I have 50.00 to buy 50 animals. Chickens are .50, cows are 2.50,and birds are 5.00. How many of each animal can i purchase?
3 answers
Divide each of those numbers by 50
Then you must balance it all out
So 50/.5=100
50/2.5=20
50/5=10
So 50/.5=100
50/2.5=20
50/5=10
number of chickens --- x
number of cows ------- y
number of birds -------50-x-y
.5x + 2.5y + 5(50-x-y) = 50
times 10
5x + 25y + 50(50-x-y) = 500
5x + 25y + 2500 - 50x - 50y = 500
-45x -25y = -2000
9x + 5y = 400
we need integer solutions , where both x and y must be between 0 and 50
clearly (0,80) would be a solution to our equation, but not permitted since we only have a total of 50 animals.
BUT, I also know that the slope of the line is -9/5
so if we decrease the y by 9 and increase the x by 5 we get more integer solutions to our equation:
0 80
5 71
10 62
15 53
20 44
25 35
30 26
35 17
40 8
45 -1
but remember that 50-x-x also has to be a positive whole number
the only combination that works is
x =40, y = 8
then 50-x-y = 2
so we have 40 chickens
8 cows
2 birds
check:
40+8+2 = 50
.5(40) + 2.5(8) +5(2) = 50
Yeahhh!!!
number of cows ------- y
number of birds -------50-x-y
.5x + 2.5y + 5(50-x-y) = 50
times 10
5x + 25y + 50(50-x-y) = 500
5x + 25y + 2500 - 50x - 50y = 500
-45x -25y = -2000
9x + 5y = 400
we need integer solutions , where both x and y must be between 0 and 50
clearly (0,80) would be a solution to our equation, but not permitted since we only have a total of 50 animals.
BUT, I also know that the slope of the line is -9/5
so if we decrease the y by 9 and increase the x by 5 we get more integer solutions to our equation:
0 80
5 71
10 62
15 53
20 44
25 35
30 26
35 17
40 8
45 -1
but remember that 50-x-x also has to be a positive whole number
the only combination that works is
x =40, y = 8
then 50-x-y = 2
so we have 40 chickens
8 cows
2 birds
check:
40+8+2 = 50
.5(40) + 2.5(8) +5(2) = 50
Yeahhh!!!