Question 174531
120g of fat, 220g of carbohydrates, and 80g of protein. 
_______________ A______B_______C
Fat____________ 10 g____ 4g_____ 12 g
Carbohydrates___11 g___77 g_____0g
Protein_________4 g_____1 g_____16 g 
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A)10A+4B+12C=120
  11A+77B   =220
   4A+ 1B+16C=80
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B)using cramers rule
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find determinant of main matrix{{{(matrix(3,3,10,4,12,11,77,0,4,1,16))}}}using column 3---> 12{{{(matrix(2,2,11,77,4,1))}}}-0{{{(matrix(2,2,10,4,4,1))}}}+16{{{(matrix(2,2,10,4,11,77))}}}---->12(-297)+16(726)=8052
:
A=det{{{(matrix(3,3,120,4,12,220,77,0,80,1,16))}}}/8052---->so using the 3rd column--->det = 12{{{(matrix(2,2,220,77,80,1))}}}-0+16{{{(matrix(2,2,120,4,220,77))}}}--->12(-5940)+16(8360)=62480
:{{{highlight(A=62480/8052=7.76)}}}
:
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B=det{{{(matrix(3,3,10,120,12,11,220,0,4,80,16))}}}/8052---->so using the 3rd column--->det = 12{{{(matrix(2,2,11,220,4,80))}}}-0+16{{{(matrix(2,2,10,120,11,220))}}}---->0+16(880)=14080
:{{{highlight(B=14080/8052=1.75)}}}
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C=det{{{(matrix(3,3,10,4,120,11,77,220,4,1,80))}}}/8052---->using column 1--->det= 10{{{(matrix(2,2,77,220,1,80))}}}-11{{{(matrix(2,2,4,120,1,80))}}}+4{{{(matrix(2,2,4,120,77,220))}}}--->10(5940)-11(200)+4(-8360)--->59400-2200-33440=
23760
:{{{highlight(C=23760/8052=4.37)}}}
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C)if C was no longer available we would have the equation
:
10A+4B+0C=120
11A+77B+0C=220
 4A+  B+0C=80
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we cannot use Cramers rule because there would be no unique solution as the determinant you would be dividing by would be zero.
:according to this graph there is no unique solution to this scenario.
It appears you could reach 2 of the conditions in any order but not all 3 at once.
{{{graph(300,300,-100,50,-100,50,(-5/2)x+30,(-1/7)x+220/77,-4x+80)}}}