Dear Teacher:
If the farmer bought x pigs, y ewes and z hens, the equations are...
x + y + z = 100
10x + 5y + .5z = 100
And
x + y = 100 − z
10x + 5y = 100 − .5z
And
x = .9z − 80
y = 180 − 1.9z
Then z must be a multiple of 10 and
80/.9 < z < 180/1.9
88.88 < z < 94.74
So
z = 90 hens
x = 81 − 10 = 1 pig
y = 180 − 171 = 9 ewes
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