Sarah would like to make a 6lb 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 50/50 mixture? show your work!

                 2 lbs of 80/20 and 4 lbs of 50/50 mixtures

Step-by-step explanation:

Let A represent the quantity of the 80/20 mixture.  Let B represent the quantity of the 50/50 mixture.

Let's solve based on peanuts:

First, set up the equations:

Now, solve the system above using substitution: A + B = 6  ⇒  B = 6 - A

0.8A + 0.5(6 - A) = 0.6(6)

0.8A + 3 - 0.5A = 3.6

           3 + 0.3A = 3.6

                 0.3A = 0.6

                       A = 2

Lastly, plug in A (above) into the Quantity equation to solve for B:

A + B = 6

(2) + B = 6

       B = 4

SIDE NOTE: You could have solved based on almonds by setting up the system as:

which would give the same answer (A = 2, B = 4)

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step-by-step explanation:

distance formula is d=√(x2-x1)^2   +   (y2-y1)^2