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19 December, 20:41

Sarah would like to make a 6 lb nut mixture that is 60% peanuts and 40% almonds. She has several pounds of a mixture that is 80% peanuts and 20% almonds and several pounds of a mixture that is 50% peanuts and 50% almonds.

a. What is the system that models this situation?

b. What is the solution to the system? How many pounds of the 80/20 mixture? How many pounds of the 50/50 mixture?

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  1. 19 December, 23:26
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    The system of equations is (8x + 5Y = 36) and (2X + 5Y = 24).

    The solution is 2 pounds of 80/20 mixture and 4 pounds of 50/50 mixture.

    Step-by-step explanation:

    Let's assume they use "X" lbs of 80/20 mixture and "Y" lbs of 50/50 mixture to make 6 lbs of 60/40 mixture.

    So peanuts would be 80% of X, 50% of Y, and 60% of 6 lbs.

    Mathematically, (80% of X) + (50% of Y) = 60% of 6.

    80X + 50Y = 60*6

    8X + 5Y = 36

    Similarly almonds would be 20% of X, 50% of Y, and 40% of 6 lbs.

    Mathematically, (20% of X) + (50% of Y) = 40% of 6.

    20X + 50Y = 40*6

    2X + 5Y = 24

    So, the system to model this situation is (8X + 5Y = 36) and (2X + 5Y = 24).

    To solve this system, we can subtract second equation from first equation.

    (8X + 5Y) - (2X + 5Y) = 36 - 24

    6X = 12

    X = 2

    We can plug this x=2 into any equation to solve for y.

    8 (2) + 5Y = 36

    16 + 5Y = 36

    5Y = 36 - 16 = 20

    Y = 4

    So, they use 2 pounds of 80/20 mixture and 4 pounds of 50/50 mixture.
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