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?
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
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!!!
To determine how many of each animal you can purchase, we need to solve a system of equations based on the given information. Let's assign variables to the number of each animal:
Let C represent the number of chickens.
Let O represent the number of cows.
Let B represent the number of birds.
We can create the following equations based on the given information:
Equation 1: C + O + B = 50 (since you want to buy a total of 50 animals)
Equation 2: 0.5C + 2.5O + 5B = 50 (since the total cost of the animals should be $50)
To solve this system, we can use substitution or elimination. Let's solve it using elimination:
Multiply Equation 1 by 5 to make the coefficients of O the same in both equations:
5(C + O + B) = 5(50) -> 5C + 5O + 5B = 250
Now, subtract Equation 2 from the modified Equation 1 to eliminate O:
(5C + 5O + 5B) - (0.5C + 2.5O + 5B) = 250 - 50
4.5C + 2.5O = 200
We now have two equations:
Equation 1: C + O + B = 50
Equation 3: 4.5C + 2.5O = 200
From Equation 1, we can rewrite it as B = 50 - C - O and substitute it into Equation 3:
4.5C + 2.5O = 200
4.5C + 2.5O = 200
4.5C + 2.5(50 - C - O) = 200
4.5C + 125 - 2.5C - 2.5O = 200
2C + 125 - 2.5O = 200
2C - 2.5O = 75 (this is now Equation 4)
We now have two equations:
Equation 4: 2C - 2.5O = 75
Equation 3: 4.5C + 2.5O = 200
Solve this system of equations to find the values of C and O.