Question 178699: for 100 crowns, a certain farmer bought some horses, goats, and sheep. For a horse, he paid 3 1/2 crowns, for a goat, 1 1/3 crowns, and for a sheep, 1/2 crown. Assuming he bought at least 7 of each and that he bought 100 animals in all, how many horses, goats, and sheep did he buy?
many thanks for your help.
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! Let x=number of horses
y=number of goats
z=number of sheep
Now we are told that:
x+y+z=100---------------------------eq1
and
(7/2)x+(4/3)y+(1/2)z=100 multiply each term by 6
21x+8y+3z=600---------------------------eq2
TWO EQUATIONS AND THREE UNKNOWNS USUALLY MEANS SOME TRIAL AND ERROR
multiply eq1 by 3 and subtract it form eq2 and we get:
18x+5y=300-----------------eq1a
Solve eq1a for y:
subtract 18x from each side
5y=300-18x divide each term by5
y=60-18x/5------------------------------eq1b
We know two very important things about this problem:
(1) We cannot have negative animals----we deal in positive numbers
(2) We cannot have fractions of animals---we deal in whole numbers
Now we are told that:
60-18x/5>=7 subtract 100 from each side
-18x/5>=7-60 or
-18x/5>=-53 bultiply each side by 5
-18x>=-265 divide each side by -18 (inequality sign changes)
x<= 14.7 or
x<=14 and we are told that x>=7, so
7<=x<=14
Now, by trial and error, we look at eq1b and determine which values of x between 7 and 14 (inclusive) will yield a whole number for y, starting with 7.
By inspection, we can see that in order for 18x to be divisible by 5, x must end in either a 0 or 5, so the only value for x that works between 7 and 14 is 10.
From eq1b:
If x=10, y=24
Substitute these values into eq1:
10+24+z=100
z=66
So, we have:
Horses=10
Goats=24
Sheep=66
Substitute these values into eq2 to make sure the money checks:
(21x+8y+3z=600)
21*10+8*24+3*66=600
210+192+198=600
600=600
Hope this helps---ptaylor
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