algebra question

Question

Gonzalez Manufacturing borrowed ​$36000. Part of the money was borrowed at

4​%, part at 6​%, and part at 8%. The annual interest was ​$2240​, and the total amount borrowed at 4​% and 6​% was twice the amount borrowed at 8​%.

Use Gaussian elimination or​ Gauss-Jordan elimination to find the amount borrowed at each rate

Answer 1

X+Y+z = 36000

0.04x + 0.06y + 0.08z = 2240

x+y = 2z

third equation says (x+y)/2 = z

(1/2)(x+y) = z

substitutes:

x+y+ (1/2)(x+y) = 36000

2x + 2y + x + y = 72000 <-- multiplies by 2

3x + 3y = 72000

x+y = 24000 <--- labels this equation ALPHA

0.04x + 0.06y + 0.08(1/2)(x+y) = 2240

0.04x + 0.06y + 0.04(x+y) = 2240

0.04x + 0.06y + 0.04x + 0.04y = 2240

0.08x + 0.10 y = 2240

8x + 10y = 224000

4x+5y = 112000

4x + 5(24000-x) = 112000 <--- substitutes equation ALPHA

4x + 120000 - 5x = 112000

-x = -8000

x= 8000

Then y = 16000 per equation ALPHA

Finally z = (1/2)(x+y) = 12000

check: 4% of 8000 + 6% of 16000 + 8% of 12000 =

320 + 960 + 960 = 2240

$ 8000 invested at 4%

$16000 invested at 6%

$12000 invested at 8%

Now for the Matrices... keep reading!!!

x y z

1 1 1 36000

0.04 0.06 0.08 2240

1 1 -2 0

==================

row2*100

x y z

1 1 1 36000

4 6 8 224000

1 1 -2 0

==================

row2/2

x y z

1 1 1 36000

2 3 4 112000

1 1 -2 0

==================

-2*row1 + row2;

-row1+row3

x y z

1 1 1 36000

0 1 2 40000

0 0 -3 -36000

row 3 says -3z = -36000

z = 12000

row 2 says y + 2z = 40000

y + 2(12000) = 40000

y + 24000 = 40000

y = 16000

'

the total per row 1 must be 36000,

so x = 8000