6z = 36800…(3).

The augmented matrix of this linear system is A =

1

1

1

7300

1

1

-1

-500

2

5

6

36800

To determine the values of x, y, z , we will reduce A to its RREF as under:

Add -1 times the 1st row to the 2nd row

Add -2 times the 1st row to the 3rd row

Interchange the 2nd row and the 3rd row

Multiply the 2nd row by 1/3

Multiply the 3rd row by -1/2

Add -4/3 times the 3rd row to the 2nd row

Add -1 times the 3rd row to the 1st row

Add -1 times the 2nd row to the 1st row

Then the RREF of A is

1

0

0

1200

0

1

0

2200

0

0

1

3900

Hence x = 1200,y = 2200 and z = 3900. Thus, the person invested $ 1200 at 4 %, $ 2200 at 10 % and $ 3900 at 12 %.

1

1

1

7300

1

1

-1

-500

2

5

6

36800

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