SOLUTION: A farmer keeps some animals on a strict diet. Each animal is to recieve 12 grams of protein and 6.5 grams of carbohydrates. The farmer uses two food mixes with nutrients as shown i
Algebra ->
Expressions-with-variables
-> SOLUTION: A farmer keeps some animals on a strict diet. Each animal is to recieve 12 grams of protein and 6.5 grams of carbohydrates. The farmer uses two food mixes with nutrients as shown i
Log On
Question 536737: A farmer keeps some animals on a strict diet. Each animal is to recieve 12 grams of protein and 6.5 grams of carbohydrates. The farmer uses two food mixes with nutrients as shown in the table. How many grams of each mix should be used to provide the correct nutrients for each animal.
Protiens Carbohydrates
Mix A 12% 9%
Mix B 15% 5%
Please for the gentleman or lady that solves my
problem please, give an explanation as to why you did what
you did in each step. Thank you and have a blessed day-Tray Tillmon Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A farmer keeps some animals on a strict diet.
Each animal is to receive 12 grams of protein and 6.5 grams of carbohydrates.
The farmer uses two food mixes with nutrients as shown in the table.
How many grams of each mix should be used to provide the correct nutrients for each animal.
:
Proteins | Carbohydrates
Mix A: 12%| 9%
Mix B: 15%| 5%
:
Write an equation for each
Prot eq: .12A + .15B = 12
Carb eq: .09A + .05B = 6.5
:
We can use elimination here, multiply the 2nd equation by -3, add to the 1st eq
.12A + .15B = 12
-.27A - .15B = -19.5
----------------------add eliminates B, find A
-.15A = -7.5
A =
A = 50 grams of Mix A
:
Find B using the 1st equation
.12(50) + .15B = 12
6 + .15B = 12
.15B = 12 - 6
.15B = 6
B =
B = 40 grams of mix B
:
:
Check these solutions in the 2nd original equation
.09A + .05B = 6.5
.09(50) + .05(40) =
4.5 + 2.0 = 6.5; confirms our solution of A=50, B=40
:
:
