You can
put this solution on YOUR website! Sol: Suppose he bought x cows, y pigs and z chickens,then we have two equations:
x + y + z =100 ...(1)
10 x+ 3 y + 0.5 z = 100 or
20 x + 6 y + z = 200 ...(2)
This is a linear system of 2 equations in 3 variables.
Usually,there are infinitely many solutions.However, x,y,z are nonnegative
integers , so we mayget reasonal solutions for the system.
Let's consider from the number of the most expensive animal,ie x(chick).
By (2),we see that 20x <= 200, so 0 <= x <= 10.
Form (2)-(1): 19 x + 5y = 100, or 19 x = 100 -5y = 5 (20-y),
so 5 is a divisor of 19x and so x must be a multiple of 5.
Since x<=10, the only choices for x are 0,5, or 10.
Case i) When x = 0, we have y + z = 100 and
6y + z= 200
Use z = 100-y, we have 6y + 100-y = 200, so y = 20, z =80.
Case ii) When x = 5, we have y + z = 95 and
6y + z= 100
So, 5y = 5, y= 1 and then z = 94.
Case iii) x=10, then 20x =200 ,hence
y + z = 90 and
6y + z= 0
then 5y = 90, y = -18 (negative invalid)
Therefore, there are two possibilities:
(1) He bought 0 cows, 20 pigs and 80 z chickens
(2) He bought 5 cows, 1 pig and 94 chickens
You can
put this solution on YOUR website!let number of cows be x, no. of pigs be y,
no. of chicks = 100- (x+y)
10x + 3 y + 0.5(100-(x+y)) = 100
9.5x + 2.5y = 100-50 = 50
19x + 5y = 100
as theer is no other equation, this problem becomes unsolvable.
Had there been two kinds of animals, it might have been solved.
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