Help with a word problem

Question

 A total of $12,000 is invested in two funds  paying 9% and 11% simple interest. If the yearly and trust is $1180, how much of the $12,000 is invested at each rate?  

Answer 1

I would solve this by creating a matrix.

 

If we let x=$ amount invested at 9%

             y=$ amount invested at 11%

 

We get the following two equations:

 

Total amount is $12,000.  So:  x + y = 12,000

 

Interest was at 9% and 11%, total of $1,180.  So, 0.09x + 0.11 y = 1,180

 

This give us the following matrix Iteration 0:

 

1          1          =          12000

0.09     0.11      =           1180

 

Want to get the value in R2, C1, to equal 0.  So, do:  -9R1+100R2 = new R2; which gives Iteration 1 of:

 

1          1          =          12000

0          2          =          10000

 

Next, we want the value in R1, C2, to equal 0.  So, we do:  -R2+2R1 = new R1; which gives Iteration 2 of:

 

2          0          =          14000

0          2          =          10000

 

Then, we do the following row reductions:

 

(1/2)R1 = new R1

(1/2)R2 = new R2

 

Which leaves us with the solution matrix of:

 

1          0          =          7000

0          1          =          5000

 

So, (x,y) is:  (7000,5000)

 

Which means that $7,000 was invested at 9%; $5,000 invested at 11%

 

Check work # 1:   7,000 + 5,000 = 12,000     [Check]

Check work # 2:   0.08 * 7,000 + 0.11 * 5,000 = 1,180     [Check]