Notam x -> pretul cartii, y-> pretul creionului, z -> pretul caietului
7z + 4x = 26
7x + 4z + 11y = 40
4x+7z = 26
7x+11y+4z = 40
4x+7z = 26
7x+4z = 40 - 11y
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11x+11z = 66-11y => 11x+11y+11z = 66 => 11(x+y+z) = 66 => x+y+z = 6
x+y+z = 6| (i)*(-4), (ii)*(-7)
4x+0y+7z = 26 (i) +
7x+11y+4z = 40 (ii) +
x+y+z = 6
/ -4y + 3z = 2
/ 4y - 3z = -2
y = k => 3z = 2+4k =>[tex]z = \frac{2+4k}{3} [/tex]
[tex]x+k+\frac{2+4k}{3}=6 \Rightarrow 3x+3k+2+4k = 18 \Rightarrow 3x = 16-7k \Rightarrow \\ \Rightarrow x = \frac{16-7k}{3} \\ \\ \Rightarrow \\ x = \frac{16-7k}{3} \\ y=k \\ z= \frac{2+4k}{3}[/tex]
Daca k = 1 => x = 3; y = 1; z = 2 Deci, o carte costa 3 lei, un creion costa 1 leu, iar un caiet costa 2 lei.
=> 4 carti + 9 creioane + 2 caiete = 4x+9y+2z = 12+9+4 = 21+4 = 25
Daca k = 2 =>
[tex]x = \frac{2}{3} ,y=2,z= \frac{10}{3} \\ \Rightarrow 4* \frac{3}{2}+ 9*2 + 2* \frac{10}{3} = \frac{36+108+40}{6} = \frac{184}{6} = 30,(6) [/tex]
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Sunt o infinitate de solutii.....
