Respuesta :
Answer:
Step-by-step explanation:
suppose the number of Xavier sets = x
suppose the number of Yvonne sets = y
suppose the number of Zena sets = z
Blue Black Red
Xavier(x) 1 1 0
Yvonne(y) 2 3 1
Zena(z) 4 5 1
refills 11 14 3
[tex]1x+2y+4z=11\\1x+3y+5z=14\\0x+1y+1z=3[/tex]
[tex]\left[\begin{array}{ccc}1&2&4\\1&3&5\\0&1&1\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}11\\14\\3\end{array}\right] \\\\R_2\rightarrow R_2\rightarrow R_1\\\\\left[\begin{array}{ccc}1&2&4\\0&1&1\\0&1&1\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}11\\3\\3\end{array}\right]\\\\ R_3\rightarrow R_3\rightarrow R_2\\\\[/tex]
[tex]\left[\begin{array}{ccc}1&2&4\\0&1&1\\0&0&0\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}11\\3\\0\end{array}\right]\\\\x+2y+4z=11\\y+z=3\\0=0\\\\y=3-2\\x=11-2(3-2)-4z\\=11-6+2z-4z=5-4z[/tex]
value of x and y depend upon the value of z
suppose
[tex]z=0,\,y=3,\,x=5-2z=5\\z=1,\,y=2,\,x=5-2=3\\z=2,\,y=1,\,x=5-4=1\\z=3,\,y=0,\,x=5-(2\times 3) =-1(not\,possible)[/tex]
Hence following ways company package to sell
5 xavier, 3 yvonnes , o zena or
3 xavier, 2 yvonnes, 1 zena or
1 xavier, 1 yvonnes, 2 zena