a steer for 10 thalers, a calf for 2 thalers, a cow for 5 thalers, a sheep for 1/2 thaler.
2007-10-16 23:55:41 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
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2007-10-17 00:05:31 · answer #1 · answered by Jun Agruda 7 · 2⤊ 0⤋
10x+2y+5z+1/2w=100
x+y+z+w=100
1)w must be even
2)w=100-(x+y+z)
so w/2= 50-(x+y+z)/2
19x+3y+9z=100
100-19 x must be a multiple of 3
x= 1 Ok and x= 4 Ok
For x=4
3y+9z=24
y+3z=8 so y=8-3z so z can be 1 or2
For z= 1 y=5 and x+y+z= 10 so w/2 = 45 and w =90Solution
for z=2 y=2 x+y+z=8 so w/2=46 and w=92 Solution
You can test x=1
x=steer
y=calf
z=cow
w=sheep
2007-10-17 08:05:28 · answer #2 · answered by santmann2002 7 · 0⤊ 0⤋
4 steer for 40 thalars,5 calf for 10 thalars,1 cow for 5 thalars,90 sheeps for 45 thalars
2007-10-17 00:39:08 · answer #3 · answered by mani m 1 · 0⤊ 0⤋
17 thalers altogether. but i do not understand ur question.
2007-10-17 00:00:07 · answer #4 · answered by freesportsgirl 1 · 0⤊ 0⤋
that guy is going to have a good time with al those animals!
GO ROCKIES!
2007-10-16 23:58:57 · answer #5 · answered by ? 6 · 0⤊ 0⤋