Beginning 2799

Three friends were playing bullets. They did not have the same number of marbles at the start of the game. They had them in a ratio of 2:7:5, while Mišo and Jano had a total of 77 bullets. How many marbles did their friend Peter have at the beginning? Could everyone have the same number of bullets at the end of the game?

Result

p =  77


p =

Step-by-step explanation:


m+j = 77
7j = 2p
5p = 7m
s = m+j+p
d = s/3

m+j = 77
7·j = 2·p
5·p = 7·m
s = m+j+p
d = s/3

j+m = 77
7j-2p = 0
7m-5p = 0
j+m+p-s = 0
3d-s = 0

Pivot: Row 1 ↔ Row 5
3d-s = 0
7j-2p = 0
7m-5p = 0
j+m+p-s = 0
j+m = 77

Row 4 - 1/7 · Row 2 → Row 4
3d-s = 0
7j-2p = 0
7m-5p = 0
m+1.28571p-s = 0
j+m = 77

Row 5 - 1/7 · Row 2 → Row 5
3d-s = 0
7j-2p = 0
7m-5p = 0
m+1.28571p-s = 0
m+0.28571p = 77

Row 4 - 1/7 · Row 3 → Row 4
3d-s = 0
7j-2p = 0
7m-5p = 0
2p-s = 0
m+0.28571p = 77

Row 5 - 1/7 · Row 3 → Row 5
3d-s = 0
7j-2p = 0
7m-5p = 0
2p-s = 0
p = 77

Row 5 - 1/2 · Row 4 → Row 5
3d-s = 0
7j-2p = 0
7m-5p = 0
2p-s = 0
0.5s = 77


s = 77/0.5 = 154
p = 0+s/2 = 0+154/2 = 77
m = 0+5p/7 = 0+5 · 77/7 = 55
j = 0+2p/7 = 0+2 · 77/7 = 22
d = 0+s/3 = 0+154/3 = 51.33333333

d = 154/3 ≈ 51.333333
j = 22
m = 55
p = 77
s = 154

Our linear equations calculator calculates it.
d=s/3=51.333/N



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